Class 11 Physics

• Circular Motion Very Short Answer Questions
• Newtonian Gravitation And Planetary Motion Very Short Answer Questions
• Statics Very Short Answer Questions

• Transmission Of Heat Very Short Answer Questions
• Vector Very Short Answer Type Questions
• Viscosity And Surface Tension Very Short Answer

• Expansion Of Solid And Liquids Short Answer Questions
• WBCHSE Class 11 Physics Change Of State Of Matter Short Answer Questions
• WBCHSE Class 11 Physics Circular Motion Short Answer Questions
• WBCHSE Class 11 Physics Friction Short Answer Type Questions
• WBCHSE Class 11 Physics Newton Law Of Motion Short Answer Questions
• WBCHSE Class 11 Physics Newtonian Gravitation And Planetary Motion Short Answer Questions

• WBCHSE Class 11 Physics On Elasticity Short Questions And Answers
• WBCHSE Class 11 Physics One-Dimensional Motion Short Answer Questions
• WBCHSE Class 11 Physics Simple Harmonic Motion Short Answer Type Questions
• WBCHSE Class 11 Physics Statics Short Answer Questions
• WBCHSE Class 11 Physics Transmission Of Heat Short Answer Questions
• WBCHSE Class 11 Physics Viscosity And Surface Tension Short Answer Type Questions

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies

System Of Particles Motion

Rotation Of Rigid Bodies Introduction: A rigid body is a continuous distribution of particles of definite masses within an extended volume; these particles do not change their relative positions with respect to one another while rotating about a fixed axis.

Let us consider a rigid body R. The constituent particles of the body are A, B, C, … For a pure rotational motion of this rigid body, each constituent particle undergoes a circular motion. The characteristic of these circular motions is that the centres of all the circles lie on a single fixed straight line.

System of particles motion

This line is normal to the plane of each of the circular paths and is called the axis of rotation of the rigid body.

Rotation Of Rigid Bodies Definition: The axis of rotation is defined as a fixed straight line that passes normally through the centre of the circular path followed by a rotating particle, or by any constituent particle of a rotating rigid body. The axis of rotation may pass through the body itself or may lie entirely outside the body.

WBCHSE Class 11 Physics Rotation Of Rigid Bodies System Of Particles Motion

Couple Torque

Couple Definition: Two equal, parallel and opposite forces, having different lines of action, acting simultaneously on a body, constitute a couple.

A pair of parallel and opposite forces F, act on a body at points A and B. Then, (F, F) is a couple acting on the body. The perpendicular distance between the lines of action of the two forces is called the arm of the couple. A couple tends to set up a rotational motion.

Torque Definition: The tendency of rotational motion, set up in a body by a couple, is called the moment of the couple or torque. The torque is given by the product of one of the forces of the pair and the arm of the couple.

The direction of the torque is given by the direction of advance of a right-handed screw when turned in the direction of rotation. Turning on or off a water tap, screwing or unscrewing the cap of a bottle, using a screwdriver, etc., are associated with torques applied with our fingers thereby setting up a rotational motion.

Torque Vector Form: Torque is a vector quantity. The vector representation for the relation between torque and force is \(\vec{\tau}=\vec{r} \times \vec{F}\)

Torque and the vector product τ = rF sinø

Torque is the vector product between the force vector \(\vec{F}\) and vector \(\vec{r}\). \(\vec{\tau}=\vec{r} \times \vec{F}\)

Unit And Dimension Of Torque:

CGS System: dyn · cm

SI: N · m

Dimension of torque = dimension of force x dimension of length = MLT^-2 x L = ML²T^-2

System of particles motion

Torque And Pure Rotation: Torque due to a couple can produce rotational motion only. The resultant of the two forces applied at points A and B is zero, i.e., F – F = 0. As the resultant force is zero, no change occurs in its translational motion. But their lines of action are separate. So, only rotation is set up in this case. Such a rotation without translation is called pure rotation.

Moments Of The Two Forces Of A Couple: A point o is taken on the line BC as shown. Moment of force F applied at point A with respect to O = Fx CO. Also, force F at point B sets up the moment about O as FxBO. Hence, the algebraic sum of these two moments = F x CO + F x BO = F(CO+BO) = F x BC. But F x BC is the torque generated by the couple.

The algebraic sum of moments of the two forces of a couple, about a point, is equal to the moment of the couple (often called torque) about that point.

Moment Of Force And Torque Are Identical: From the above discussion, it is clear that torque alone can produce pure rotational motion. However, it is often observed that pure rotation can also be produced by a single force only. For example, a door can be opened by applying a single force on the door panel.

However, even in this case, an equal and opposite reaction force on the door panel is developed at the hinges. This reaction, along with the applied force, constitutes a couple and exerts a torque. Hence, moment of a force and torque are two identical physical quantities.

WBCHSE Class 11 Physics System Of Particles Motion

Work Done By A Couple: We know that two forces of equal magnitude constitute a couple. When a couple produces rotation in a body, the sum of the work done by the two forces of the couple is the measure of the work done by the couple. Suppose a couple (F, F) is acting on a body.

AB is the arm of this couple. So, the moment of the couple or torque, τ = F x AB.

Suppose the body is rotated by an angle θ under the influence of the couple about point O. As a result, point A is shifted to A₁ and B to B₁. If θ is very small, then we can assume that the arcs AA₁ and BB₁ are almost straight lines.

Now, AA₁ = AO · θ and BB₁ = BO · θ

Work done by the force acting on the point A, W₁ = F · AA₁ = F · AO · θ

Similarly, work done by the force acting on the point B, W₂ = F · BB₁ = F · BO · θ

So, the total work done by the couple,

W = W₁ + W₂ = F · AO · θ + F · BO · θ

= F · (AO+BO) · θ = F · AB · θ = τ · θ

= torque x angular displacement

Hence, the amount of work done does not depend on the position of the axis of rotation. For one complete rotation of the body, the angular displacement is 2π. So, for n complete rotations the angular displacement will be 2πn.

Hence, for n complete rotations, the work done by the couple, W = 2πx torque

WBCHSE Class 11 Physics System Of Particles Motion

Relation Between Torque And Angular Acceleration

Moment Of Interia Or Rotational Interia: When a force is applied to a body, a linear acceleration is produced in that body. Similarly, when a torque is applied to a body, an angular acceleration is produced in it. So it can be said that torque plays the same role in rotational motion as that of force in the case of linear motion. Hence, torque is the rotational analogue of force.

Suppose the body PQR is revolving with a uniform angular acceleration about the axis AB. The body is assumed to be made up of innumerable point masses m₁, m₂, m₃, …, etc. These point masses are at distances r₁, r₂, r₃, … etc. respectively from the axis of rotation AB. In the case of pure rotation, the axis of rotation remains fixed and the angular acceleration of each point mass remains the same.

But due to the difference in distances of the point masses from the axis of rotation, their linear accelerations are different. If the linear acceleration of the particle m₁ is a₁, then a₁ = r₁α and the force acting on it is F₁ = m₁a₁ – m₁r₁α.

The moment of force F₁ about the axis of rotation, G₁ = force x perpendicular distance of the particle from the axis of rotation

⇒ \(F_1 r_1=m_1 r_1^2 \alpha\)

In this way, the moment of force can be found for every particle. The couple or torque acting on the entire rigid body is the algebraic sum of the moments of the forces acting on individual particles.

Hence, torque \(\tau=G_1+G_2+\cdots=m_1 r_1^2 \alpha+m_2 r_2^2 \alpha+\cdots\)

= \(\left(m_1 r_1^2+m_2 r_2^2+\cdots\right) \times \alpha=\sum_i m_i r_i^2\)

[m_i is the mass of the i-th particle and r_i is its perpendicular distance from the axis of rotation] = Iα …(1)

Here, \(I=\sum_i m_i r_i^2\)….(2)

= moment of inertia of the body about the axis of rotation

So, \(I=\frac{\tau}{\alpha}\)

i.e., moment of inertia = \(\frac{\text { torque }}{\text { angular acceleration }}\)

System of particles motion

Definition Moment Of Inertia: A body about an axis of rotation is defined as the torque acting on the body divided by the corresponding angular acceleration thus generated about the same axis of rotation.

In calculus, equation (2) can be represented as I = \(\int r^2 d m\)…(3)

Unit And Dimension Of Moment Of Inertia:

CGS System: g · cm²

SI: kg · m²

Dimension of moment of inertia = dimension of mass x (dimension of distance)² = ML²

WBCHSE Class 11 Physics System Of Particles Motion

Some Important Points About Moment Of Inertia:

1. Moment of inertia not only depends on the mass of a body but also depends on the perpendicular distance of the particles constituting the body from the axis of rotation, i.e., on the distribution of mass of the body.

2. In case of translational motion, force = mass x acceleration (F = ma)

Again, in case of rotational motion, torque = moment of inertia x angular acceleration (τ = lα)

Hence, the equation τ = lα is the rotational analogue of the equation F = ma. Moreover, we know that rotational analogues of force and linear acceleration are torque and angular acceleration, respectively.

So, comparing the above two equations, we can say that the rotational analogue of the mass of a body is its moment of inertia. Hence, the moment of inertia in rotational motion plays the same role as the mass in the case of translational motion.

3. The moment of inertia of a rigid body about a specific axis does not depend on the total mass \(\left(M=\sum_i m_i\right)\) of the body but on the distribution of mass of the constituent particles i.e., \(\sum_i m_i r_i^2\) of the body.

As the distribution of masses from the axis of rotation changes, the moment of inertia is due to the change of the axis of rotation in its position. Except in those cases, the moment of inertia of the rigid body about a specific axis of rotation. It can safely be assumed to be a scalar quantity.

Concept Of Moment Of Inertia: it has been said that the moment of inertia in rotational motion plays the same role as the mass in translational motion. It is evident from the following discussion.

• We know that the mass of a body in translational motion can be called its translational inertia. This is because mass is nothing but the hindrance that is generated in a body to resist any change in its translational motion.
• In the case of rotational motion, a body is compelled to change its state of motion when an external torque (rotational analogue of force) acts on it.
• In the absence of external torque, the body either remains at rest or executes uniform circular motion. It means that the moment of inertia of a body can be called its rotational inertia.
• It resists any change in the rotational motion of the body. To sum up it can be said that the relation between moment of force (torque) and moment of inertia is similar to the relation between force and mass.
• It is clear that the more the moment of inertia of a body about an axis, the more the torque necessary to rotate the body about that axis or to stop the body from rotating.

Two Important Theorems Regarding Moment Of Inertia: A regular-shaped body usually has some axis of symmetry. When the body rotates about such an axis, it undergoes just a spinning motion; during this spin, the entire body remains confined in the same region of space. A few examples of such axes of symmetry are:

1. Circular Ring Or Circular Disc: The axis passing through the centre of the circle and perpendicular to its plane is the axis of symmetry.
2. Sphere: Any diameter is an axis of symmetry.
3. Right Circular Cylinder: The axis passing through the centres of the two circular faces is the axis of symmetry.

Now, it should be mentioned that the symmetry axis is not the only possible axis of rotation of a rigid body; a body may rotate about any other axis as well.

• For example, the diurnal motion of the earth (a sphere) is about its diameter, which is of course an axis of symmetry. In addition, the Earth rotates around the distant sun the axis of rotation passing through the sun is certainly not an axis of symmetry of the Earth. Earth has a different moment of inertia about that axis also.
• From the above discussion, it is evident that a rigid body may rotate about many possible axes. Fortunately, it is not necessary to tabulate the formulae for moments of inertia corresponding to all those axes.
• The following two theorems help us to find the moments of inertia of a body about some special axes of rotation, provided that the expression for the moment of inertia about a symmetry axis is known beforehand.

1. Parallel-Axes Theorem: This theorem is applicable for a body of any shape.

The moment of inertia (I) of a rigid body about any axis is equal to the sum of its moment of inertia (I_cm) about a parallel axis through its centre of mass and the product of the mass (M) of the body with the square of the perpendicular distance (d) between the two axes.

The mathematical form of the theorem, I = I_cm + Md² …(1)

WBCHSE Class 11 Physics System Of Particles Motion

Parallel-Axes Theorem Explanation: Let a body is composed of an infinite number of straight-line segments parallel to the z-axis. The masses of the segments are m₁, m₂, m₃,….. The part at which the body intercepts the xy-plane is shown.

Let, the total mass of the body M is concentrated at that intersection and mass m₁ is at a distance r from the z-axis

Now, the moment of inertia of \(m_1\) about z-axis, \(I_1=m_1 r^2=m_1\left(x_1^2+y_1^2\right)\)

= \(m_1\left\{\left(x_{\mathrm{cm}}+x_1^{\prime}\right)^2+\left(y_{\mathrm{cm}}+y_1^{\prime}\right)^2\right\}\)

= \(m_1\left(x_{\mathrm{cm}}^2+y_{\mathrm{cm}}^2\right)+2 m_1\left(x_{\mathrm{cm}} x^{\prime}+y_{\mathrm{cm}} y^{\prime}\right)\) + \(m_1\left(x_1^{\prime 2}+y_1^{\prime 2}\right)\)

Therefore, the moment of inertia of the whole body about the z-axis,

I = \(\left(x_{\mathrm{cm}}^2+y_{\mathrm{cm}}^2\right) \sum m_i+2 x_{\mathrm{cm}} \sum_i m_i x_i^{\prime}\)

+ \(2 y_{\mathrm{cm}} \sum_i m_i y_i^{\prime}+\sum_i m_i\left(x_i^{\prime 2}+y_i^{\prime 2}\right) \cdots(2)\)…(2)

[where \(m_i=\) mass of the i th particle]

Now, \(\left(x_{\mathrm{cm}}^2+y_{\mathrm{cm}}^2\right)=d^2 and \sum m_i=M\).

Again \(\frac{\sum_i m_i x_i^{\prime}}{\sum_i m_i}\) and \(\frac{\sum_i m_i y_i^{\prime}}{\sum_i m_i}\) are x and y-coordinates respectively of the mean position of the particles of mass \(m_i\) about the centre of mass of the body.

∴ \(\frac{\sum_i m_i x_i^{\prime}}{\sum_i m_i}=0=\frac{\sum_i m_i y_i^{\prime}}{\sum_i m_i}\)

Hence, \(\sum_i m_i x_i{ }^{\prime}=0=\sum_i m_i y_i{ }^{\prime}\)

The last term of equation (2) is the moment of inertia of the body about A B.

Hence, \(\sum_i m_i\left(x_i^{\prime 2}+y_i^{\prime 2}\right)=I_{\mathrm{cm}}\)

∴ I = \(I_{\mathrm{cm}}+M d^2\)

System of particles motion

2. Perpendicular-Axes Theorem: The moment of inertia (I_Z) of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moments of inertia (I_x + I_y) of the lamina about two mutually perpendicular axes lying on the plane of the lamina and intersecting each other at the point through where the perpendicular axis passes.

The mathematical form of the theorem, I_x + I_y = I_z ….(3)

WBCHSE Class 11 Physics System Of Particles Motion

Perpendicular-Axes Theorem Explanation: Let the plane lamina be composed of an infinite number of particles and the masses of the particles are m₁, m₂, m₃,…… Let the distance of the particle of mass m₁ from the axes x, y and z be y₁, x₁ and d₁ respectively.

Now, the moments of inertia of the particle of mass m₁ about x, y and z-axes, \(I_{x_1}=m y_1^2, I_{y_1}=m_1 x_1^2, I_{z_1}=m_1 d_1^2\)

Therefore, the moments of inertia of the whole lamina about x, y and z-axes,

⇒ \(I_x=\sum_i m_i y_i^2, I_y=\sum_i m_i x_i^2, I_z=\sum_i m_i d_i^2\)

∴ \(I_x+I_y=\sum_i m_i\left(x_i^2+y_i^2\right)=\sum_i m_i d_i^2=I_z\)

It is to be noted that this theorem of perpendicular axes is applicable only for plane sheets of small thicknesses.

Determination Of Moment Of Inertia Of Some UniForm Symmetrical Objects:

1. Moment Of Inertia Of A Uniform Rod About The Perpendicular Axis To Its Length Passing Through Its Centre Of Mass: Let PQ be a uniform rod of mass m and length l. The centre of mass is at the midpoint O of the rod.

Considering O as the origin (0,0) and the x-axis along the length of the rod, the position coordinates of the points P and Q are (-\(\frac{1}{2}\), o) and (\(\frac{1}{2}\), o) respectively. The moment of inertia about the axis CD passing through the point O and perpendicular to the rod is to be determined.

Mass per unit length of the rod = \(\frac{m}{l}\)

Let us consider a small segment dx which is at a distance x from point O.

So, the mass of length dx = (\(\frac{m}{l}\)dx)

Moment of inertia of this small segment dx about CD = (\(\frac{m}{l}\)dx)x²

Hence, the moment of inertia of the whole rod about CD,

⇒ \(I_{C D}=\int_{-l / 2}^{L / 2} \frac{m}{l} x^2 d x=\frac{m}{l}\left[\frac{x^3}{3}\right]_{-\frac{l}{2}}^{\frac{l}{2}}\)

= \(\frac{m}{3 l}\left[\left(\frac{l}{2}\right)^3-\left(-\frac{l}{2}\right)^3\right]\)

= \(\left(\frac{m}{3 l} \cdot \frac{3}{4}\right)=\frac{1}{12} m l^2\)

WBCHSE Class 11 Physics System Of Particles Motion

2. Moment Of Inertia Of A Uniform Rod About The Perpendicular Axis To Its Length Passing Through One End Of The Rod (Application Of Parallel-Axes Theorem): Suppose, the mass of the rod = m, length of the rod = l. Moment of inertia of A the rod about the axis CD passing through its centre of mass and perpendicular to its length, I_CD = \(\frac{1}{12}\)ml²

Now by the parallel axes theorem we can write,

⇒ \(I_{A B}=I_{C D}+m\left(\frac{l}{2}\right)^2\)

= \(\frac{1}{12} m l^2+\frac{1}{4} m l^2=\frac{1}{3} m l^2\)….(1)

3. Moment Of Inertia Of A Uniform Rectangular Lamina About An Axis Parallel To Its Length And Breadth Passing Through Its Centre Of Mass: Suppose, the mass of the lamina = m, length = l, breadth = b. The centre of mass of the lamina is O. The moment of inertia of the lamina about CD parallel to its breadth and passing through O is to be determined.

The mass per unit area of the rectangular lamina = \(\frac{m}{l b}\). Let us imagine a small rectangular strip of width dr at a distance r from CD

Area of this strip = bdr

Mass of this strip = bdr = \(\frac{m}{l b}\)bdr = \(\frac{m}{l}\)dr

Therefore, moment of inertia of the whole lamina about the axis parallel to its breadth and passing through the centre of mass,

Similarly, moment of inertia of the lamina about an axis parallel to its length and passing through the centre of mass, I_y = \(\frac{1}{12}\)mb²

System of particles motion

4. Moment Of Inertia Of A Uniform Rectangular Lamina About An Axis Perpendicular To Its Plane Passing  Through Its Centre Of Mass (Application Of Perpendicular-Axes Theorem): Suppose, the mass of the lamina = m; length of the lamina = l; breadth of the lamina = b

Suppose O be the centre of mass of the lamina. OX and OY are the two axes lying on the plane of the lamina, mutually perpendicular to each other. The axis OZ is perpendicular to the plane of the lamina.

WBCHSE Class 11 Physics System Of Particles Motion

We know, the moment of inertia of the lamina about an axis passing through its centre of mass and parallel to its length, I_x = \(\frac{1}{12}\)mb²

The moment of inertia of the lamina about an axis passing through its centre of mass and parallel to its breadth, I_y = \(\frac{1}{12}\)ml²

Now, by the perpendicular-axes theorem we can write,  \(I_z =I_x+I_y=\frac{1}{12} m b^2+\frac{1}{12} m R^2\)

= \(\frac{1}{12} m\left(b^2+R^2\right)\)….(2)

5. Moment Of Inertia Of A Ring About An Axis Passing Through Its Centre And Perpendicular To The Plane Of The Ring: The mass of the circular ring is m and the radius of the ring is r, whose centre is O. The moment of inertia about AB passing through the point O and perpendicular to the plane of the ring to be calculated.

Let us imagine a small element of length dx on the circumference of the ring.

Mass per unit length of the ring = \(\frac{m}{2 \pi r}\)

Mass of that small element = \(\frac{m}{2 \pi r}\) dx

The moment of inertia of the element of length dx about AB = \(\left(\frac{m}{2 \pi r} d x\right) r^2=\frac{m r}{2 \pi} d x\)

∴ The moment of inertia of the ring about AB, \(I=\int_0^{2 \pi r} \frac{m r}{2 \pi} d x=\frac{m r}{2 \pi}[2 \pi r-0]=m r^2\)

6. Moment Of Inertia Of A Ring About Its Diameter (Application Of Perpendicular Axes Theorem): Let AB and CD be the axes along two mutually perpendicular diameters of the ring.

Now, by the theorem of perpendicular axes we can write, a moment of inertia of the ring about the axis AB + moment of inertia of the ring about the axis CD = moment of inertia of the ring about an axis through the centre of the ring O) and perpendicular to its plane,

i.e., I_AB + I_CD = mr² [where m = mass of the ring, r = radius of the ring]

For symmetry of the ring, I_AB + I_CD = I (say)

∴ I + I = mr² or, I = \(\frac{m r^2}{2}\) …..(3)

So, moment of inertia of a ring about its diameter = \(\frac{m r^2}{2}\)

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

7. Moment Of Inertia Of A Circular Disc About An Axis Passing Through Its Centre And Perpendicular To The Plane Of The Disc: P is a circular disc of mass m and radius r with centre O. The moment of inertia about AB passing through the point O and perpendicular to the plane of the disc is to be calculated.

Mass per unit area of the disc = \(\frac{m}{\pi r^2}\)

Let us imagine an annular ring of width dx at a distance x (x < r) from the centre of the disc.

Area of this annular ring = \(\left(\frac{m}{\pi r^2}\right) 2 \pi x d x=\frac{2 m}{r^2} x d x\)

Therefore, a moment of inertia of this annular ring about

AB = \(\left(\frac{2 m}{r^2} x d x\right) x^2=\frac{2 m}{r^2} x^3 d x\)

Hence, moment of inertia of the whole disc about AB,

I = \(\int_0^r \frac{2 m}{r^2} x^3 d x=\frac{2 m}{r^2}\left[\frac{x^4}{4}\right]_0^r\)

= \(\frac{2 m}{4 r^2}\left[r^4-0\right]=\frac{m r^2}{2}\)

8. Moment Of Inertia Of A Circular Disc About Its Diameter (Application Of Perpendicular-Axes Theorem): Since the disc is symmetrical with respect to all diameters, its moment of inertia about every diameter is the same.

System of particles motion

Let AB and CD be the axes along two mutually perpendicular diameters of the circular disc.

Now, by the perpendicular-axes theorem, we can write, a moment of inertia of the disc about the axis AB + moment of inertia of the disc about the axis CD = moment of inertia of the disc about an axis through the centre of the disc O and perpendicular to its plane,

i.e., I_AB + I_CD = \(\frac{m r^2}{2}\)

[where, m = mass of the circular disc, r = radius of the circular disc]

For symmetry of the disc I_AB + I_CD = I (say)

∴ I+I= \(\frac{m r^2}{2}\)

or, \(I=\frac{m r^2}{4}\)……(4)

So, moment of inertia of a circular disc about its diameter = \(\frac{3 r^2}{4}\)

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

9. Moment Of Inertia Of A Circular Disc About A Tangent In The Plane Of The Disc (Application Of Parallel-Axes Theorem): Let CD be a tangent in the plane of the circular disc and AB be an axis along the diameter parallel to CD.

Let the mass of the disc be m and its radius is r.

By parallel-axes theorem, we can write, a moment of inertia of the disc about CD = moment of inertia of the disc about AB+mr²

i. e., \(I_{C D}=I_{A B}+m r^2\)

= \(\frac{m r^2}{4}+m r^2\)

= \(\frac{5}{4} m r^2\)…(5)

So, the moment of inertia of a circular disc about a tangent on the plane of the disc = \(\frac{5}{4}\)mr².

10. Moment Of Inertia Of A Circular Disc About A Tangent Perpendicular To The Plane Of The Disc (Application Of Parallel-Axes Theorem): Let CD be a tangent to the circular disc perpendicular to its plane and AB be an axis passing through the centre O of the disc and parallel to CD.

Let the mass of the disc be m and its radius is r.

By parallel-axes theorem, we can write, a moment of inertia of the disc about CD = moment of inertia of the disc about AB+mr²

i.e., \(I_{C D}=I_{A B}+m r^2\)

= \(\frac{m r^2}{2}+m r^2\)

= \(\frac{3}{2} m r^2\)….(6)

So, the moment of inertia of a circular disc about a tangent perpendicular to its plane = \(\frac{3}{2}\)mr².

System of particles motion

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Radius Of Gyration: Notice from the above table, that in all cases moment of inertia of an extended body rotating about a specific axis depends not on total mass but on the mass distribution of the body from that very axis.

We shall now find a measuring way in which the mass of a rotating rigid body is related to the moment of inertia. For this, a new parameter, the radius of gyration (it) is introduced.

We notice that in all cases Moment of inertia can be expressed as I = Mk² form, where k has the dimension of length. ‘ k’ is a geometric property of the body and axis of rotation.

We know that if a point mass M is at a distance k from the axis of rotation, its moment of inertia, I = Mk². From this, we can define radius of gyration.

∴ k = \(\sqrt{\frac{I}{M}}\)

∴ I = Mk²

Radius Of Gyration Definition: If the whole mass of a body is assumed to be concentrated at a point such that the moment of inertia of the whole body equals the moment of inertia of that point, then the radial distance of the point from the axis of rotation is called the radius of gyration.

Radius Of Gyration Example: Moment of inertia of a solid sphere about its diameter is, I = \(\frac{2}{5}\) Mr². So, its radius of gyration with respect to its diameter,

k = \(\sqrt{\frac{\frac{2}{5} M r^2}{M}}=\sqrt{\frac{2}{5}} r\)

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Rotational Kinetic Energy

Let PQR be a rigid body, revolving about the axis AB with angular velocity co. Due to this motion, the body possesses some kinetic energy. This kinetic energy is called rotational kinetic energy.

The rigid body can be assumed as an aggregate of a number of particles. Let the masses of the particles be m₁, m₂, m₃,….., etc., at distances r₁, r₂, r₃,….., etc., respectively from the axis of rotation AB. Since the body is rigid, the angular velocity of the constituent particles is the same, i.e., ω. However, due to the difference in their distances from the axis of rotation, the linear velocities of different particles are different.

Let the linear velocity of the particle of mass m₁ be v₁

So, v₁ = ωr₁

System of particles motion

So, the kinetic energy of the particle of mass \(=\frac{1}{2} m_1 v_1^2=\frac{1}{2} m_1 \omega^2 r_1^2\)

Similarly, the kinetic energy of the particle of mass = \(\frac{1}{2} m_2 v_2^2=\frac{1}{2} m_2 \omega^2 r_2^2\) and so on.

In this way, adding the kinetic energies of all particles, the kinetic energy of the entire rigid body is obtained.

So, the rotational kinetic energy of the body

= \(\frac{1}{2} m_1 \omega^2 r_1^2+\frac{1}{2} m_2 \omega^2 r_2^2+\frac{1}{2} m_3 \omega^2 r_3^2+\cdot\)

= \(\frac{1}{2} \omega^2\left(m_1 r_1^2+m_2 r_2^2+m_3 r_3{ }^2+\cdots\right)\)

= \(\frac{1}{2} \omega^2 \sum_i m_i r_i^2\) = \(\frac{1}{2} \omega^2 I\) (where, \(I=\sum_i m_i r_i^2\))

[r_i = the perpendicular distance of i-th particle from the axis of rotation, m_i = mass of the i -th particle of the rigid body] = \(\frac{1}{2}\) Iω²

Here, I = \(\sum_i m_i r_i^2\) = moment of inertia of the body about the axis of rotation.

Comparing torque with force, it is seen that the moment of inertia in rotational motion plays the same role as that played by the mass in translational motion. Now, it is also seen that the same conclusion can be drawn by comparing the translational kinetic energy \(\frac{1}{2}\)mv² with the rotational kinetic energy \(\frac{1}{2}\) Iω².

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Relation Between Rotational Kinetic Energy With Work Done: Let a force F on a body of mass m cause a change in its kinetic energy (Δk) only. If the body under this force is displaced linearly from the initial position x₁ to the final position x₂, then work done by the force,

W = \(\int_{x_1}^{x_2} F d x\)

According to the work-kinetic energy theorem, Δk = W

i.e., \(\frac{1}{2} m v_2^2-\frac{1}{2} m v_1^2=\int_{x_1}^{x_2} F d x\)….(1)

[where v₁ = initial speed, v₂ = final speed]

Now consider a torque z on a body of moment of inertia I (about a certain axis of rotation) causes a change in its rotational kinetic energy (Δk_r) only. If the body under this torque is displaced from the initial angular position θ₁ to the final angular position θ₂ by the torque,

W = \(\int_{\theta_1}^{\theta_2} \tau d \theta\)

Just like above equation (1) now we can relate work and rotational kinetic energy as below: Δk_r = W

i.e., \(\frac{1}{2} I \omega_1^2-\frac{1}{2} I \omega_2^2=\int_{\theta_1}^{\theta_2} \tau d \theta\)….(2)

[where ω₁ = initial angular speed, ω₂ = final angular speed] when τ is constant, W = τ(θ₂ – θ₁)

Therefore, power, i.e., the rate at which the work is done, P = \(\frac{W}{dt}\) = τω

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Angular Momentum

The rotational analogues of the mass (m) of a body and its linear velocity (v) are moment of inertia (I) and angular velocity (ω), respectively. Hence, the rotational analogue of the linear momentum (mv) of the body is Iω. This physical quantity is called the angular momentum (L) of the body.

Angular Momentum Definition: The dynamical property generated in a body under rotational motion, due to the moment of inertia about an axis and angular velocity, is called the angular momentum of the body about that axis.

Angular momentum is measured by the product of moment of inertia and angular velocity, i.e., L = Iω.

Since I is a scalar and ω is an axial vector, angular momentum L is also an axial vector whose direction is along the axis of rotation, and in the direction of ω.

Unit And Dimension Of Angular Momentum:

CGS System: g · cm² · s^-1

SI: kg · m² · s^-1

Dimension of L = dimension of I x dimension of ω = ML² x T^-1 = ML²T^-1

Relation Between Linear Momentum And Angular Momentum: Suppose a body is revolving with an angular velocity ω about an axis. If m₁, m₂, m₃,…. are the constituent particles of that body and they are at distances r₁, r₂, r₃,…. respectively from the axis of rotation, then the moment of inertia of the body,

I = \(m_1 r_1^2+m_2 r_2^2+m_3 r_3^2+\cdots=\sum_i m_i r_i^2\)

In the case of pure rotation, the angular velocity of each particle becomes equal to the angular velocity of the body.

So, the angular momentum of the body,

L = \(I \omega=\sum_i m_i r_i^2 \cdot \omega=\sum_i m_i r_i v_i\) (because \(v_i=\omega r\))

= \(\sum_i r_i \times m_i v_i=\sum_i r_i \times p_i\)

[p_i = m_iv_i = linear momentum of i-th particle]

For the particles, the quantities r₁ x m₁v₁, r₂ x m₂v₂,…… etc., can be called the moments of linear momentum, or in brief, moments of momentum (in analogy with the moment of force).

So, the angular momentum of a body about an axis is the algebraic sum of the moments of linear momentum about the same axis, of all particles constituting the body.

Thus, for a particle rotating about a circle of radius r and having a linear momentum p, the angular momentum will be L = rp.

Vector Representation: The vector representation for the relation between linear and angular momentum is \(\vec{L} = \vec{r} \times \vec{p}\). This is often referred to as the defining equation of \(\vec{L}\).

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

We know the vector representation for the relation between linear velocity and angular velocity is \(\vec{v}=\vec{\omega} \times \vec{r}\).

If \(\vec{v}\) and \(\vec{\omega}\) are replaced by \(\vec{p}\) and \(\vec{L}\), respectively, the geometric form for the relation of \(\vec{L}\), \(\vec{p}\) and \(\vec{r}\) is obtained.

Relation Between Angular Momentum And Torque: In case of rotational motion, when a torque is applied to a body, an angular acceleration is produced in it. If the initial angular velocity of the body is ω₁ and its angular velocity after time t is ω₂, then the angular acceleration of the body,

α = \(\frac{\omega_2-\omega_1}{t}\)

Again, torque = moment of inertia x angular acceleration

or, \(\tau=I \alpha=I \times \frac{\omega_2-\omega_1}{t}=\frac{I \omega_2-I \omega_1}{t}\)

or, \(\tau t=I \omega_2-I \omega_1\)

Hence, torque x time = change in angular momentum of the body during that interval

This is the relation between torque and angular acceleration. From this relation, it is evident that a change in angular momentum takes place about the axis along which the torque acts on the body.

We know that in the case of translational motion, Ft = mv – mu and the rotational analogue of this equation is τt = Iω₂  – Iω₁. The quantity Ft is known as the impulse of force. Similarly, the quantity τt is known as the angular impulse or the impulse of torque.

Law Of Conservation Of Angular Momentum: Suppose the moment of inertia of a body changes from I₁ to I₂ in time t. In this case, the equation τt = Iω₂  – Iω₁ changes to τt = I₂ω₂  – I₁ω₁ Now, if no external torque acts on the body, i.e., if τ = 0, then from the equation, τt = I₂ω₂  – I₁ω₁ we get, I₂ω₂  – I₁ω₁ = 0, or, τt = I₂ω₂  = I₁ω₁

It means that the final angular momentum of the body is equal to its initial angular momentum, i.e., the angular momentum is conserved.

Law: if the net external torque on a body is zero, the angular momentum of the body rotating about an axis always remains conserved.

So, this law is nothing but the rotational analogue of the law of conservation of linear momentum.

Again we know, \(\frac{dL}{dT}\) = τ_ext

From this, it is clear that, if total external torque acts on a body is zero; its angular velocity decreases with the increase of its moment of inertia and vice versa i.e., angular momentum remains constant.

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Related Experiments And Practical Examples:

1. A man is sitting on a turntable holding a pair of dumbbells of equal mass, one in each hand with his arms out-stretched while the turntable rotates with a definite angular velocity, If the man suddenly draws the dumbbells towards his chest, the speed of rotation of the turntable is found to increase.

System of particles motion

• This is due to the fact that when the man draws the dumbbells towards his chest, the moment of inertia of the man about the axis of rotation decreases and his angular velocity increases due to conservation of angular momentum.
• If the man again stretches his arms, his angular velocity decreases due to an increase in moment of inertia, and the turntable consequently rotates slowly.

2. In a diving event, when a competitor dives from a high platform or springboard into water, he keeps his legs and arms outstretched and starts descending with less angular velocity, After that he curls his body by rolling the legs and arms inwards, his moment of inertia decreases.

• As angular momentum is conserved, his angular velocity goes on increasing rapidly. As a result, his body begins to spin rapidly and before reaching the surface of the water, he can perform a good number of somersaults.
• In the case of skating on the surface of ice or during the performance of acrobatics, the principle of conservation of angular momentum can be applied in a similar way.

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Angular Momentum Numerical Examples

Example 1. If the radius of the earth decreases by \(\frac{1}{2}\)%, then what will be the change in the length of a day? Assume that the earth is a uniform sphere and its moment of inertia, I = \(\frac{2}{5}\)MR², where M and R are the mass and the radius of the earth.
Solution:

If the mass of a solid sphere remains unaltered, then its moment of inertia ∝ (radius)².

Here, the changed radius \(=\frac{100-\frac{1}{2}}{100} R=\frac{199}{200} R\).

So, if the moment of inertia of the earth for its present radius R is I and the moment of inertia for its changed radius is I’, then

⇒ \(\frac{I}{I^{\prime}}=\frac{R^2}{\left(\frac{199 R}{200}\right)^2}=\left(\frac{200}{199}\right)^2\)

If the present angular velocity of the earth is ω and its changed angular velocity is ω’, then according to the principle of conservation of angular momentum,

⇒ \(I \omega=I^{\prime} \omega^{\prime}\)

or, \(\omega^{\prime}=\frac{I \omega}{I^{\prime}}\)

or, \(\frac{2 \pi}{T^{\prime}}=\frac{I}{I^{\prime}} \times \frac{2 \pi}{T}\)

or, \(T^{\prime}=\frac{I^{\prime}}{I} \cdot T=\left(\frac{199}{200}\right)^2 \times 24=23.76 \mathrm{~h}\)

∴ The length of the day will decrease by (24-23.76) = 0.24 h = 14 min 24 s

System of particles motion

Example 2. A solid sphere of mass 1 kg and of radius 10 cm is rotating about one of its diameters with an angular; velocity of π rad · s^-1. Calculate the kinetic energy of the sphere by using the relevant formula.
Solution:

Let the moment of inertia of the sphere about its diameter I = \(\frac{2}{5}\)MR², M = mass of the sphere and R = radius of the sphere.

The kinetic energy of the body = rotational kinetic energy of the body

= \(\frac{1}{2} I \omega^2=\frac{1}{2} \times \frac{2}{5} M R^2 \cdot \omega^2\)

= \(\frac{1}{5} \times 1000 \times(10)^2 \times \pi^2\)

= \(197392.09 \mathrm{erg} .\)

Example 3. A thin rod of length l and mass m per unit length is rotating about an axis passing through the midpoint of its length and perpendicular to it. Prove that its kinetic energy \(\frac{1}{24}\) mω²l³ = ω = angular velocity of the rod.
Solution:

Kinetic energy of the rod = \(\frac{1}{2}\) mω²

According to the problem,

I = \(\frac{1}{12}\)Ml² [M = mass of the rod = ml]

= \(\frac{1}{12}\) x ml x l² = \(\frac{m l^3}{12}\)

∴ Kinetic energy of the rod = \(\frac{1}{2} \times \frac{m l^3}{12} \times \omega^2=\frac{1}{24} m \omega^2 l^3 .\)

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Example 4. Calculate the moment of inertia of a solid cylinder of I length 10 cm and of radius 20 cm about its own axis. The density of the material of the cylinder = 9 g · cm^-3.
Solution:

L= length of the cylinder, R = radius of the cylinder and M = mass of the cylinder

= volume of the cylinder x density

=  πR²L X density

= π x (20)² x 10 x 9 g

Moment of inertia of a solid cylinder about its own axis,

I = \(\frac{1}{2} M R^2\)

I = \(\frac{1}{2} \times \pi \times(20)^2 \times 10 \times 9 \times(20)^2\)

= \(22.6 \times 10^6 \mathrm{~g} \cdot \mathrm{cm}^2\)

Example 5. A solid sphere of diameter 2 cm and of mass 20 g is rolling with a velocity of 3 cm · s^-1. What is the total kinetic energy of the sphere?
Solution:

Let M = mass of the sphere, R = radius of the sphere, V = linear velocity of the sphere, I = \(\frac{2}{5}\)MR² (moment of inertia of the sphere about its diameter), ω = \(\frac{V}{R}\)

Total kinetic energy of the sphere = translational kinetic energy + rotational kinetic

= \(\frac{1}{2} M V^2+\frac{1}{2} I \omega^2=\frac{1}{2} M V^2+\frac{1}{2} \times \frac{2}{5} M R^2\left(\frac{V}{R}\right)^2\)

= \(\frac{1}{2} M V^2+\frac{1}{5} M V^2=\frac{7}{10} M V^2=\frac{7}{10} \times 20 \times(3)^2\)

= \(126 \mathrm{erg}\)

System of particles motion

Example 6. A stone of mass m tied with a thread Is rotating along a horizontal circular path (force of gravity is neglected). The length of the thread decreases gradually in such a manner that the angular momentum of the stone remains constant with respect to the centre of the circle. If the tension in the thread Is T = Arⁿ, where A = constant, r = instantaneous radius of the circle, then find the value of n.
Solution:

If the instantaneous angular velocity of the stone is w, then angular momentum,

L = Iω = mr²ω = constant (according to the problem)

or, ω = \(\frac{L}{m r^2}\)

Here the tension in the thread provides the necessary centripetal force for rotation.

So, T = \(A r^n=m \omega^2 r=m \cdot \frac{L^2}{m^2 r^4} r=\frac{L^2}{m} r^{-3}\)

= \(A r^{-3} \quad\left(A=\frac{L^2}{m}=\text { constant }\right)\)

∴ n=-3 .

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Example 7. Two ends of a uniform rod weighing W, are placed on supports so that the rod remains horizontal. If a support at one end is suddenly removed, what will be the force exerted on the horizontal rod by the support at the other end?
Solution:

Let the length of the rod = l cm, its weight = W = Mg, where M is the mass of the rod. When the support at one end is removed suddenly, the centre of gravity of the rod falls downwards with an acceleration a. Let R = reaction force at the end with the support. Hence, if the C.G. now falls with an acceleration a, the rod will turn about the point P.

The torque on the rod = Mg · \(\frac{l}{2}\)

Also, Mg – R = Ma or, \(a=\frac{M g-R}{M}\)

Here moment of inertia, I =  \(\frac{1}{3}\)Ml² = moment of inertia of the rod about the perpendicular axis passing through the end of the rod and the angular acceleration, α = \(\frac{a}{V / 2}=\frac{2 a}{l}\)

∴ \(\frac{1}{3} M l^2 \alpha=M g \frac{l}{2}\) (because \(\tau=I \alpha\))

or, \(\frac{1}{3} M R^2 \cdot \frac{2 a}{l}=M g \frac{l}{2} \text { or, } \frac{2}{3} a=\frac{g}{2} \text { or, } \frac{2}{3}\left(\frac{M g-R}{M}\right)=\frac{g}{2}\)

R = \(\frac{M g}{4}=\frac{W}{4}\)

Therefore, when one support is removed, the support at the other end will exert a reaction force of \(\frac{W}{4}\)

System of particles motion

Example 8. A rod of length L and M is attached with a hinge on a wall at point O. After releasing the rod from its vertical position OA, when it comes to position OA’, what is the reaction on point O of the rod by the hinge?

Solution:

Let, the angular velocity of the rod at the horizontal position OA’ is ω.

∴ At that instant its kinetic energy = \(\frac{1}{2} I \omega^2=\frac{1}{2} \cdot \frac{M L^2}{3} \cdot \omega^2=\frac{M L^2 \omega^2}{6}\)

The centre of mass of the rod shifts down by \(\frac{L}{2}\) from OA to OA’.

So, decrease in potential energy of the rod = Mg\(\frac{L}{2}\)

According to the kinetic energy conservation law, \(M g \frac{L}{2}=\frac{M L^2 \omega^2}{6} \quad \text { or, } \omega=\sqrt{\frac{3 g}{L}}\)…(1)

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Two forces act on the rod at position OA’

1. Gravitational force (Mg) vertically downward direction and
2. Reaction force (n) of the hinge

Let, the horizontal and the vertical n components of n are n_x and n_y respectively; the horizontal and the vertical components of the acceleration of the centre of mass of the rod area a_x and a_y respectively.

∴ According to \(M g-n_y=M a_y\)….(2)

and \(n_x=M a_x=M \omega^2 \cdot \frac{L}{2}\)

(because \(a_x=\) centripetal acceleration)

= \(M \cdot \frac{3 g}{L} \cdot \frac{L}{2}=\frac{3}{2} M g\)

[putting the value of ω from equation (1)]

The rod starts to rotate due to the action of torque created by n_y and Mg.

If the angular acceleration of the rod is α, \(M g \cdot \frac{L}{2}=I \alpha=\frac{M L^2}{3} \alpha\)

∴ \(\alpha=\frac{3 g}{2 L}\)

The acceleration along the vertical direction, \(a_y=\frac{L}{2} \alpha=\frac{3 g}{4}\)

Putting the value of ay in equation (2) we get, \(M g-n_y=\frac{3 M g}{4} \text { or, } n_y=\frac{M g}{4}\)

∴ n = \(\sqrt{n_x^2+n_y^2}=\sqrt{\left(\frac{3}{2} M g\right)^2+\left(\frac{M g}{4}\right)^2}=\frac{\sqrt{37}}{4} M g\)

System of particles motion

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Motion of A Mass Suspended From A Rope Wrapped Around A Solid Cylinder

Let a solid cylinder of mass M and radius R is kept in such a way that it can rotate freely about its axis XX’. A mass m is suspended from a rope wrapped around the cylinder. It is then released from rest and for this, the cylinder begins to rotate about XX’.

Two forces act on the suspended mass m

1. Its weight (vertically downward) and
2. Tension of the rope T (vertically upward).

1. Acceleration Of The Attached Mass m: If the linear acceleration directed downward of mass m is a, then, mg – T = ma…..(1)

If the moment of inertia and angular acceleration about the axis of rotation is I and α respectively.

⇒ \(\tau=I \alpha=T R\)

∴ \(I \frac{a}{R}=T R \quad \text { or, } T=\frac{I a}{R^2}\)

From equation (1) we get, \(m a=m g-\frac{I a}{R^2}\)

or, \(m a+\frac{I a}{R^2}=m g\) or, \(a\left[m+\frac{l}{R^2}\right]=m g\)

or, \(a=\frac{g}{1+\frac{l}{m R^2}}\)….(2)

The moment of inertia about the axis of the cylinder.

I = \(\frac{1}{2} M R^2\)

∴ \(a=\frac{g}{1+\frac{1}{2} \frac{M R^2}{m R^2}}=\frac{g}{1+\frac{M}{2 m}}\)

2. Angular Acceleration Of The Cylinder: We know the angular acceleration

= \(\frac{\text { linear acceleration }}{\text { radius }}\)

Hence, \(\alpha=\frac{a}{R}\)

From equation (2) we get., \(\alpha=\frac{g}{1+\frac{1}{m R^2}}\)

3. Tension Of The Thread: From equation (1) we get,

T = \(m g-m a=m g-\frac{m g}{1+\frac{1}{m R^2}}\)

= \(m g\left[1-\frac{1}{1+\frac{1}{m R^2}}\right]=m g\left[\frac{\frac{l}{m R^2}}{1+\frac{l}{m R^2}}\right]=\frac{m g}{\frac{m R^2}{l}+1}\)

The moment of inertia about the axis of the cylinder,

I = \(\frac{1}{2} M R^2\)

∴ \(T=\frac{m g}{\frac{2 m R^2}{M R^2}+1}=\frac{m g}{1+\frac{2 m}{M}}\)

System of particles motion

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Mixed Motion

If a body undergoes translation and rotation simultaneously, then its motion is called a mixed motion. As for exam¬ ple, we will now discuss the rolling of a body without slipping.

Downward Rolling Of A Body Without Slipping On An Inclined Plane: Let a body (say a cylinder or a sphere) of mass M and radius R roll down a plane without slipping inclined at an angle θ with the horizontal. We shall now find an expression for the linear acceleration (a) of the centre of mass of the body rolling down the plane. After that, the value of friction will also be calculated.

1. Linear Acceleration Of The Centre Of Mass Of The Body: Different forces with their components are shown.

Applying Newton’s 2nd law of motion along the inclined plane, we get, Mgsinθ – f = Ma…(1)

Here f is the static friction which generates torque and the angular acceleration. If the moment of inertia about an axis passing through the centre of mass are I and α respectively, then the torque acting on the body,

⇒ \(\tau=I \alpha=f R\)

∴ f = \(\frac{I \alpha}{R}=\frac{I a}{R^2}\)

(because \(\alpha=\frac{a}{R}\))

From the equation we get, \(M g \sin \theta-\frac{I a}{R^2}=M a\)

or, \(g \sin \theta=a+\frac{I a}{M R^2} or, a=\frac{g \sin \theta}{1+\frac{I}{M R^2}}\)

Special cases: In case of solid cylinder, I = \(\frac{1}{2} M R^2\)

∴ a = \(\frac{g \sin \theta}{1+\frac{\frac{1}{2} M R^2}{M R^2}}=\frac{2}{3} g \sin \theta\)

In case of solid sphere, I = \(\frac{2}{5} M R^2\)

∴ a = \(\frac{g \sin \theta}{1+\frac{\frac{2}{5} M R^2}{M R^2}}=\frac{5}{7} g \sin \theta\)

In the case of a hollow cylinder, I = MR²

∴ a = \(\frac{g \sin \theta}{1+\frac{M R^2}{M R^2}}=\frac{1}{2} g \sin \theta\)

In case of hollow sphere, \(I=\frac{2}{3} M R^2\)

∴ a = \(\frac{g \sin \theta}{1+\frac{\frac{2}{3} M R^2}{M R^2}}=\frac{3}{5} g \sin \theta\)

2. Friction acting on the body: From equation (2) we get, f = \(\frac{I a}{R^2}\)

Special cases: In the case of solid cylinders, \(f=\frac{1}{2} M R^2 \times \frac{1}{R^2} \times \frac{2}{3} g \sin \theta=\frac{1}{3} M g \sin \theta\)

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

In the case of solid spheres,

∴ \(f=\frac{2}{5} M R^2 \times \frac{1}{R^2} \times \frac{1}{2} g \sin \theta=\frac{1}{2} M g \sin \theta\)

In the case of hollow cylinders,

∴ \(f=M R^2 \times \frac{1}{R^2} \times \frac{1}{2} g \sin \theta=\frac{1}{2} M g \sin \theta\)

In the case of hollow spheres,

∴ \(f=\frac{2}{3} M R^2 \times \frac{1}{R^2} \times \frac{3}{5} g \sin \theta=\frac{2}{5} M g \sin \theta\)

System of particles motion

WBCHSE Class 11 Physics Rotation Of Rigid Bodies

Comparison Between Linear And Rotational Motions

In the discussion of rotational motion, some physical quantities and numerical formulae, that we have already dealt with, are the rotational analogues of some physical quantities and numerical formulae of linear motion. These quantities and formulae are given below:

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies

Linear And Rotational Motions Numerical Examples

Example 1. A particle of mass m is projected at an angle of 45° with the horizontal. At the highest point of its motion (h), what will be its angular momentum; concerning the point of projection?
Solution:

At any point, the horizontal component of the velocity of the particle = \(v_x=v \cos 45^{\circ}=\frac{v}{\sqrt{2}}\); the vertical velocity at the highest point = 0

If the time taken by the particle to reach the highest point is t, then 0 = vsin45°- gt

or, t = \(\frac{\nu}{\sqrt{2} g}\)[initial vertical velocity = vsin45°]…(1)

If the maximum height attained is h, then

h = \(\nu \sin 45^{\circ} \cdot t-\frac{1}{2} g t^2=\frac{v}{\sqrt{2}} \cdot \frac{\nu}{\sqrt{2} g}-\frac{1}{2} g \cdot \frac{v^2}{2 g^2}\)

= \(\frac{v^2}{2 g}-\frac{v^2}{4 g}=\frac{v^2}{4 g}\)….(2)

∴ Angular momentum of the particle with respect to the point of projection

= \(m v_x \times h=\frac{m \nu}{\sqrt{2}} \cdot \frac{v^2}{4 g}=\frac{m v^3}{4 \sqrt{2} g}\)

From equation (2) we get, \(v^2=4 g h \text { or, } v=2 \sqrt{g h}\)

∴ The angular momentum of the particle about the point of projection.

= \(\frac{m}{4 \sqrt{2} g} \cdot 8 \cdot(g h)^{3 / 2}=m h \sqrt{2 g h} .\)

System of particles motion

Example 2. Initially, a sphere of radius r is rotating with an angular velocity ω about its own horizontal axis. When the sphere falls on a surface (coefficient of friction μ), it begins to skid first and then starts rotating without skidding.

1. What will be the final linear velocity of its centre of mass?
2. How much distance will the sphere cover before reaching this velocity?

Answer:

1. Let the mass of the sphere be m. Then its moment of inertia about the axis of rotation, I = \(\frac{2}{5} m r^2\)

The moment of frictional force (μmg) resists the rotational motion of the sphere. If the angular retardation is α, then \(\mu m g r=I \alpha=\frac{2}{5} m r^2 \alpha\)

or, \(\alpha=\frac{5 \mu g}{2 r}\)…(1)

Due to this, if the angular velocity of the sphere becomes ω’ in time t, then \(\omega^{\prime}=\omega-\alpha t=\omega-\frac{5 \mu g t}{2 r}\)

The speed of a rotating point on the upper surface of the sphere, \(v^{\prime}=\omega^{\prime} r=\left(\omega-\frac{5 \mu g t}{2 r}\right) r\)…..(2)

Again, due to frictional force μmg, if the sphere skids over the surface with an acceleration a, then μmg = ma or, a = μg

∴ The linear velocity of the centre of mass of the sphere in time t, v = 0 + at= μgt…..(3)

The condition of rotational motion of the sphere without skidding is, v = v’. If the values of these two velocities become the same in time t, the sphere will undergo pure rotation.

From equations (2) and (3) we get, \(\mu g t =\left(\omega-\frac{5 \mu g t}{2 r}\right) r=\omega r-\frac{5 \mu g t}{2}\)

or, \(\frac{7}{2} \mu g t =\omega r \quad \text { or, } \quad t=\frac{2 \omega r}{7 \mu g}\)

∴ From equation (3) we get, \(\nu=\mu g \times \frac{2 \omega r}{7 \mu g}=\frac{2}{7} \omega r.\)

2. Distance covered, x = \(\frac{1}{2} a t^2=\frac{1}{2} \mu g\left(\frac{2 \omega r}{7 \mu g}\right)^2=\frac{2}{49} \cdot \frac{r^2 \omega^2}{\mu g} .\)

System of particles motion

Example 3. A small sphere of radius r at rest begins to slide down the surface of a hemispherical bowl from the brim of the bowl. When the sphere reaches the bottom of the bowl, what fraction of its total energy will be converted into translational kinetic energy and what fraction into rotational kinetic energy?
Solution:

Let the initial position of the small sphere be A

The velocity of the sphere, when it reaches the point B = V

∴ At the point B, translational kinetic energy of the sphere = \(K_t=\frac{1}{2} m V^2\) and rotational kinetic energy of the sphere

⇒ \(K_r =\frac{1}{2} I \omega^2\)

= \(\frac{1}{2} \times \frac{2}{5} m r^2 \times \omega^2\)

= \(\frac{1}{5} m V^2\)

∴ Total energy of the sphere, \(K =K_t+K_r\)

= \(\frac{1}{2} m V^2+\frac{1}{5} m V^2=\frac{7}{10} m V^2\)

∴ The ratio of the translational kinetic energy to the total kinetic energy, \(\frac{K_t}{K}=\frac{\frac{1}{2} m V^2}{\frac{7}{10} m V^2}=\frac{5}{7} \).

Again, the ratio of the rotational kinetic energy to the total kinetic energy, \(\frac{K_r}{K}=\frac{\frac{1}{5} m V^2}{\frac{7}{10} m V^2}=\frac{2}{7}\)

So, \(\frac{5}{7}\) part of the total energy will be converted into translational kinetic energy and \(\frac{2}{7}\) part into rotational kinetic energy.

 

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies Useful Relations For Solving Examples

Torque, \(\vec{tau}\) (\(\vec{r} \times \vec{F}\) = (\(\vec{F}\) = force applied on the body, \(\vec{r}\) = position vector of the point of application of the force with respect to the origin)

Torque (τ) = moment of inertia (I) x angular acceleration (α); here, I = \(\sum_i m_i r_i^2\) = moment of inertia of the body about its axis of rotation.

Work done by the couple, W = τ · θ = torque x angular displacement

For n complete revolutions, work done by a couple, W = 2πnx torque

If the mass of an extended body is M and its moment of inertia about any axis of rotation is I, the radius of gyration,

K = \(\sqrt{\frac{I}{M}} \quad \text { or, } \quad I=M k^2\)

Rotational kinetic energy of a body = \(\frac{1}{2}\)Iω²

Angular momentum (I) = moment of inertia (I) x angular velocity (ω)

System of particles motion

If a particle revolves along a circular path of instantaneous radius vector \(\vec{r}\) and if the linear momentum of the particle is \(\vec{p}\), then the angular momentum of the particle, \(\vec{L} = \vec{r} \times \vec{p}\)

Torque x time = change in angular momentum of the body during that time = angular impulse or impulse of torque

∴ \(\vec{\tau}_{\text {ext }}=\frac{d \vec{L}}{d t}\)

The rate of change of angular momentum of a body is equal to the external torque acting upon the body.

Power, P = τω = torque x angular velocity

The total kinetic energy of rolling = translational kinetic energy+rotational kinetic energy

= \(\frac{1}{2}\)mv² + \(\frac{1}{2}\)Iω²

 

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies Very Short Answer Type Questions

Question 1. What is the unit of angular momentum?
Answer: g · cm² · s^-1

Question 2. State whether the length of a day will increase or decrease if the radius of the earth becomes half of its present value keeping its mass constant.
Answer: Decrease

Question 3. What is the dimension of angular momentum?
Answer: ML²T^-1

Question 4. What is the vector relation of linear momentum and angular momentum?
Answer: \(\vec{L}=\vec{r} \times \vec{p}\)

Question 5. A girl is standing at the centre of a rotating horizontal platform with her hands drawn inwards. What will happen if she stretches her hands horizontally?
Answer: The platform will rotate slowly

Question 6. Write down the dimension of torque.
Answer: ML²T^-2

Question 7. When we turn on a tap we apply a _____ on it with the help of our fingers.
Answer: Couple

Question 8. What is the CGS unit of moment of inertia?
Answer: g · cm²

Question 9. Write down the expression of the moment of inertia of a circular disc (mass = m, radius = r) about the perpendicular axis passing through its centre.
Answer: \(\frac{1}{2}\)mr²

Question 10. Two spheres have equal masses and their external radii are the same. One of them is solid and the other hollow. Which one will have a greater radius of gyration?
Answer: Hollow sphere

System of particles motion

Question 11. Is the radius of gyration a constant quantity?
Answer: No

Question 12. What is the moment of inertia of a solid sphere (radius = r, mass = m ) about an axis passing through any of its diameters?
Answer: \(\frac{2}{5}\)mr²

Question 13. What is the radius of gyration of a solid sphere with respect to its diameter?
Answer: \(\sqrt{\frac{2}{5}} R\)

Question 14. What is the kinetic energy of a rotating body about its axis of rotation?
Answer: \(\frac{1}{2}\)Iω²

Question 15. What is needed to produce pure rotation?
Answer: Torque

Question 16. Write down the vector equation relating torque and angular momentum.
Answer: \(\vec{\tau}=\frac{d \vec{L}}{d t}\)

Question 17. If the moment of inertia of a body rotating about an axis is increased, state whether its angular velocity increases or decreases when no external torque acts on the body.
Answer: Decrease

Question 18. If the ice at the polar regions were to melt completely, state whether the length of a day would increase or decrease.
Answer: Increase

Question 19. What is the rotational analogue of the impulse of a force?
Answer: Angular impulse or impulse of a torque

Question 20. Which physical quantity is represented by the product of the moment of inertia and angular velocity?
Answer: Angular momentum (L = Iω]

Question 21. What is the relation between torque and moment of inertia?
Answer: Torque = moment of inertia x angular acceleration

Question 22. Torque x time = change of ______ of the body in that time.
Answer: Angular momentum

Question 23. Rotational analogue of force is _________
Answer: Torque

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The moment of inertia of a circular ring about a given axis is more than the moment of inertia of a circular disc of the same mass and same size, about the same axis.

Statement 2: The circular ring is hollow; so its moment of inertia is more than circular disc which is solid.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 2.

Statement 1: If the earth shrinks (without change in mass) to half its present size, the length of the day would become 6 hours.

Statement 2: As the size of the earth changes, its moment of inertia changes.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

System of particles motion

Question 3.

Statement 1: Many great rivers flow towards the equator. The sediments that they carry increase the time of rotation of the earth about its own axis.

Statement 2: The angular momentum of the earth about its rotation axis is conserved.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 4.

Statement 1: The mass of a body cannot be considered to be concentrated at the centre of mass of the body for the purpose of computing its moment of inertia.

Statement 2: For then the moment of inertia of every body about an axis passing through its centre of mass would be zero.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 5.

Statement 1: The moment of inertia of a uniform disc and solid cylinder of equal mass and radius about an axis passing through the centre and perpendicular to the plane will be the same.

Statement 2: Moment of inertia depends upon the distribution of mass from the axis of rotation, i.e., the perpendicular distance from the axis.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 6.

Statement 1: The angular velocity of a rigid body in motion is defined for the whole body.

Statement 2: All points on a rigid body performing pure rotational motion are having same angular velocity.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 7.

Statement 1: The moment of inertia about an axis passing through the centre of mass is minimum.

Statement 2: The Theorem of the parallel axis can be applied only to a two-dimensional body of negligible thickness.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 8.

Statement 1: In the rotational plus translational motion of a rigid body different particles of the rigid body may have different velocities but they will have the same accelerations.

Statement 2: The translational motion of a particle is equivalent to the translational motion of the rigid body

Answer: 4. Statement 1 is false, statement 2 is true.

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies Match The Columns

Question 1. If the radius of Earth is reduced to half without changing its mass, then match the following columns.

Answer: 1. C, 2. D, 3. B

Question 2. From a uniform disc of mass M and radius R, a concentric disc of radius R/2 is cut out, For the remaining annular disc: I₁ is the moment of inertia about axis ‘1’, I₂ about ‘2’, I₃ about ‘3′ and l₄ about ‘4‘. Axes ‘1’ and ‘2’ are perpendicular to the disc and ‘3’ and ‘4’ are in the plane of the disc. Axes ‘2’, ‘3’ and ‘4’ intersect at a common point.

Answer: 1. C, 2. A, 3. A, 3. B

Question 3. A solid sphere is rotating about an axis as shown. An insect follows the dotted path on the circumference of the sphere

Answer: 1. B, 2. C, 3. A, 4. C

System of particles motion

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies Comprehension Type Questions And Answers

Question 1. Two discs A and B are mounted co-axially on a vertical axle. The discs have moments of inertia I and 2I, respectively, about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential energy of a spring compressed by a distance x₁. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x₂. Both the discs rotate in the clockwise direction

1. The ratio \(\frac{x_1}{x_2}\) is

1. 2
2. \(\frac{1}{2}\)
3. √2
4. \(\frac{1}{\sqrt 2}\)

Answer: 3. √2

2. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is

1. \(\frac{2 I \omega}{3 t}\)
2. \(\frac{9 I \omega}{2 t}\)
3. \(\frac{9 I \omega}{4 t}\)
4. \(\frac{3 I \omega}{2 t}\)

Answer: 1. \(\frac{2 I \omega}{3 t}\)

3. The loss of kinetic energy during the above process is

1. \(\frac{I \omega^2}{2}\)
2. \(\frac{I \omega^2}{3}\)
3. \(\frac{I \omega^2}{4}\)
4. \(\frac{I \omega^2}{6}\)

Answer: 2. \(\frac{I \omega^2}{3}\)

Question 2. A uniform solid sphere is released from the top of a fixed inclined plane of inclination 30° and height h. It rolls without sliding.

1. The acceleration of the centre of the sphere is

1. \(\frac{3 g}{5}\)
2. \(\frac{4 g}{5}\)
3. \(\frac{4 g}{7}\)
4. \(\frac{3 g}{7}\)

Answer:  4. \(\frac{3 g}{7}\)

2. The speed of the point of contact of the sphere with the inclined plane, when the sphere reaches the bottom of the incline, is

1. \(\sqrt{2 g h}\)
2. \(\sqrt{\frac{10 g h}{7}}\)
3. \(zero\)
4. \(2 \sqrt{2 g h}\)

Answer: 3. \(zero\)

3. The time taken by the sphere to reach the bottom is

1. \(\sqrt{\frac{2 h}{g}}\)
2. \(\sqrt{\frac{70 h}{9 g}}\)
3. \(\sqrt{\frac{25 h}{18 g}}\)
4. \(\sqrt{\frac{25 h}{6 g}}\)

Answer: 2. \(\sqrt{\frac{25 h}{18 g}}\)

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies Integer Answer Type Questions

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. A cube of mass 2 kg is held stationary against a rough wall by a force F = 40 N passing through centre C. Find the perpendicular distance of normal reaction between wall and cube from point C. Side of the cube is 20 cm. Take g = 10 m · s^-2.
Answer: 5

Question 2. A wheel has an angular acceleration of 2.0 rad · s^-2 and an initial angular speed of 1.0 rad · s^-1 What will be the angular displacement (in radian) in 2 s?
Answer: 6

Question 3. A sphere of radius 2 m rolls on a plank. The accelerations of the sphere and the plank are indicated. What is the value of angular acceleration (in rad · s^-2)?
Answer: 3

System of particles motion

Question 4. For two rings of radii R and nR are made up of the same material. The ratio of moments of inertia about axes passing through their centres is 1: 8. What should be the value of n?
Answer: 2

 

• Expansion Of Solid And Liquids
• Friction
• Measurement And Dimension Of Physical Quantity
• Newtonian Gravitation And Planetary Motion
• Simple Harmonic Motion
• Statics

• Transmission Of Heat
• Viscosity And Surface Tension
• Circular Motion
• Newtonian Gravitation And Planetary Motion
• First And Second Law Of Thermodynamics
• Change Of State Of Matter
• Hydrostatics

Doppler Effect In Sound

Doppler Effect Notes for Class 11 WBCHSE

Effect Of Relative Motion Between A Source Of Souno And A Listener When a train approaches a station sounding its horn, the pitch of the sound seems to be higher to a listener standing on the platform.

Again, when the train passes the platform and moves away from the station, the same sound seems to be of a lower pitch to the listener.

• On the other hand, if the source of sound is at rest and the listener approaches it or moves away from it, the pitch of the sound appears to be higher or lower, respectively. These phenomena are known as the Doppler effect.
• We know that the pitch of sound is determined by its frequency. So, an apparent change in the frequency of sound is caused by the relative motion of the source of sound and the listener and this is known as the Doppler effect.

Understanding Doppler Effect in Sound and Light

Doppler effect: The apparent change in the pitch of a note due to the relative motion of the source of sound and the listener is called the Doppler effect.

Doppler shift: Let n be the actual frequency of a source of sound and n’ be the apparent frequency of it due to the relative motion of the source and the listener. The apparent change in frequency, i.e., (n’ -n) is called the Doppler shift. The apparent increase or decrease of the pitch of sound refers to positive or negative Doppler shifts, respectively.

Experimental demonstration: A source of sound S which can emit continuous sound of the same frequency (for example, a whistle or a bell driven by a battery) is taken. It is tied to one end of a strong thread of length about 2 m to 3 m and whirled at a high speed along a circle ABCDA with P as the centre.

• Under this condition, a listener at O can easily recognise the successive increase and decrease of the pitch of the sound.
• This increase or decrease of the pitch is due to the Doppler effect. When the source S along the arc ABC approaches the listener the pitch of the sound is increased. When along CDA the source recedes from the listener, the pitch of the sound is decreased.

Doppler Effect In Sound Calculation Of Apparent Frequency And Doppler Shift

Let the velocity of sound in the air be V and the actual frequency of the source of sound be n.

So, the wavelength of the sound wave emitted from the source in air, \(\lambda=\frac{V}{n}\)…(1)

Effect of the motion of the listener: Suppose, the listener O is approaching a stationary source of sound S with velocity u₀. Since the source is at rest, n number of waves extend over the distance V in the direction of the listener, i.e., n number of waves extend over the distance traversed by the sound wave in unit time.

Examples of Doppler Effect in Everyday Life

• So, the length of the sound wave that reaches the ears of the listener at O is given by \(\frac{V}{n}\). Thus, the wavelength of the sound remains unchanged. So it may be said that, n ∝ V.
• In this case, the velocity of sound relative to the listener is not V, rather the apparent velocity of sound relative to u₀ (the velocity of the listener) increases and becomes V’ = V+ u₀. So, the frequency of the sound heard by the listener also increases.

Effect of the motion of the source: Suppose, the source S is approaching the listener O at rest with velocity u_s.

• Since the listener is at rest, the velocity of sound relative to him is, V i.e., in this case, the velocity of sound remains unchanged. So, it may be said that, \(n \propto \frac{1}{\lambda}\).
• Now in this case, n number of waves do not extend over the distance V in the direction of the listener; rather due to the velocity of the source (u_s), the distance covered by n number of waves = V- u_s.
• Thus the apparent wavelength of the sound wave decreases and becomes \(\lambda^{\prime}=\frac{V-U_s}{n}\). Hence, the frequency of the sound heard by the listener increases.

Effect of the motion of both the listener and the Source: If the listener and the source both are in motion, the apparent frequency of the sound heard by the listener is given by,

⇒ \(n^{\prime}=\frac{V^{\prime}}{\lambda^{\prime}}=\frac{V+u_o}{\frac{V-u_s}{n}} \text { or, } n^{\prime}=\frac{V+u_o}{V-u_s} \times n\) ….(2)

∴ Doppler shift = \(n^{\prime}-n=\frac{u_s+u_o}{V-u_s} \times n\)…(3)

Frequency And Doppler Shift Special cases:

Source at rest and listener In motion: In this case, u_s = 0. So, from equations (2) and (3),

n’ = \(\frac{V+u_o}{V} \times n \quad \text { and } n^{\prime}-n=\frac{u_o}{V} \times n\)

Listener at rest and source in motion: In this case, u₀ = 0. So, from the equations (2) and (3),

n’ = \(\frac{V}{V-u_s} \times n \quad \text { and } n^{\prime}-n=\frac{u_s}{V-u_s} \times n\)

Frequency And Doppler Shift Discussions:

1. It is very important to use positive and negative signs properly while putting the values of u₀ and u_s in the equations (2) and (3). The rule, that is followed, is:

1. If the listener moves towards the source, u₀ is positive.
2. If the source moves towards the listener, u_s is positive.
   • Conversely:
     1. If the listener moves away from the source, u₀ is negative.
     2. If the source moves away from the listener, u_s is negative.

In general, it may be said that if u₀ and u_s are inclined at an angle θ with the direction of motion of sound, u₀cosθ and u_scosθ are to be placed in equations (2) and (3).

2. It is evident that if there Is no relative motion between the source and the listener, then u_s = -u₀. In that case, no Doppler effect takes place [equations (2) and (3)]. If the distance between the source and the listener decreases with respect to time, the apparent pitch of the sound increases and vice versa.

3. While calculating apparent frequency and Doppler shift it has been assumed that the velocities of both the source and the listener are less than that of sound in air, i.e., u_s < V and u₀ < V. If the velocity of either the source or the listener exceeds the velocity of sound (i.e., it becomes supersonic), the nature of Doppler effect becomes entirely different.

4. Effect of wind: Let the velocity of wind be v. If the wind blows in the direction of motion of sound, v is positive. Then the apparent velocity of sound relative to the listener at rest = V+ v. So, in the calculation of the Doppler effect, the velocity of sound V is to be replaced by V+ v. In that case, equations (2) and (3) become,

n’ = \(\frac{V+v+u_o}{V+v-u_s} \times n\)…(4)

and \(n^{\prime}-n=\frac{u_s+u_o}{V+v-u_s} \times n\)…(5)

It is obvious that if the wind blows in the opposite direction, -v is to be placed instead of v in equations (4) and (5).

5. Doppler effect in case of echo: Let the source of sound S be moving towards the stationary reflector R with velocity u_s. The velocity of the listener O is u₀ in the same direction.

In this case for the echo produced by the reflector R, S’ is the apparent source of sound which is the image of the principal source S. The listener O is approaching the apparent source S’ and the apparent source S’ is also approaching the listener O. So, u₀ and u_s are both positive.

∴ Apparent frequency n’ = \(\frac{V+u_o}{V-u_s} \times n\)…(6)

Here, both the source S and the observer O are moving towards the reflector. If any of them moves in the opposite direction, i.e., recedes from the reflector, u_s or u₀ in equation (6) is replaced by -u_s or -u₀.

Again, as a special case, if the positions of the source of sound and the listener always remain the same (for example, the horn of a car and a passenger of the same car), then u_s = u₀.

Then according to equation (6), \(n^{\prime}=\frac{V+u_o}{V-u_o} \times n\)

Formation of beats due to original sound and its echo: If the difference between the actual frequency and the apparent, frequency due to the Doppler effect (i.e., n-n’ or n’-n) is less than 10 Hz, then the original sound and its echo are superposed and beats are formed.

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NEET Foundation	Class 12 Physics	NEET Physics

Doppler Effect In Sound Frequency And Doppler Shift Numerical Examples

Example 1. The frequency of the whistle of a train Is 512 Hz. The train crosses a station at a speed of 72km · h^-1. Calculate the frequency of the sound heard by a listener, standing on the platform, before and after the train crosses the station. Neglect the effect of wind. The velocity of sound is 336 m · s^-1.
Solution:

Given

The frequency of the whistle of a train Is 512 Hz. The train crosses a station at a speed of 72km · h^-1.

The velocity of sound is 336 m · s^-1.

Velocity of the train = 72 km · h^-1

= \(\frac{72 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= 20 m · s^-1

When the train approaches the station, the distance covered by 512 sound waves =336-20 =316 m

∴ Apparent wavelength, \(\lambda^{\prime}=\frac{316}{512} \mathrm{~m}\)

∴ Apparent frequency due to the Doppler effect

= \(\frac{\text { velocity of sound }(V)}{\lambda^{\prime}}\)

= \(\frac{336}{\frac{316}{512}}=\frac{336 \times 512}{316}=544.4 \mathrm{~Hz}\)

Again, when the train recedes from the station, the distance covered by 512 sound waves = 336 + 20 = 356 m

∴ Apparent wavelength, \(\lambda^{\prime}=\frac{356}{512} \mathrm{~m}\)

∴ Apparent frequency = \(\frac{V}{\lambda^{\prime}}=\frac{336}{\frac{356}{512}}=\frac{336 \times 512}{356}=483.2 \mathrm{~Hz}\)

Alternative Method:

According to the question, the listener is at rest and the source is in motion. So, the apparent frequency to the listener,

n’ = \(\frac{V}{V-u_s} \times n\)

In the first case, V = \(336 \mathrm{~m} \cdot \mathrm{s}^{-1}, n=512 \mathrm{~Hz}\)

and \(u_s=72 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{72 \times 1000}{60 \times 60}=20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ \(n^{\prime}=\frac{336}{336-20} \times 512=\frac{336}{316} \times 512=544.4 \mathrm{~Hz}\)

In the second case, V = \(336 \mathrm{~m} \cdot \mathrm{s}^{-1}, n=512 \mathrm{~Hz}\)

and \(u_s=-20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ \(n^{\prime}=\frac{336}{336-(-20)} \times 512=\frac{336 \times 512}{356}=483.2 \mathrm{~Hz}\)

Example 2. A sound of frequency 512 Hz Is emitted from a stationary source. A train running at a speed of 72 km · h^-1 passes the source. What will be the frequency of the sound heard by a passenger of the train before and after passing the source? Neglect the effect of wind. The velocity of sound is 336 m · s^-1.
Solution:

Given

A sound of frequency 512 Hz Is emitted from a stationary source. A train running at a speed of 72 km · h^-1 passes the source.

Velocity of the train = 72 km · h^-1

= \(\frac{72 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}=20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

When the train approaches the source, the velocity of sound relative to the passenger,

V’ = 336 + 20 = 356 m · s^-1

Since the source is at rest, the wavelength of sound remains the same.

∴ Apparent frequency due to the Doppler effect

= \(\frac{V^{\prime}}{\lambda}=\frac{V^{\prime}}{\frac{V}{n}}=\frac{V^{\prime} n}{V}=\frac{356 \times 512}{336}=542.5 \mathrm{~Hz}\)

Again when the train recedes from the source, the velocity of sound relative to the passenger V’ = 336 – 20 = 316 m · s^-1

∴ Apparent frequency = \(\frac{V^{\prime} n}{V}=\frac{316 \times 512}{336}=481.5 \mathrm{~Hz}\)

Practice Questions on Doppler Effect for Class 11

Example 3. When a train approaches a listener, the apparent frequency of the whistle is 100Hz, while the frequency appears to be 50 Hz when the train recedes. Calculate the frequency when the listener is in the train.
Solution:

Given

When a train approaches a listener, the apparent frequency of the whistle is 100Hz, while the frequency appears to be 50 Hz when the train recedes.

If the listener is in the train, he will listen to the actual frequency of the whistle (n).

If V is the velocity of sound, u_s is the velocity of the source and u₀ is the velocity of the listener,

the apparent frequency, \(n^{\prime}=\frac{V+u_o}{V-u_s} \times n\)

In the given problem, in both cases, the listener is at rest. So, u₀ = 0

In the first case, the motion of the train is towards the listener. So, u_s is positive.

Again in the second case, the train recedes from the listener. So, u_s is negative.

Therefore, in the two cases we have, \(100=\frac{V+0}{V-u_s} \times n\)

or, \(\frac{V}{V-u_s} \times n=100\)….(1)

and \(50=\frac{V+0}{V-\left(-u_s\right)} \times n\)

or, \(\frac{V}{V+u_s} \times n=50\)….(2)

From equation (1), \(\frac{n}{100}=\frac{V-u_s}{V}=1-\frac{u_s}{V} \quad \text { or, } \frac{u_s}{V}=1-\frac{n}{100}=\frac{100-n}{100}\)

From equation (2), \(\frac{n}{50}=\frac{V+u_s}{V}=1+\frac{u_s}{V} \quad \text { or, } \frac{u_s}{V}=\frac{n}{50}-1=\frac{n-50}{50}\)

∴ \(\frac{100-n}{100}=\frac{n-50}{50} \quad \text { or, } 100-n=2 n-100\)

or, \(3 n=200 \quad \text { or, } n=66 \frac{2}{3} \mathrm{~Hz}\)

Example 4. Two engines pass each other in opposite directions. One of them blows a whistle of frequency 540 Hz. Find the frequencies heard by a passenger sitting on the other engine before and after passing each other. The velocity of both engines = 72 km · h^-1; velocity of sound =340 m · s^-1.
Solution:

Given

Two engines pass each other in opposite directions. One of them blows a whistle of frequency 540 Hz.

Velocity of the engine,

u = \(72 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{72 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}=20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Velocity of sound, V = 340 m · s^-1;

Actual frequency, n = 540 Hz

At the time of approaching each other, the velocity of the whistle, u_s = +20 m · s^-1

The velocity of the passenger on the other engine, u₀ = +20 m · s^-1

∴ Apparent frequency,

n’ = \(\frac{V+u_o}{V-u_s} \times n\)

= \(\frac{340+20}{340-20} \times 540=\frac{360}{320} \times 540=607.5 \mathrm{~Hz}\)

Again at the time of receding from each other, u_s = -20 m · s^-1 and u₀ = -20 m · s^-1

∴ Apparent frequency,

n’ = \(\frac{V+u_o}{V-u_s} \times n\)

= \(\frac{340-20}{340-(-20)} \times 540=\frac{320}{360} \times 540=480 \mathrm{~Hz}\)

Example 5. A car travelling at a speed of 36 km · h^-1 sounds its horn of frequency 500 Hz. It is heard by the driver of another car which Is travelling behind the first car In the same direction with a velocity of 20 m · s^-1. Another sound Is heard by the driver of the second car after reflection from a bridge ahead. What will be the frequencies of the two sounds heard by the driver of the second car? Sound travels in air with a speed of 340m · s^-1.
Solution:

Given

The driver of the second car O is the listener. The first car S and its image S’ due to reflection on the bridge ahead are two sources of sound.

The direction of motion of O is towards S and S’.

So velocity of the listener, u₀ = + 20 m · s^-1

Here, S is receding from O. So, the velocity of the source S,

u₀ = -36 km · h^-1

= \(-\frac{36 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}\)=\(-10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Apparent frequency to the listener due to S,

n’ = \(\frac{V+u_o}{V-u_s} \times n\)

= \(\frac{340+20}{340-(-10)} \times 500\)

= \(\frac{360}{350} \times 500=514.3 \mathrm{~Hz}\)

On the other hand, S’ is approaching O.

So, velocity of the source S’, u_s = +10 m · s^-1

∴ Apparent frequency to the listener due to S’,

n’ = \(\frac{V+u_o}{V-u_s} \times n=\frac{340+20}{340-10} \times 500\)

= \(\frac{360}{330} \times 500=545.5 \mathrm{~Hz}\)

So the driver of the second car will hear the sounds of frequencies 514.3 Hz and 545.5 Hz.

Problems On Doppler Effect for Class 11

Example 6. A whistle, emitting a sound of frequency 440 Hz, is tied to a thread of length 1.5 m and rotated with an angular velocity of 20 rad · s^-1 on a horizontal plane, Calculate the range of frequency of the sound heard Velocity of sound in air =330 m · s^-1
Solution:

Given

A whistle, emitting a sound of frequency 440 Hz, is tied to a thread of length 1.5 m and rotated with an angular velocity of 20 rad · s^-1 on a horizontal plane,

Linear velocity of the whistle (u)

= angular velocity x radius of the circular path

= 20 x 1.5 = 30 m · s^-1

The velocity of the whistle at that position of the circular path where the whistle approaches the listener.

u_s = +30m · s^-1

∴ Apparent frequency, \(n_1 =\frac{V}{V-u_s} \times n=\frac{330}{330-30} \times 440\)

= \(\frac{330}{300} \times 440=484 \mathrm{~Hz}\)

Again, the velocity of the whistle at that position of the circular path where the whistle recedes from the listener, u’s =-30m-s_1

Apparent frequency, \(n_2 =\frac{V}{V-u_s^{\prime}} \times n=\frac{330}{330-(-30)} \times 440\)

= \(\frac{330}{360} \times 440=403.3 \mathrm{~Hz}\)

So, the range of frequency is 403.3 Hz to 484 Hz.

Example 7. Each of the two persons has a whistle of frequency 500 Hz. One person is at rest at a particular place and the second person recedes from him with a velocity of 1.8 m · s^-1. If both of them blow whistles, how many beats will be heard by each of them? The velocity of sound = 330 m · s^-1.
Solution:

Given

Each of the two persons has a whistle of frequency 500 Hz. One person is at rest at a particular place and the second person recedes from him with a velocity of 1.8 m · s^-1. If both of them blow whistles,

Both of them listen to the sound of frequency 500 Hz for their own whistles.

In the first case, let us suppose that the first person is listening to the sound coming from the whistle of the second person.

Here the velocity of the listener, u₀ = 0

Since the source is receding, the velocity of the source, u_s = -1.8 m · s^-1

∴ Apparent frequency,

n’ = \(\frac{V}{V-u_s} \times n=\frac{330}{330-(-1.8)} \times 500=497.29 \mathrm{~Hz}\)

∴ Number of beats per second = n-n’ = 500-497.29

= 2.71 ≈ 3

In the second case, let us suppose that the second person is listening to the sound coming from the whistle of the first person.

Here the velocity of the source, u_s = 0

Since the listener is receding,

The velocity of the listener, u₀ = -1.8 m · s^-1

∴ Apparent frequency,

n’ = \(\frac{V+u_o}{V} \times n=\frac{330-1.8}{330} \times 500=497.27 \mathrm{~Hz}\)

∴ Number of beats per second = n-n’ =500-497.27

= 2.73 ≈ 3

Example 8. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. What is the ratio of the velocity of train B to that of train A?
Solution:

Given

A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren.

Let, u_A be the velocity of train A.

∴ Frequency of siren relative to the passenger of train A \(n_1=n\left(\frac{V+u_A}{V}\right)\)

∴ \(5\left(\frac{V+u_A}{V}\right)=5.5\)….(1)

Similarly, the frequency of the siren relative to the passenger of train B,

⇒ \(n_2=n\left(\frac{V+u_B}{V}\right) \quad\left[u_B=\text { velocity of train } B\right]\)

∴ \(5\left(\frac{V+u_B}{V}\right)=6\)…(2)

From equation (1), 5.5 V- 5 V = 5 u_A

∴ V = \(\frac{5}{0.5} u_A\)….(3)

From equation (2), 6V-5V = 5u_B

∴ V = 5u_B……(4)

Comparing equations (3) and (4), \(\frac{5}{0.5} u_A=5 u_B\)

∴ \(\frac{u_B}{u_A}=\frac{1}{0.5}=\frac{2}{1}\)

∴ \(u_B: u_A=2: 1\)

Example 9. A railway track and a road are mutually perpendicular. A train Is approaching the railway crossing at a speed of 80 km/h. When the train h at a distance 1 km from the crossing it blows a whistle of frequency 400 Hz. Which frequency of sound will he hear from a man on a road at a distance of 600 m from the crossing? velocity of sound =330 m · s^-1
Solution:

Given

A railway track and a road are mutually perpendicular. A train Is approaching the railway crossing at a speed of 80 km/h. When the train h at a distance 1 km from the crossing it blows a whistle of frequency 400 Hz.

Train (S) is moving with a velocity 80 km/h along SA. At the time of blowing the whistle, the distances of the train and a man (O) from the crossing (A) are 1 km and 600 m respectively.

∴  SA = 1 km = 1000 m; OA = 600 m

If u_S be the velocity of the train, the component along SO is u_S cosθ.

∴ \(\cos \theta=\frac{S A}{S O}=\frac{S A}{\sqrt{S A^2+A O^2}}=\frac{100}{\sqrt{(1000)^2+(600)^2}}=0.8575\)

∴ \(u_S \cos \theta=80 \times \frac{5}{18} \times 0.8575=19.05 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Apparent frequency of sound, n’ = \(\frac{V}{V-u_S \cos \theta} n\)

Here, V = velocity of sound = 330 m · s^-1 and actual frequency, n = 400 Hz

∴ n’ = \(\frac{330}{330-19.05} \times 400=424.5 \mathrm{~Hz}\)

Doppler Effect In Sound and Light

Like sound, the Doppler effect also takes place in the case of light waves. When a source of light and an observer are in relative motion, an apparent change in the frequency, i.e., the wavelength of light is perceived in the eyes of the observer.

It means that the colour of the light is found to change in the eyes of the observer. This Doppler effect of light is of two kinds.

Redshift: When the source and the observer recede from each other, the wavelength of the light apparently increases. This apparent increase of wavelength due to the Doppler effect is called redshift. This is a displacement of the Lines in the spectra towards the red end of the visible spectrum (i.e., towards a longer wavelength).

Blueshift: When the source and the observer approach each other, the wavelength of the light apparently decreases. This apparent decrease of wavelength is called blue shift. This is a displacement of the lines in the spectra towards the blue end of the visible spectrum (i.e., towards a shorter wavelength).

Calculation of apparent wavelength and Doppler shift: Let the velocity of light in a vacuum be c, the original frequency of a monochromatic light be n and the original wavelength of the light, \(\lambda=\frac{c}{n}\).

Now suppose that the source of this monochromatic light and an observer approach each other with relative velocity u. In that case, n number of waves produced per second do not occupy the distance c, rather the distance occupied by those n number of waves = c- u

So, the apparent wavelength is \(\lambda^{\prime}=\frac{c-u}{n}=\frac{c}{n}\left(1-\frac{u}{c}\right) \quad \text { or, } \lambda^{\prime}=\left(1-\frac{u}{c}\right) \lambda\)…(1)

∴ Doppler shift of wavelength = \(\lambda^{\prime}-\lambda=-\frac{u}{c} \lambda\)….(2)

Doppler Effect Of Light Special cases:

1. When the source approaches the observer, u is positive in equations (1) and (2). So, λ’ is less than λ, i.e., the wavelength apparently decreases. It is the phenomenon of blue shift.
2. When the source recedes from the observer, u is negative in equations (1) and (2). So, λ’ is greater thanλ, i.e., the wavelength apparently increases. It is the phenomenon of redshift.

Doppler Effect Of Light Discussions:

1. Velocity of light in vacuum, c = 3 x 10⁸ m · s^-1. If the relative velocity of the source and the observer is very small, i.e., u<<c or, \(\frac{u}{c} \ll 1\), then according to equation (1), λ’ ≈ λ. In this case, the Doppler effect is practically absent. So when a train or motor car lighting their headlights approaches an observer, red or blue shift becomes negligible, i.e., no apparent change in the colour of light takes place.

2. On the other hand, if the relative velocity of the source and the observer is nearly equal to the velocity of light, i.e., u<<c or, \(\frac{u}{c} \approx 1\) then according to equation (1), λ’  ≈ 0. But in practical cases the wavelength never becomes zero or nearly zero. So it is evident that if u ≈ c, equation (1) is no longer applicable.

In that case, the corrected form of equation (1) will be \(\lambda^{\prime}=\lambda \sqrt{\frac{c-u}{c+u}}\) and the associated frequency will be, \(n^{\prime}=n \sqrt{\frac{c+u}{c-u}}\).

The theory of relativity is essential to reach these equations which is out of the current syllabus.

Applications of Doppler Effect in Science

3. The Doppler effect of light differs remarkably from that of sound. The relative velocity of sound increases or decreases when the listener is in motion with respect to the source of sound.

• On the other hand, the velocity of light, as proposed in the theory of relativity, is always equal to c, whatever be the relative velocity of the observer and the source.
• Hence Doppler effect of light takes place for the relative velocity u of the source and the observer. Here, it is not necessary to consider the velocities of the source of light and the observer separately.

Application of Doppler effect of light:

1. Velocity of Slots: The velocity of a distant star relative to the earth can be determined by measuring the Doppler shift of the light that comes to the Earth from the star. Generally, for a star or galaxy of stars, this shift is a red shift from which it is understood that these stars are receding gradually from the earth (velocity of separation is 10 km · s^-1 to 300km ·  s^-1). This observation supports the theory of an expanding universe.

2. Binary Star: For a star, if both red shift and blue shift are observed, it may be concluded that the star is actually a binary star. The two stars are revolving around a common axis.

3. Quasar: The redshift of these stars is very high. Calculations show that these stars are receding from the earth at a tremendously high speed (in some cases tire speed becomes 80% of the speed of light).

4. Rotation of the sun about its own axis: As the sun rotates about its axis, one side faces the earth while the other side goes behind it. So, a blue shift for one side and a rod shift for the other side are observed. From the calculations of these shifts, it is found that the speed of rotation of the sun is about 2 km · s^-1.

5. Doppler radar: It is used to measure the speed of an aeroplane flying at a high speed. The waves emitted from a Doppler radar are reflected from a flying aeroplane. By measuring the Doppler shift of the reflected radar waves, the speed of the aeroplane can be determined.

6. Measurement of high temperature: A blue shift is observed in the light emitted by a hot substance when the molecules of the substance move towards the observer. Whereas a red shift is observed when the molecules move away from the observer.

Measuring this shift, the expression of rms velocity of die molecules is determined. Next from the expression of rms velocity = \(\sqrt{\frac{3 R T}{M}}\), the temperature T is obtained (M = Mass of 1 mol of the substance).

Doppler Effect In Sound Doppler and Light Conclusion

The apparent change in the pitch of a note due to relative motion between a source of sound and a listener is called the Doppler effect.

If n is the actual frequency of a source of sound and n’ is the apparent frequency of it due to the relative motion of the source and the listener, the apparent change in frequency (n’ -n) is called Doppler shift.

• No Doppler effect can be noticed by an observer in the absence of any relative motion between the source and observer.
• When the source and the observer recede from each other, an apparent increase in the wavelength of light Is observed. This apparent increase of wavelength due to the Doppler effect is called redshift. This is a displacement of the lines in a spectrum towards the red end (i.e., towards a longer wavelength).
• When the source and the observer approach each other, an apparent decrease in the wavelength of light is observed. This apparent decrease of wavelength is called blue shift. This is a displacement of the lines in a spectrum towards the blue end (i.e., towards a shorter wavelength).

Doppler Effect In Sound Doppler Effect Of Light Useful Relations For Solving Numerical Examples

If the listener and the source are both in motion, the apparent frequency of the sound heard by the listener is,

n’ = \(\frac{V+u_o}{V-u_s} n\)

where V = velocity of sound in air, n = actual frequency of source of sound, u_s = velocity of the source of sound.

Doppler shift, n’ -n = \(\frac{u_s+u_o}{V-u_s} n\)

Sign convention:

1. If the listener moves towards the source, u₀ is positive,
2. If the source moves towards the listener, u_s is positive.
   • Conversely,
     1. If the listener moves away from the source, u₀ is negative,
     2. If the source moves away from the listener, u_s is negative.

If the velocity of wind is v and if the wind blows in the direction of motion of sound, then n’ = \(\frac{V+v+u_o}{V+v-u_s} n\)

If the velocity of wind is v and if the wind blows in the opposite direction of motion of sound, then n’ = \(\frac{V-v+u_o}{V-v-u_s} n\)

A car moving towards a stationary reflector with velocity u₀ blows a horn of frequency n. The frequency of echo heard by the passenger of the car will be

n’ = \(\frac{V+u_0}{V-u_0} n\)

A stationary car blows a horn of frequency n. If a reflecting surface moves towards the car with velocity u_s, the frequency of echo heard by the passenger of the car is

n’ = \(\frac{V+u_s}{V-u_s} n\)

Some Important Cases of the Doppler Effect

Doppler Effect In Sound Doppler Effect Of Light Very Short Answer Type Questions

Question 1. Which property of sound undergoes an apparent change due to the Doppler effect?
Answer: Pitch

Question 2. When a source moves at a speed greater than that of sound, will the Doppler effect hold?
Answer: No

Question 3. Will there be a Doppler effect for sound when the source and listener move at a right angle to the line joining them?
Answer: No

Question 4. When a source of sound approaches a stationary listener, the sound appears to be _______ to the listener.
Answer: Sharper

Question 5. When a listener approaches a source of sound, the sound appears to be ______ to the listener.
Answer: Sharper

Question 6. The apparent change of frequency of sound due to the Doppler effect is called Doppler ______
Answer: Shift

Question 7. If there is a relative motion between a source of light and an observer, a change of colour of the light appears in the eyes of the observer. What is the name of this phenomenon?
Answer: Doppler effect of light

Question 8. When a source of light and an observer recede from each other, the apparent change in the wavelength of light is called _______ shift.
Answer: Red

Doppler Effect In Sound Doppler Effect Of Light Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The intensity of sound waves changes when the listener moves towards or away from the stationary source.

Statement 2: The motion of the listener causes the apparent change in wavelength.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 2.

Statement 1: When there is no relative velocity between the source and observer, then the observed frequency is the same as emitted.

Statement 2: The velocity of sound when there is no relative velocity between the source and observer is zero.

Answer: 3. Statement 1 is true, statement 2 is false.

Doppler Effect In Sound Doppler Effect Of Light Match Column 1 With Column 2

Question 1. Source has frequency f. Source and observer both have the same speed. The apparent frequency observed by the observer matches the following:

Answer: 1. C, 2. A, 3. B, 4. C

Doppler Effect In Sound Doppler Effect Of Light Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A source 5 of an acoustic wave of frequency v₀ = 1700 Hz and a receiver R are located at the same point. At the instant t = 0, the source starts from rest to move away from the receiver with a constant acceleration ω. The velocity of sound in air is v = 340 m · s^-1.

1. If ω = 10 m · s^-2, the apparent frequency that will be recorded by the stationary receiver at t = 10 s will be

1. 1700 Hz
2. 1.35 Hz
3. 850 Hz
4. 1.27 Hz

Answer: 2. 1.35 Hz

2. if ω = 0 for t> 10 s, the apparent frequency recorded by the receiver at t = 15 s will be

1. 1700Hz
2. 1310Hz
3. 850Hz
4. 1.23kHz

Answer: 2. 1310Hz

3. If ω = 10 m · s^-2, the apparent frequency that will be recorded by the stationary receiver just at the instant when the source is exactly 1 km away from the receiver will be

1. 1700 Hz
2. 1310 Hz
3. 850 Hz
4. 1.26 kHz

Answer: 4. 1.26 kHz

Question 2. A small source of sound vibrating at a frequency 500 Hz is rotated in a circle of radius (100/π)cm at a constant angular speed of 5.0 revolutions per second, The speed of sound in air is 330 m · s^-1, An observer (A) is situated at a great distance on a straight line perpendicular to the plane of the circle, through its centre. Another observer (B) is at rest at a great distance from the centre of the circle but nearly in the same plane. After some time the source of sound comes to rest after reaching the centre of the circle, At that time, another observer (C) moves towards the source with a constant speed of 20 m · s^-1, along the radial line to the centre.

1. The apparent frequency of the source heard by A will be

1. Greater than 500 Hz
2. Smaller than 500 Hz
3. Always 500 Hz
4. Greater for half the circle and smaller during the other half

Answer: 3. Always 500 Hz

2. The minimum and the maximum values of the apparent frequency heard by B will be

1. 455 Hz and 535 Hz
2. 485 Hz and 515 Hz
3. 485 Hz and 500 Hz
4. 500 Hz and 515 Hz

Answer: 2. 485 Hz and 515 Hz

3. The change in the frequency of the source heard by C will be

1. 6%
2. 3%
3. 2%
4. 9%

Answer: 1. 6%

Doppler Effect In Sound Doppler Effect Of Light Integer Answer Type Questions

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. The frequency of the sound of a car horn as perceived by an observer towards whom the car is moving differs from the frequency of the horn by 2.5%. Assuming that the velocity of sound in air is 320 m · s^-1, find the velocity (in m · s^-1) of the car.
Answer: 8

Question 2. A man is watching two trains, one leaving and the other coming in with equal speeds of 4m · s^-1. If they sound their whistles, each of frequency 240 Hz, find the number of beats heard by the man (velocity of sound in air =320 m · s^-1).
Answer: 6

Question 3. The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is 2% of the natural frequency of the source. If the velocity of sound in air is 300 m · s^-1, find the velocity (in m · s^-1) of the source.
Answer: 3

Question 4. A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f₀. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m· s^-1.
Answer: 7

Doppler Effect In Sound Doppler Effect Of Light Short Answer Type Questions

Question 1. Show that, the change in frequency of sound during the motion of the source towards the audience is more than that when the audience moves towards the source with the same velocity.
Answer:

Natural frequency of the source = n;

velocity of sound = V;

velocity of the source = v and

velocity of the audience = u.

So, apparent change in frequency = Doppler shift (Δn)

= \(\frac{\nu+u}{V-\nu} n\)

When the source of sound remains stationary, v = 0.

When the audience moves towards the source, \(\Delta n_1=\frac{u}{V} n\)

When the audience is stationary, u = 0.

when the source moves towards the audience with the same velocity (i.e., v= u), \(\Delta n_2=\frac{u}{V-u} n\)

Clearly, Δn₂ > Δn₁

Question 2. A car is moving with a speed of 72 km · h^-1 towards a roadside source that emits sound at a frequency of 850 Hz. The car driver listens to the sound while approaching the source and again while moving away from the source after crossing it. If the velocity of sound is 340 m · s^-1, the difference of the two frequencies, the driver hears is

1. 50 Hz
2. 85 Hz
3. 100 Hz
4. 150 Hz

Answer:

Given

A car is moving with a speed of 72 km · h^-1 towards a roadside source that emits sound at a frequency of 850 Hz. The car driver listens to the sound while approaching the source and again while moving away from the source after crossing it. If the velocity of sound is 340 m · s^-1,

Velocity of the car, u₀ = 72 km · h^-1 = 20 m · s^-1

When the car is approaching the source, the apparent frequency is,

∴ \(n_1=\left(\frac{V+u_o}{V}\right) n=\left(\frac{340+20}{340}\right) 850=900 \mathrm{~Hz}\)

When the car moves away from the source, then appar¬ent frequency is,

∴ \(n_2=\left(\frac{V-u_o}{V}\right) n=\left(\frac{340-20}{340}\right) 850=800 \mathrm{~Hz}\)

∴ n₁ -n₂ = 900 – 800 = 100 Hz

The option 3 is correct.

Doppler Effect Formulas Explained

Question 3. A train is moving with a uniform speed of 33 m/s and an observer is approaching the train with the same speed. If the train blows a whistle of frequency 1000 Hz and the velocity of sound is 333 m/s, then the apparent frequency of the sound that the observer hears is

1. 1220 Hz
2. 1099 Hz
3. 1110 Hz
4. 1200 Hz

Answer:

Given

A train is moving with a uniform speed of 33 m/s and an observer is approaching the train with the same speed. If the train blows a whistle of frequency 1000 Hz and the velocity of sound is 333 m/s,

Here, the velocity of the source, u_s = 33 m/s

The velocity of the observer, u₀ = 33 m/s,

Now, velocity of sound, v = 333 m/s

As the source and the observer are approaching each other,, the apparent frequency,

n’ = \(\frac{\nu+u_0}{\nu-u_2} \cdot n=\frac{333+33}{333-333} \times 1000=1220 \mathrm{~Hz}\)

The option 1 is correct.

Question 4. A train is moving on a straight track with a speed 20 m · s^-1. It is blowing its whistle at the frequency of 1000Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 m · s^-1) close to

1. 6%
2. 12%
3. 18%
4. 24%

Answer:

Given

A train is moving on a straight track with a speed 20 m · s^-1. It is blowing its whistle at the frequency of 1000Hz.

Velocity of the listener, v₀ = 0 ;

The velocity of the train, v_s = 20 m/s

So, apparent frequency when the train moves towards the listener,

⇒ \(n_1=\frac{v}{v-\nu_s} n=\frac{320}{320-20} \times 1000=1066.7 \mathrm{~Hz}\)

Again, apparent frequency when the train moves away from the listener,

⇒ \(n_2=\frac{v}{\nu-\left(-v_s\right)}=\frac{320}{320+20} \times 1000=941.2 \mathrm{~Hz}\)

Hence, the percentage change in the apparent frequency

= \(\frac{n_1-n_2}{n} \times 100=\frac{1066.7-941.2}{1000} \times 100\)

= 12.55% ≈ 12%

The option 2 is correct.

Question 5. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at a frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 x 10⁸ m · s^-1)

1. 10.1 GHz
2. 12.1GHz
3. 17.3 GHz
4. 15.3 GHz

Answer:

Given

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at a frequency 10 GHz.

Speed of the observer = \(\frac{c}{2}\)

∴ Doppler effect in relevant with the theory of relativity as follows:

v’ = \(\nu \sqrt{\frac{1+\beta}{1-\beta}}\)

β = \(\frac{\text { speed of observer }}{\text { speed of light }}=\frac{1}{2}\)

∴ \(\nu^{\prime} =10 \sqrt{\frac{1+\frac{1}{2}}{1-\frac{1}{2}}}=10 \sqrt{3}=10 \times 1.73\)

= 17.3 GHz

The option 3 is correct.

Question 6. A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km · h^-1. He finds that traffic has eased and a car moving ahead of him at 18 km · h^-1 is honking at a frequency of 1392 Hz. If the speed of sound is 343 m · s^-1, the frequency of the honk as heard by him will be

1. 1332 Hz
2. 1372 Hz
3. 1412 Hz
4. 1454 Hz

Answer:

Given

A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km · h^-1. He finds that traffic has eased and a car moving ahead of him at 18 km · h^-1 is honking at a frequency of 1392 Hz. If the speed of sound is 343 m · s^-1,

Speed of motorcyclist, u₀ = 36 km · h^-1 = 10 m · s^-1

Speed of car, u_s = 18 km · h^-1 = 5 m · s^-1

Now, the fundamental frequency of honk, v = 1392 Hz

∴ \(\lambda^{\prime}=\frac{V+u_s}{n}=\frac{343+5}{1392}=0.25 \mathrm{~m}\)

Hence, the frequency of honking as heard by the cyclist

= \(\frac{V+V_0}{\lambda^{\prime}}=\frac{343+10}{0.25}=1412 \mathrm{~Hz}\)

The option 3 is correct.

Question 7. A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15m · s^-1. Then the frequency of sound that the observer hears in the echo reflected from the cliff is (take the velocity of sound in air = 330 m · s^-1

1. 800 Hz
2. 838 Hz
3. 885 Hz
4. 765 Hz

Answer:

Given

A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15m · s^-1.

Apparent frequency of echo

n’ = \(n\left(\frac{v}{v-v_s}\right)=800\left(\frac{330}{330-15}\right)\)

= \(\frac{800 \times 330}{315}=838 \mathrm{~Hz}\)

Question 8. Due to the Doppler effect, the shift in wavelength observed is 0.1 Å, for a star producing a wavelength of 6000 Å. The velocity of recession of the star will be

1. 20 km s^-1
2. 2.5 km s^-1
3. 10 km s^-1
4. 5 km s^-1

Answer:

Doppler shift, \(\Delta \lambda=\lambda \frac{v}{c}\)

∴ v = \(c \frac{\Delta \lambda}{\lambda}=\left(3 \times 10^8\right) \times \frac{0.1}{6000}\)

= \(5 \times 10^3 \mathrm{~m} / \mathrm{s}=5 \mathrm{~km} / \mathrm{s}\)

The option 4 is correct.

Question 9. A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.

1. What is the frequency of the whistle for a platform observer when the train

1. Approaches the platform with a speed of 10 m/s,
2. Recedes from the platform with a speed of 10 m/s.

2. What is the speed of sound in each case if the speed of sound in still air is 340 m/s.

Answer:

1. Here, n = 400Hz; v = 340m/s

When a train approaches the platform, v_s = 10m/s

So, required frequency, = \(\frac{v}{v-v_s} \times n=\frac{340 \times 400}{340-10}=412.12 \mathrm{~Hz}\)

When the train recedes from the platform, v_s = 10 m/s

Hence, required frequency = \(\frac{v}{v+v_s} \times n=\frac{340 \times 400}{340+10}=388.6 \mathrm{~Hz}\)

2. The speed of sound in both the cases is same.

Question 10. Once Amit was going to his house. He was listening to music on his mobile with earphones while crossing the railway line and he did not hear the sound of an approaching train though the train was blowing the horn. A person nearby ran towards him and pushed away just as the train reached there. Amit realised his mistake and thanked the person.

1. Describe the value possessed by the person.
2. Name the phenomenon of change in frequency of sound when there is relative motion between the observer and source of sound.

Answer:

1. The values possessed by the person are—
   1. Presence of mind,
   2. General awareness,
   3. Good understanding,
   4. Prompt decision-making ability,
   5. Concern for other people’s safety and well-being,
   6. Helping and caring nature.
2. The phenomenon is Doppler’s effect on sound.

Doppler Effect In Sound and Light  Question and Answers

Doppler Effect Question and Answers for Class 11

Question 1. Each of the two men A and B is carrying a source of sound of frequency. If A approaches B with a velocity u,

1. How many beats per second will be heard by A and
2. How many beats per second will be heard by B? (Velocityofsound =c)

Answer:

Given

Each of the two men A and B is carrying a source of sound of frequency.

Apparent frequency, n’ = \(\frac{c+u_o}{c-u_s} \times n\)

The distance between A and B is decreasing. So, the velocity of the listener u₀ and that of the source u_s, are both positive.

1. When A is the listener, velocity of the listener, u₀ = u; velocity of the source B, u_s = 0

So, \(n_A=\frac{c+u}{c} \times n\)

Evidently, n_A > n

∴ Number of beats per second

= \(n_A-n=\left(\frac{c+u}{c}-1\right) \times n=\frac{u}{c} n .\)

2. When B is the listener, velocity of the listener, u₀ = 0; velocity of the source A, u_s = u

So, \(n_B=\frac{c}{c-u} \times n\)

Evidently, \(n_B>n\)

∴ Number of beats per second

= \(n_B-n=\left(\frac{c}{c-u}-1\right) \times n\)=\(\frac{u}{c-u} n\)

Question 2. A car Is approaching a hill at a high speed. At that time, if the horn of the car is blown, the driver hears the echo sharper than the original sound. Explain the reason.
Answer:

Given

A car Is approaching a hill at a high speed. At that time, if the horn of the car is blown, the driver hears the echo sharper than the original sound.

The original sound after being reflected from the hill approaches the car and the driver listens to the echo. So, in this case, the listener is moving towards the source of the echo. So, due to the Doppler effect, the echo is of a higher apparent frequency. Thus, it appears to be sharper than the original sound to the driver.

Question 3. Certain characteristic wavelength in the light from a galaxy has a longer wavelength compared to that from a terrestrial source. Is the galaxy approaching or receding?
Answer:

The galaxy is receding. It can be concluded from the increase in wavelength, i.e., decrease in frequency due to the Doppler effect, that the distance between the source of light (the galaxy) and the observer (the earth) is gradually increasing. Hence, the galaxy is receding.

Question 4. Show that the apparent frequency f’ of a source of sound moving with a speed vs towards a stationary receiver is \(f^{\prime}=\frac{f c}{c-v_s}\), where c is the velocity of sound and f is the frequency.
Answer:

If the source of sound approaches the stationary listener with velocity v_s, the number of sound waves f produced per second occupies the distance c – v_s.

∴ Apparent wavelength, \(\lambda^{\prime}=\frac{c-v_s}{f}\)

So, apparent frequency, f’ = \(\frac{c}{\lambda^{\prime}}=\frac{f c}{c-v_s}\)

Question 5. Two sources, each emitting a sound of wavelength A, are kept at a fixed distance. How many beats will be heard by a listener moving with a velocity u along the line joining the two sources?
Answer:

Given

Two sources, each emitting a sound of wavelength A, are kept at a fixed distance.

If n is the frequency and v is the velocity of sound, the apparent frequency to a listener in motion for a stationary source,

n’ = \(\frac{v+u}{v} \times n\)

The velocities of the listener for the two sources in question are -u and +u.

So, \(n_1^{\prime}=\frac{v-u}{v} \times n \text { and } n_2^{\prime}=\frac{v+u}{v} \times n\)

∴ Number of beats per second

= \(n_2^i-n_1^{\prime}=\frac{2 u}{v} \times n=\frac{2 u}{\frac{v}{n}}=\frac{2 u}{\lambda} .\)

Example 6. A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has a frequency of 2f. If v is the velocity of sound, what will be the velocity of the car?
Answer:

Given

A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has a frequency of 2f. If v is the velocity of sound,

If u_s and u₀ are the velocities of the source of sound and the listener respectively,

the frequency of the echo, f’ = \(\frac{v+u_o}{v-u_s} \times f\)

Here, u_s = u₀ = u(say) and f’ = 2f

∴ 2f = \(\frac{v+u}{v-u} \times f \text { or, } v+u=2(v-u) \text { or, } 3 u=v \text { or, } u=\frac{v}{3} \text {. }\)

Example 7. What should be the velocity of a source of sound so that the apparent frequency to a listener will be half the actual frequency of the source? The velocity of sound in air = v.
Answer:

If n is the actual frequency, apparent frequency = \(\frac{n}{2}\).

Here the listener is stationary, i.e., u₀ = 0

Again decrease of frequency means that the source is receding from the listener. So the velocity of the source us is negative.

Therefore, from the relation n’ = \(\frac{v+u_o}{v-u_s} \times n\), we have

⇒ \(\frac{n}{2}=\frac{v}{v+u_s} \times n \text { or, } v+u_s=2 v \text { or, } u_s=v\)

i.e., the source is receding from the listener with the velocity of sound.

Examples of Doppler Effect Questions with Answers

Question 8. What should be the velocity of a source of sound so that the apparent frequency to a listener will be twice the actual frequency of the source? The velocity of sound in air = v.
Answer:

Apparent Frequency, \(n^{\prime}=\frac{v+u_o}{v-u_s} \times n\)

In this case n’ = 2n. The listener is stationary; so u₀ = 0.

The apparent frequency is higher, i.e., the source is approaching the listener. So, the velocity of the source u_s is positive. Therefore,

2n = \(\frac{v}{\nu-u_s} \times n \text { or, } v=2\left(\nu-u_s\right) \text { or, } u_s=\frac{v}{2}\)

i.e., the source is approaching the listener with half the velocity of sound.

Question 9. What should be the velocity of a listener so that the apparent frequency of the sound coming from a stationary source to him will be twice the actual frequency? The velocity of sound in air = v.
Answer:

Apparent frequency, \(n^{\prime}=\frac{v+u_o}{v-u_s} \times n\)

In this case velocity of the source, u_s= 0.

Since the apparent frequency is higher, the listener is approaching the source, i.e., the velocity of the listener u₀ is positive.

Again, n’ = 2n.

∴ 2n = \(\frac{v+u_o}{v} \times n \text { or, } 2 v=v+u_o \text { or, } u_o=v\)

i.e., the listener is approaching the stationary source with the velocity of sound.

Question 10. Doppler effect gives an idea of a continuously expanding universe—explain.
Answer:

Doppler effect gives an idea of a continuously expanding universe

Light coming from the distant star can be analysed with the spectrometer. The experiment shows that the wavelength of a spectral line for a light source situated for away from Earth is greater than that for the same light source on Earth i.e., frequency is comparatively low.

From this apparent decrease in frequency, known as the redshift in the Doppler effect, we conclude that all distant stars are receding from Earth which indicates that the Universe is continuously expanding.

Question 11. A listener moving with constant velocity passes a stationary source. Draw a graph to show the change of apparent frequency of the source to the listener with time. The actual frequency of the source is n.
Answer:

Given

A listener moving with constant velocity passes a stationary source.

While approaching the stationary source apparent frequency will be,

n’ = \(\frac{V+u_0}{V} n\)…(1)

[velocity of sound in air = V, velocity of listener = u₀]

While receding the stationary source apparent frequency will be,

n” = \(\frac{V-u_0}{V} n\)…(2)

from equations (1) and (2) it is clear that nf and n” remain constant with time and also n’ > n> n”. The change of apparent frequency with time is shown below.

Here OS denotes the time taken by the listener to pass by the source. After that, the listener continually moves away from the source.

Question 12. Both of a sound source and a listener are approaching each other with the same speed \(\frac{c}{10}\) (speed of sound in air =c). What will be the percentage of apparent increase or decrease in frequency of sound?
Answer:

Given

Both of a sound source and a listener are approaching each other with the same speed \(\frac{c}{10}\) (speed of sound in air =c).

Apparent frequency,

n’ = \(\frac{c+u_0}{c-u_S} n\)

So, the apparent increase in frequency,

n’ – n = \(\frac{u_0+u_S}{c-u_S} n\)

∴ Percentage of change in frequency

= \(\frac{n^{\prime}-n}{n} \times 100=\frac{u_0+u_S}{c-u_S} \times 100\)

= \(\frac{\frac{c}{10}+\frac{c}{10}}{c-\frac{c}{100}} \times 100=\frac{2}{9} \times 100=22.2 \%\)

Doppler Effect Formulas: Q&A Guide

Question 13. A band of music at a frequency f is moving towards a wall at a speed v_b. A motorist is following the band with a speed v_m. If v is the speed of sound, obtain an expression for the beat frequency heard by the motorist.
Answer:

Given

A band of music at a frequency f is moving towards a wall at a speed v_b. A motorist is following the band with a speed v_m. If v is the speed of sound

Two separate sounds will be heard by the motorist, one is direct from the band and the other is the echo from the wall.

Apparent frequency of direct sound, \(f_1=f\left(\frac{v+v_m}{v+v_b}\right)\)

Apparent frequency of the echo, \(f_2=f\left(\frac{v+v_m}{v-v_b}\right)\)

Therefore, the beat frequency heard by the motorist,

n = \(f_2-f_1=f\left(v+v_m\right) \cdot\left[\frac{1}{v-v_b}-\frac{1}{v+v_b}\right]\)

= \(\frac{2 f v_b\left(v+v_m\right)}{v^2-v_b^2}\)

Question 14. Why Doppler effect is clearly realised in the case of sound but not in the case of light waves?
Answer:

Our auditory system is more sensitive to realise a small change in audible frequency. On the other hand, our visual system is not so sensitive to detect such a small change in the visible frequency of light. In our daily life, the magnitude of our relative velocity of us with a light source on earth is not high enough to observe such a noticeable change in frequency, i.e., the Doppler effect.

Doppler Effect In Sound And Light Multiple Choice Questions

Doppler Effect MCQs for Class 11 WBCHSE

Question 1. A source of sound with a frequency of 256 Hz is moving with a velocity of v towards a wall and an observer is stationary between the source and the wall. When the observer is between the source and the wall

1. He will hear beats
2. He will hear no beats
3. He will not get any sound
4. He will get the sound of the same frequency

Answer: 2. He will hear no beats

Question 2. The frequency of a progressive wave may change due to

1. Reflection
2. Refraction
3. Interference
4. Doppler effect

Answer: 4. Doppler effect

Question 3. A train blowing a whistle of frequency 1000 Hz is moving with a uniform velocity from west to east. The apparent frequency of the sound of the whistle to a stationary listener is 990 Hz. The position of the listener relative to the train is

1. On the north
2. On the south
3. On the east
4. On the west

Answer: 4. On the west

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 4. A source of sound and a listener are moving in the same direction with the same velocity. If the actual frequency of sound is 200 Hz, the apparent frequency of the sound to the listener is

1. 200 Hz
2. Less than 200 Hz
3. Greater than 200 Hz
4. None of these

Answer: 1. 200 Hz

Question 5. A bus is moving towards a huge wall with a velocity of 5 m · s^-1. The driver sounds a horn of frequency 200 Hz. The beat frequency heard by the passenger will be

1. 4
2. 6
3. 8
4. 2

Answer: 2. 6

Practice MCQs on Doppler Effect in Sound

Question 6. A motor car sounding a horn is approaching a large reflector. If the frequency of the horn is 1000 Hz, the frequency of the echo to the driver will be

1. 1000 Hz
2. Less than 1000 Hz
3. Greater than 1000 Hz
4. None of these

Answer: 3. Greater than 1000 Hz

Question 7. An observer standing on a railway crossing receives frequencies of 2.2 kHz and 1.8 kHz when the train approaches and recedes from the observer. The speed of the sound in air is 300 m · s^-1. The velocity of the train (in m · s^-1)

1. 60
2. 30
3. 90
4. 70

Answer: 2. 30

Question 8. A whistle producing sound waves of frequency 9500 Hz and above is approaching a stationary person with speed v m · s^-1. The velocity of sound in air is 300 m · s^-1. If the person can hear frequencies upto a maximum of 10000 Hz, the maximum value of v up to which he can hear the whistle is

1. 15√2 m · s^-1
2. 15/√2 m · s^-1
3. 15 m · s^-1
4. 30 m · s^-1

Answer: 3. 15 m-s^-1

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Question 9. What do you understand by the red shift of a star or of a galaxy of stars?

1. They are gradually receding from the earth
2. They are gradually approaching the earth
3. They are stationary
4. None of the above

Answer: 1. They are gradually receding from the earth

Question 10. When a source of light and an observer approach each other, the apparent change of the wavelength of light is called

1. Redshift
2. Violet shift
3. Blueshift
4. Black shift

Answer: 3. Blueshift

In this type of questions, more than one options are correct.

Question 11. State in which of the following cases, an observer will not see any Doppler effect?

1. Both the source and observer remain stationary but a wind blows.
2. The observer remains stationary but the source moves in the same direction and with the same speed as the wind.
3. The source remains stationary but the observer and the wind have the same speed away from the source.
4. The source and the observer move directly against the wind but both with the same speed.

Answer:

1. Both the source and observer remain stationary but a wind blows.

4. The source and the observer move directly against the wind but both with the same speed.

Practice MCQs on Doppler Effect in Light

Question 12. Consider a source of sound S and an observer P. The sound source is of frequency n₀. The frequency observed by P is found to be n₁ if P approaches S at speed v and S is stationary, n₂ if S approaches P at speed v and P is stationary and n₃ if each of P and S has speed \(\frac{v}{2}\) towards one another. Which of the following conclusions is correct?

1. n₁ = n₂ = n₃
2. n₁< n₂
3. n₃ > n₀
4. n₃ lies between n₁ and n₂

Answer:

2. n₁< n₂

3. n₃>n₀

4. n₃ lies between n₁ and n₂

Sample MCQs on Frequency Shift Due to Doppler Effect

Question 13. An observer A is moving directly towards a stationary sound source while another observer B is moving away from the source with the same velocity. Which of the following conclusions are correct?

1. The average of frequencies recorded by A and B is equal to the natural frequency of the source.
2. The wavelength of waves received by A is less than that of waves received by B.
3. The wavelength of waves received by two observers will be the same.
4. Both observers will observe the wave travelling at same speed.

Answer:

1. The average of frequencies recorded by A and B is equal to natural frequency of the source.

3. The wavelength of waves received by two observers will be the same.

Question 14. Two cars, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these cars blows a whistle of frequency f₁. An observer in the other car hears the frequency of the whistle to be f₂. The speed of sound in still air is v. Correct statement(s) is are:

1. If the wind blows from the observer to the source, f₂ > f₁
2. If the wind blows from the source to observer, f₂>f₁
3. If the wind blows from the observer to the source, f₂<f₁
4. If the wind blows from the source to the observer, f₂<f₁

Answer:

1. If the wind blows from the observer to the source, f₂ > f₁
2. If the wind blows from the source to the observer, f₂>f₁
3. If the wind blows from the observer to the source, f₂<f₁

Nature Of Vibration – Free Or Natural Vibration

WBBSE Class 11 Nature of Vibration Notes

A vibrating body always moves to and fro about an equilibrium position. At the instant when it is at the equilibrium position, there are no forces acting on the body. However, the body does not stop because of its inertia of motion.

As soon as it crosses the equilibrium position, a force acts on it. This force is always directed toward the equilibrium position and is called the restoring force.

If no force other than the restoring force acts on the body, or the effect of other forces is negligible, the body can vibrate without interruption, i.e., its vibration is a free or natural vibration. The amplitude of this vibration remains unchanged with the passage of time.

Understanding Free and Forced Vibration

Free Or Natural Vibration Definition: If the effect of forces other than the restoring force is negligible on a vibrating body, its motion is called a free or natural vibration.

A body undergoing free vibration has a definite frequency, i.e., the body executes a fixed number of vibrations in unit time. It is called its natural frequency. The natural frequency (n₀) depends on the density, shape elasticity, etc. of the vibrating body. For example:

A simple pendulum: \(n_0=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{g}{L}}\); so if g is constant, the natural frequency of a simple pendulum depends on its length L.

An elastic spring: \(n_0=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\) = here the spring constant k is related to the elasticity of the material of the spring.

It may be said that the natural frequency of the spring depends on the mass (m) hanging at the end of the spring and the elasticity of the material of the spring. Every vibrating body such as a tuning fork, the string of a musical instrument, etc. has its characteristic natural frequency.

Nature Of Vibration – Damped Vibration

Damped Vibration Definition: If resistive forces act on a vibrating body in addition to the restoring force, its amplitude gradually diminishes. After some time, the body comes to rest at its position of equilibrium. This type of vibration is called damped vibration. The resistive effect is called damping.

• In fact, free vibration does not exist in real life. All vibrating bodies ultimately come to rest after some time, i.e., all vibrations are damped. Different types of resistive forces act on different types of vibrating bodies.
• For a simple pendulum, the resistive force is the viscous force of air; in a moving coil galvanometer, the resistive force is electromagnetic damping. The vibrating body has to work against these resistive forces. So its energy decreases. When the energy of the body becomes zero, it comes to rest.

Graphical representation of free and damped vibration: The characteristics of free and damped vibrations can be understood easily with the help of displacement time graphs. If the vibration of a body is free, it will vibrate forever with its amplitude unchanged. In real life, the amplitude gradually diminishes and ultimately the body comes to rest.

Types of Vibration in Physics

Beneficial examples of damped vibration: Damping plays a beneficial role in our modern-day life. One such application of damped oscillation is the car suspension system. It makes use of damping to make our ride less bumpy and more comfortable by counteracting and hence reducing the vibrations of the car when it is on the road for optimal passenger comfort, the system is critically damped or slightly underdamped.

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Different types of damped motion:

1. If the damping is very weak, the body vibrates with almost its natural frequency. For example, if the bob of the simple pendulum is heavy enough, then the damping becomes insignificant. So the pendulum continues to oscillate for a long time. The time period and frequency of such a pendulum are almost equal to those of a free pendulum.

2. If the damping is stronger, the vibration of a body does not continue for a long time. For example, if a light piece of wood is used as the bob of a simple pendulum and is allowed to oscillate, it comes to rest after a few oscillations. In this case, the time period becomes very large, i.e., the frequency of vibration is much less than the natural frequency.

3. In case of very strong damping, the vibrating body comes back to its position of equilibrium from its displaced position and stops there. So the body cannot move past its position of equilibrium. The body does not vibrate at all. This is called overdamped motion or aperiodic motion.

4. There is a particular state of damping, between small damping and overdamping, for which the body returns to its equilibrium position in the least time, but cannot travel past its position of equilibrium. This is called critical damping, In practical cases, if we want to stop the vibration of a body quickly, its damping is kept close to the state of critical damping.

5. If the damping is less than critical damping, the body oscillates with decreasing amplitude. This is known as underdamping.

Decrement of amplitude In damped vibration: From the graph for damped vibration we get, A₁ = initial amplitude of vibration, A₂ = amplitude of vibration after one complete oscillation, and A₃ = amplitude of vibration after two complete oscillations.

Obviously, A₁ > A₂ > A₃

Two important characteristics of such a damped vibration are:

1. The amplitude of vibration decreases in a constant ratio for each complete vibration.

That is, \(\frac{A_1}{A_2}\)=\(\frac{A_2}{A_3}\)=\(\frac{A_3}{A_4}\)=\(\cdots\) = constant. This constant is called the decrement.

2. From the very initiation of motion, damping comes into play. Therefore, even the first amplitude A₁ of damped vibration is less than the amplitude A₀ of free vibration.

Equation of motion: A particle under damped har-monic vibration in one dimension is subject to two types of forces:

A restoring force: F₁ = -kx, where k is constant.

One or more resistive forces: Each resistive force is proportional to the velocity \(\left(v=\frac{d x}{d t}\right)\)of the particle, and acts in a direction opposite to that of the instantaneous velocity.

Applications of Vibration in Engineering

So, the resultant of the resistive forces is, \(F_2=-k^{\prime} v=-k^{\prime} \frac{d x}{d t}\), where k’ is another constant.

The acceleration of the particle is, \(a=\frac{d v}{d t}=\frac{d^2 x}{d t^2}\). Thus from the relation F = ma, we get, ma = F₁ + F₂

or, \(m \frac{d^2 x}{d t^2}=-k x-k^{\prime} \frac{d x}{d t}\)

or, \(m \frac{d^2 x}{d t^2}+k^{\prime} \frac{d x}{d t}+k x=0\)

or, \(\frac{d^2 x}{d t^2}+2 b \frac{d x}{d t}+\omega^2 x=0\)

Were, \(\omega=+\sqrt{\frac{k}{m}} and b=\frac{k^{\prime}}{2 m}\),

The equation (1) is known as the equation of motion of a damped SHM.

Nature Of Vibration – Damped Vibration Numerical Examples

Short Answer Questions on Vibration

Example 1. A second pendulum is shifted 4 cm away from its equilibrium position and then released. After 2 s  the pendulum is 3 cm away from its position of equilibrium. What will be the position of the pendulum after another 2s?
Solution:

Given

A second pendulum is shifted 4 cm away from its equilibrium position and then released. After 2 s  the pendulum is 3 cm away from its position of equilibrium.

Time period of a second pendulum is 2s. In the case of this damped vibration, let the change in the time period be negligible. For the given pendulum the initial amplitude of vibration A₁ = 4 cm. After one complete oscillation, the amplitude of vibration A₂ = 3 cm.

So, if the amplitude of vibration after two complete oscillations is A₃, then the decrement is:

⇒ \(\frac{A_1}{A_2}=\frac{A_2}{A_3} \text { or, } A_3=\frac{\left(A_2\right)^2}{A_1}=\frac{(3)^2}{4}=2.25 \mathrm{~cm} \text {. }\)

Example 2. After 100 complete oscillations, a pendulum’s amplitude becomes 1/3 rd of its initial value. What will be its amplitude after 200 complete oscillations? Express it as a fraction of the initial amplitude.
Solution:

Given

After 100 complete oscillations, a pendulum’s amplitude becomes 1/3 rd of its initial value.

The pendulum has a damped vibration. So, the amplitude decreases at the same rate. Since, after 100 complete oscillations the amplitude of vibration becomes 1/3 rd of the initial amplitude, after 200 complete oscillations, the amplitude will be 1/3 x 1/3 = 1/9 th of the initial amplitude.

Nature Of Vibration – Forced Vibration

Practically all vibrations are damped vibrations. The vibrating body works against different resistive forces. So its energy diminishes and the amplitude gradually decreases.

• To maintain a steady vibration, energy from an external source is needed. If energy is supplied from an external source in such a way that the rate of supply of energy exactly balances the rate of loss of energy, then the amplitude of the body remains constant.
• The value of the amplitude is similar to that of the free vibration of the body. This type of vibration is called a forced vibration. For example, if we do not wind a pendulum clock, it will stop after a while due to damping.
• When we wind the clock, we compress a spring within the clock which stores potential energy and supplies that energy continuously. The pendulum oscillates continuously with constant amplitude and time period.

External means are required not only for maintaining the vibration but also to vibrate a body that is initially at rest.

Examples of forced vibration:

1. If we strike a prong of a tuning fork, the intensity of the emitted sound is not very high so it cannot be heard from a distance. Now, if the handle of the vibrating tuning fork is made to touch the surface of a table, the tuning fork sets the table surface into forced vibration. Sound is emitted from the table also. Hence, the sound is amplified.

• It must be kept in mind that, according to the principle of conservation of energy, the total amount of energy cannot be increased. When the handle of the tuning fork is pressed against a table surface, a part of the energy from the tuning fork is transferred to the table.
• As the damping of the table is higher than that of the tuning fork, the energy transferred to the table by the tuning fork decays at a faster rate. Hence, the vibration of the table stops earlier and the intensity of the sound produced due to vibration of the table is comparatively higher.
• In this example of forced vibration, the upper surface of the table is made to vibrate by the tuning fork. The vibration of the tuning fork is the driving vibration and the vibration of the upper surface of the table is the driven vibration.

2. A thread is tied loosely between P and Q. Two pendulums C and D, having different lengths, are sus¬pended from two points between P and Q. If the pen¬dulum C is made to oscillate, it will continue its oscillation with its natural frequency. As the thread PQ is tied loosely, energy will be transferred from pendulum C to pendulum D through the thread.

• As a result, the pendulum D will begin to oscillate. As the lengths of the pendulums are different, their natural frequencies are not the same. It is found that the pen¬dulum D initially tries to oscillate at its natural frequency.
• But its vibration is damped quickly. Then pendulum D begins to oscillate at the natural frequency of C. In this example, pendulum C provides the driving vibration and the vibration of pendulum D is the driven vibration.
• In these two examples, it is to be noted that the external driving forces are neither steady nor momentary. Rather, the force originating from the vibrating body is a periodic force. In fact, the spring of a clock exerts a periodic force on the pendulum.

From the above discussions forced vibration can be defined in the following way.

Forced Vibration Definition: If an external periodic force is applied to a freely vibrating body, the body tries to maintain its vibrations at its own frequency; but after some time, the body begins to vibrate with the frequency of the applied periodic force. Such a vibration of a body is called a forced vibration.

Equation Of motion: A particle under forced harmonic vibration in one dimension is subject to three types of forces:

A restoring force: F₁ = -kx, where k is constant.

Resistive forces: The resultant of all the resistive forces acting on a particle is, \(F_2=-k^{\prime} v=-k^{\prime} \frac{d x}{d t}\)

where v is the velocity of the particle and k’ is a constant.

Externally impressed periodic force: This is of the form, \(F_3=F_0 \sin \omega t\)

where ω is the circular frequency and F₀ is the amplitude of the periodic force; both ω and F₀ are constants.

The acceleration of the particle is, a = \(\frac{d v}{d t}=\frac{d^2 x}{d t^2}\). Then from the relation F= ma, we get, ma = F₁ + F₂ + F₃

or, \(m \frac{d^2 x}{d t^2}=-k x-k^{\prime} \frac{d x}{d t}+F_0 \sin \omega t\)

or, \(m \frac{d^2 x}{d t^2}+k^{\prime} \frac{d x}{d t}+k x=F_0 \sin \omega t\)

or \(\frac{d^2 x}{d t^2}+2 b \frac{d x}{d t}+\omega_0^2 x=a_0 \sin \omega t\) where, \(\omega_0= \pm \sqrt{\frac{k}{m}}, b=\frac{k^{\prime}}{2 m}\) and \(a_0=\frac{F_0}{m}\).

The equation (1) is known as the equation of motion of a forced SHM. It is to be noted that, the natural frequency ω₀ of the particle vibration is, in general, different from the frequency co of the external periodic force. In the special case, when cω = ω₀, the phenomenon of resonance comes into play.

Comparison Between Free Vibration And Forced Vibration

Nature Of Vibration – Resonance Or Resonant Vibration

Resonance in Vibrating Systems

Usually, when a body executes a forced vibration, its amplitude and velocity remain small. In most cases, the frequency of the applied periodic force does not come close to the natural frequency of the vibrating body.

• However, when the frequency of the applied periodic force becomes equal to the natural frequency of the body, the amplitude and the velocity of the body become very large. This phenomenon is called resonance.
• It is important to keep in mind that some objects are very resonant at a particular frequency while others barely resonate. For example, a temple bell will only ring at a fixed frequency, which is its resonant frequency but a lump of jelly will vibrate at many different frequencies but will not resonate at all.

Resonance Or Resonant Vibration Definition: When the frequency of the applied periodic force matches the natural frequency of a vibrating body, its amplitude rapidly increases to a large value. This phenomenon is called resonance.

This vibration is also known as sympathetic vibration. Resonant vibration is of two types.

Amplitude resonance: In this case, the amplitude of vibration becomes maximum. At amplitude resonance die frequency of applied force is slightly less than the natural frequency i.e., \(\omega=\sqrt{\omega_0^2-2 b^2}\)

Velocity resonance: In this case, the velocity of the vibrating body becomes maximum. Velocity resonance occurs when the natural frequency is exactly equal to the frequency of applied force i.e., ω = ω₀.

Examples of resonance:

1. A thread is tied loosely between P and Q, and four simple pendulums C, D, E, and F are suspended from the thread. Pendulums C and D have the same length.

• So they have the same natural frequency. But the lengths of pendulums E and F are different from those of C and D. So they have different natural frequencies. Now if pendulum C is set into vibration, it executes SHM.
• Hence, a periodic force acts on the pendulums D, E, and F through the thread PQ. After a while, it is observed that, though E and p are set in forced vibration, their amplitudes and velocities of vibration are not large: but pendulum D vibrates with a large amplitude. This is because pendulum D resonates with pendulum C.
• As soon as pendulum D is set into resonant vibration, it applies, in mm, periodic forces on pendulums C F., and F through die thread PQ. This does not affect the vibrations of E and F very much, but pendulum C experiences resonance. In this wave, energy is alternately transferred from C to D and D to C.
• As a result, it is observed diet. when the amplitude of the vibration of pendulum C becomes very large, pendulum D almost comes to a stop; in die next moment the amplitude of pendulum D starts to increase while that of pendulum C gradually decreases.

Real-Life Examples of Vibrations

2. Resonant air column: The length of an air column contained in a tube determines die natural frequency of the column. If a vibrating tuning fork is held at the mouth of such a tube, forced vibration is set up in the air column.

• If the frequency of the tuning fork is the same as the natural frequency of the air column, then resonance occurs in the air column and a loud sound is heard. Using this phenomenon, sometimes a tuning fork is mounted on a hollow box.
• The box is so shaped that its frequency becomes equal to the natural frequency of the tuning fork. So, when the tuning fork is struck a resonance is produced and a loud sound is heard.

3. Hollow box in musical instruments: In musical instruments, such as violin, esraj, sitar, etc., the strings are stretched on a wooden hollow box or cavity The vibration of a string induces forced vibration on the wooden box and on a large mass of air inside it. This increases die intensity of die emitted sound. If resonance occurs die emitted sound is further intensified.

4. In musical instruments like violin, esraj, sitar, etc., there are a number of strings in addition to the principal string, which are adjusted or tuned to predefined notes of different frequencies.

When the principal string is played to produce a certain note, resonant vibrations may occur in some other string. This contributes to the increase in both the loudness and quality of the musical sound.

5. Helmholtz’s resonator: To detect the presence of tones of different frequencies in a note, the German scientist Helmholtz devised a resonator. It consists of a brass shell of nearly spherical shape with two openings a and b of different diameters.

• The natural frequency of die air inside the spherical shell depends on the size of the shell. The larger opening a, called hole or neck, is turned towards the source of sound while we place our ears in front of the smaller opening b, called pip.
• Now, suppose that in a note there is a tone whose frequency is equal to the natural frequency of the resonator. In that case, resonance is produced in the air inside the resonator and the tone will be heard distinctly. With the help of different resonators of known frequencies, we can detect different tones in a note.

Comparison between Forced Vibration and Resonance

Nature Of Vibration Long Answer Type Questions

WBBSE Class 11 Nature of Vibration Q&A

Question 1. Why does an empty container emit a louder sound than a water-filled container when they are struck?
Answer:

When we strike the container, its wall vibrates. These vibrations produce forced vibrations in the air or water inside the container.

• If the container is filled with water, it becomes heavy and the amplitude of vibration becomes small. But if the container is empty, it is comparatively lighter and the amplitude of vibration is large.
• As the loudness of sound is proportional to the square of the amplitude of vibration, the empty container emits a louder sound.

Question 2. Why do buildings get demolished by earthquakes?
Answer:

Buildings get demolished by earthquakes

An earthquake induces forced vibration in the walls of buildings. For intense earthquakes, the amplitudes of the forced vibrations become very high, which in many cases cross the elastic limit of the constituent materials of the buildings. So many buildings get demolished.

Question 3. A vibrating tuning fork is held at the mouth of a cylindrical tube. The tube is dipped into water. It is found that when the level of water rises to a definite height, a sound of large intensity is heard. Explain the reason behind it.
Answer:

Given

A vibrating tuning fork is held at the mouth of a cylindrical tube. The tube is dipped into water. It is found that when the level of water rises to a definite height,

The empty cylindrical tube is filled with air. The vibrating tuning fork produces forced vibration in the air column of the tube. As a result of the forced vibration of the air column, an additional sound is produced.

• The frequency of the sound produced depends on the length of the air column. By dipping the tube into water, the length of the air column can be changed.
• So, when the level of water rises to a particular height, the frequency of the air column in the tube and that of the tuning fork become equal. Then resonance takes place and a sound of large intensity is heard.

Question 4. In the presence of a resonant body, the sound produced by a body Is intensified. Is the principle of conservation of energy violated here?
Answer:

The principle of conservation of energy is not violated here.

• Suppose, a tuning fork is set into vibration. The loudness of the sound produced is not very large. If the vibrating fork is held on a hollow wooden box, the sound is intensified. In this case, the air inside the box acts as a resonant body.
• But it is found that the sound of the tuning fork lasts for a long time in the first case. But in the second case, the sound of the tuning fork is short-lived. The total dissipation of energy in both cases is the same.
• In the first case, the rate of dissipation of energy is low but continues for a long time. But in the second case, the rate of dissipation of the same amount of energy is comparatively high, and therefore the sound dies down in a short time. It does not violate the principle of conservation of energy.

Short Answer Questions on Vibration for Class 11

Nature Of Vibration Conclusion

If the effect of forces other than the restoring force acting on a vibrating body is negligible, then its vibration is called free or natural vibration.

If resistive forces act on a vibrating body in addition to the restoring force, its amplitude gradually diminishes. After some time, the body comes to rest at its position of equilibrium. This type of vibration is called damped vibration. The resistive effect is called damping.

If an external periodic force is applied to a freely vibrating body, the body tries to maintain its vibrations at its own frequency; but after some time, the body begins to vibrate with the frequency of the applied periodic force. Such a vibration of a body is called a forced vibration.

When the frequency of the applied periodic force matches the natural frequency of a vibrating body, its amplitude rapidly increases to a large value. This phenomenon is called resonance.

Nature Of Vibration Very Short Answer Type Questions

Question 1.Under the influence of which force the oscillation of a pendulum gradually dies out?
Answer: Resistive force

Question 2. Which force acts on a body during its free vibration?
Answer: Restoring force

Question 3. Which force acts on a body that vibrates freely?
Answer: Only a restoring

Question 4. A man with a wristwatch falls freely from a tall building. Will the watch give the correct time?
Answer: Yes, independent of g

Question 5. Which quantity of vibration gradually decreases during damped vibration?
Answer: Amplitude

Question 6.If a resistive force acts on a vibrating body, then its amplitude of vibration gradually ________
Answer: Decreases

Vibration Characteristics: Short Answer Questions

Question 7. What phenomenon will occur if the frequency of free vibration of a vibrating body becomes equal to the frequency of an external periodic force?
Answer: Resonance

Question 8. Which characteristic of sound increases during resonance?
Answer: Intensity

Question 9. A special case of ______ vibration is resonance.
Answer: Forced

Question 10. Which type of vibration is the vibration of the diaphragm of a microphone during a speech?
Answer: Forced

Question 11. What kind of external force has to act on a vibrating body, to occur forced vibration?
Answer: Periodic

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Nature Of Vibration Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 Is true, statement 2 is true; statement 2 Is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: A vibrating body always moves to and fro about an equilibrium position.

Statement 2: Due to inertia of motion a vibrating body does not stop at its equilibrium position.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 2.

Statement 1: The sound gets amplified when a vibrating tuning fork is made to touch the surface of a table.

Statement 2: The dimension of the table is more than that of the tuning fork.

Answer: 1. Statement 1 Is true, statement 2 is true; statement 2 Is a correct explanation for statement 1.

WBBSE Class 11 Sample Questions on Vibrational Motion

Nature Of Vibration Match The Columns

Question 1.

Answer: 1. C, 2. D, 3. A, 4. B, D

Nature Of Vibration Integer Answer Type Questions

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 0.

1. The amplitudes of the first two oscillations of a damped pendulum are 9.0 cm and 3.0 cm respectively. What will be the amplitude (in cm) of the pendulum in its third oscillation?
Answer: 1

Question 2. After 50 complete oscillations a pendulum’s amplitude becomes 1/3 of its initial vibration. If its amplitude becomes \(\frac{1}{n^3}\) of its initial vibration after 150 complete oscillations, then find the value of n.
Answer: 3

Nature Of Vibration Multiple Choice Questions And Answers

WBBSE Class 11 Vibration MCQs

Question 1. In the case of free vibration of a body, the quantity that remains constant is

1. Velocity
2. Acceleration
3. Time period
4. Phase

Answer: 3. Velocity

Question 2. During vibration, the restoring force is

1. Directly proportional to the displacement
2. Directly proportional to the velocity
3. Directly proportional to the kinetic energy
4. Directly proportional to the potential energy

Answer: 1. Directly proportional to the displacement

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 3. During the damped oscillation of a body, the force that acts is

1. Only the restoring force
2. Only the resistive force
3. The restoring force along with the resistive force
4. The restoring force along with the resistive force and the external periodic force

Answer: 3. The restoring force along with the resistive force

WBBSE Class 11 Practice Tests on Vibration Concepts

Question 4. During damped vibration, the quantity which gradually decreases is

1. Velocity
2. Phase
3. Frequency
4. Amplitude

Answer: 4. Amplitude

Question 5. In case of resonance, the characteristic property which is the same for free vibration of the body and the external periodic force is

1. Amplitude
2. Phase
3. Velocity
4. Frequency

Answer: 4. Frequency

Conceptual Questions on Vibrations for Class 11

Question 6. During the forced vibration of a particle

1. Only the restoring force acts on the particle
2. Both the restoring force and a dissipative force act on the particle
3. Both the restoring force and an external periodic force act on the particle
4. The restoring force, a dissipative force, and an external periodic force act on the particle

Answer: 4. The restoring force, a dissipative force, and an external periodic force act on the particle

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Question 7. During forced vibration, the frequency of the external periodic force is

1. Equal to
2. Less than
3. Greater than
4. Equal to, or greater than, or less than the frequency of free vibration of the body.

Answer: 4. Equal to, or greater than, or less than the frequency of free vibration of the body.

Practice MCQs on Types of Vibrations

Question 8. During forced vibration, if the frequency of free vibration of a body is equal to the frequency of the external periodic force, then the phenomenon that occurs is called

1. Beats
2. Interference
3. Resonance
4. Reverberation

Answer: 3. Resonance

Question 9. If a vibrating tuning fork is held at the open end of a tube closed at one end, then for a particular length of the tube an intense sound is heard. This phenomenon is known as

1. Beats
2. Stationary wave
3. Interference
4. Resonance

Answer: 4. Resonance