Wave Motion Question and Answers
Question 1. In the case of a transverse wave, do the particles of the medium vibrate in the u plane parallel to the plane of propagation of the wave?
Answer:
Let us assume that the wave is propagating along the z-axis. Then the plane of propagation of the wave, or the wavefronts, will be perpendicular to z -the axis, i.e., parallel to the xy-plane.
Again the direction of motion of tire vibrating particles of the medium in a transverse wave is perpendicular to z -the axis, i.e., parallel to the xy-plane. So, the particles of the medium vibrate in a plane parallel to the plane of propagation of the wave.
Question 2. The equation of a progressive wave In a stretched string Is given by, y = Asin(kx – ωt). What is the maximum particle velocity?
Answer:
Particle velocity, v = \(\frac{\partial y}{\partial t}=\frac{\partial}{\partial t}\{A \sin (k x-\omega t)\}\)
= \(-\omega A \cos (k x-\omega t)\)
Since the maximum value of a cosine function is ±1, the magnitude of maximum particle velocity is ωA.
Question 3. Regular reflection of sound takes place from a large rough reflector, but not regular reflection of light. On the other hand, regular reflection of light takes place from a small smooth reflector, but not regular reflection of sound. What Is the reason for this?
Answer:
To obtain effective reflection of a wave from a reflector, its size should be greater than the wavelength. The wavelength of audible sound in air varies from 1.5 cm to 16 in.
- On the other hand, the wavelength of visible light varies from 4 x 10-7 m to 8 x 10-7 m, i.e., the wavelength of sound is much greater than that of light. So light can be reflected from a very small reflector, but not sound.
- Again for regular reflection, the size of the notches of the surface of a reflector should be smaller than the wavelength of the wave. The wavelength of light is very small.
- So the size of the notches of the reflector should be very small, i.e., the reflector should be smooth. On the other hand, the wavelength of sound is large. So regular reflection of sound takes place even from a rough reflector.
Question 4. A sound wave travels from air to water. The angle of incidence at the surface of separation between air and water is i1 and the angle of refraction is l2. Snell’s law is applicable in this case. Determine which one of i1 and i2 is greater.
Answer:
According to Snell’s law, \(\frac{\sin i_1}{\sin i_2}\) = constant.
The value of this constant is equal to the ratio of the velocities of the wave in the two media.
i.e., \(\frac{\sin i_1}{\sin i_2}=\frac{V_1}{V_2}\)
Here, the first medium is air and the second medium is water. The velocity of sound in water is greater than that in air (i.e., V2>V1).
So, \(\sin i_2>\sin i_1 \quad \text { or, } \quad i_2>i_1\)
∴ i1 is smaller than i2.
Question 5. A balloon full of carbon dioxide and another frill of water behave as lenses for the refraction of sound. What type of lens will they resemble?
Answer:
The velocity of sound in carbon dioxide is less than that in air, i.e., when sound enters carbon dioxide from air, its velocity decreases. So in the case of refraction of sound, carbon dioxide is a denser medium relative to air.
- Hence, a balloon full of carbon dioxide, when placed in the air, behaves like a convex lens. In the case of a convex lens of glass, light rays incident on it become convergent. Similarly, sound rays passing through a balloon full of carbon dioxide become convergent.
- On the other hand, when sound enters water from the air, its velocity increases, because the velocity of sound in air is 330 m · s-1, but that in water is about 1500 m · s-1. So, in case of refraction of sound, water is a rarer medium than air.
- Hence, a balloon full of water behaves in an opposite manner. So in this case, since the sound wave moves to a rarer medium from a denser medium, the waves will diverge. Therefore, the balloon containing water will behave like a concave lens.
Question 6. In a progressive wave, the maximum velocity and maximum acceleration of a particle in the medium are v0 and a0, respectively. Find the amplitude and the frequency of the waves.
Answer:
Let the equation of a progressive wave be y = \(A \sin \omega\left(t-\frac{x}{V}\right)\)
So, particle velocity v = \(\frac{\partial y}{\partial t}=\omega A \cos \omega\left(t-\frac{x}{V}\right)\),
i.e., v0 = ωA
Again, acceleration, a = \(\frac{\partial v}{\partial t}=-\omega^2 A \sin \omega\left(t-\frac{x}{V}\right),\)
i.e., a0 = ω²A
So, amplitude, A = \(\frac{(\omega A)^2}{\omega^2 A}=\frac{v_0^2}{a_0}\)
And frequency, n = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \cdot \frac{\omega^2 A}{\omega A}=\frac{a_0}{2 \pi v_0}\).
Question 7. A plane wave of sound traveling in the air is incident upon a plane’s water surface. The angle of incidence is 60. Assuming Snell’s law to be valid for sound waves, it will be refracted into the water away from the normal. State whether the statement is true or false and explain.
Answer:
The velocity of sound in water is about 4 times that in air. So for sound waves, air is a denser medium and water is a rarer medium. In this case, the sound wave is incident on a rarer medium (water) from the denser medium (air).
- For sound waves, the critical angle in air with respect to water is about \(\sin ^{-1}\left(\frac{1}{4}\right)=14.5^{\circ}\).
- So if the angle of incidence is 60°, sound waves will not be refracted in water, rather there will be a total internal reflection from the surface of the water. So the statement is false.
Question 8. Prove that the equation \(y=4 \sin \frac{2 \pi}{15}(60 t-x) \mathrm{cm}\) represents a progressive wave.
Answer:
Here the velocity of the wave is 60 cm · s-1.
If y1 is the displacement of a particle at the point (x + 60) at time (t+1), then
⇒ \(y_1=4 \sin \frac{2 \pi}{15}[60(t+1)-(x+60)]\)
= \(4 \sin \frac{2 \pi}{15}(60 t-x)=y\)
So, the motion of a particle at a distance x cm at time t is identical to the motion of the particle at a distance (x + 60) cm at time (t + 1). In this case, the disturbance in the medium moves 60 cm in 1 s. So, the given equation is an equation of a progressive wave.
Question 9. If the wavelength of an X-ray is 3Å, what is the frequency of the wave? [1A = 10-8 cm]
Answer:
Velocity of X-ray = velocity of light
= c = 3 x 1010 cm · s-1
Wavelength, λ = 3Å = 3 x 10-8 cm
∴ Frequency, n = \(\frac{c}{\lambda}=\frac{3 \times 10^{10}}{3 \times 10^{-8}}=10^{18} \mathrm{~Hz} .\)
Question 10. Which property of waves proves that sound wave is longitudinal?
Answer:
The polarisation property of wave proves it. It is not possible for any longitudinal wave to exhibit the phenomenon of polarisation, as in longitudinal waves, the vibrating particles always move along the direction of propagation of the wave. Since sound waves do not exhibit the phenomenon of polarisation, they are longitudinal.
Question 11. Which property of waves shows that light wave is transverse?
Answer:
The polarisation property of wave proves it. Any transverse wave exhibits the phenomenon of polarisation, as the vibrating particles always move perpendicular to the direction of propagation of the wave. Since light waves exhibit the phenomenon of polarisation, they are transverse.
Question 12. If the amplitude of a progressive wave at a distance r from a point source is A, what will be the amplitude at a distance 2r?
Answer:
Intensity of a wave ∝ (amplitude)²
Again, the intensity is inversely proportional to the square of the distance from a point source,
i.e., \(\text { intensity } \propto \frac{1}{(\text { distance })^2}\)
So, \(\text { amplitude } \propto \frac{1}{\text { distance }}\)
Hence, the amplitude will be half, i.e., \(\frac{A}{2}\) at a distance 2r,
Question 13. The equation of a wave la given by y = \(A \cos ^2\left(2 \pi n t-2 \pi \frac{x}{\lambda}\right)\), Determine the amplitude, the frequency, and the wavelength of the wave.
Answer:
y = \(A \cos ^2\left(2 \pi n t-2 \pi \frac{x}{\lambda}\right)=\frac{A}{2} \cdot 2 \cos ^2\left(2 \pi n t-2 \pi \frac{x}{\lambda}\right)\)
= \(\frac{A}{2}\left[1+\cos \left(4 \pi n t-4 \pi \frac{x}{\lambda}\right)\right] \)
= \(\frac{A}{2}+\frac{A}{2} \cos \left(4 \pi n t-4 \pi \frac{x}{\lambda}\right)\)
Here, the second term, i.e., the term containing the cosine function represents the wave. Comparing this term with the general equation y = A’ cos (ωt- kx), we have,
Amplitude, A’ = \(\frac{A}{2}\)
∴ \(\omega=4 \pi n, \text { i.e., frequency }=\frac{\omega}{2 \pi}=2 n,\)
k = \(\frac{4 \pi}{\lambda}, \text { i.e., wavelength }=\frac{2 \pi}{k}=\frac{2 \pi}{\frac{4 \pi}{\lambda}}=\frac{\lambda}{2}\).
Question 14. The equation of a progressive wave Is given by y = \(A \sin 2 \pi\left(p t-\frac{x}{5}\right)\). What Is the ratio of the maximum particle velocity of the medium to the wave velocity?
Answer:
y = \(A \sin 2 \pi\left(p t-\frac{x}{5}\right)=A \sin \left(2 \pi p t-\frac{2 \pi}{5} x\right)\)
Comparing the given equation with the general equation y = Asin(ωt- kx) we have,
∴ \(\omega=2 \pi p \text { and } k=\frac{2 \pi}{5}\)
∴ Wave velocity, V = \(\frac{\omega}{k}=\frac{2 \pi p}{\frac{2 \pi}{5}}=5 p\)
Again, particle velocity,
v = \(\frac{\partial y}{\partial t}=2 \pi p A \cos \left(2 \pi p t-\frac{2 \pi}{5} t\right)\)
Maximum particle velocity, v0 = 2πpA
So, the ratio of the maximum panicle velocity to the wave velocity = \(\frac{v_0}{V}=\frac{2 \pi p A}{5 p}=\frac{2}{5} \pi \lambda\)
Question 15. Even if the ratio \(\frac{p}{\rho}\) for helium and oxygen are kept equal, why la the velocity of sound not equal hi the two gases?
Answer:
The velocity of sound, c = \(\sqrt{\frac{\gamma p}{\rho}}\), Though the value of the ratio \(\frac{p}{\rho}\) for the two gases are equal, the velocity of sound c differs due to the different values of γ. Oxygen Is a diatomic gas and helium Is a monatomic gas. The values of γ for these two gases are \(\frac{7}{5}\) and \(\frac{5}{3}\) respectively. So, the velocity of sound In them is not equal.
Question 16. Why Is the velocity of sound In the air higher In the rainy season than in winter?
Answer:
Both the temperature and the humidity of the air are higher in the rainy season than In winter. We know, the velocity of sound in air is given by,
c = \(\sqrt{\frac{\gamma p}{\rho}}=\sqrt{\frac{\gamma R T}{M}}\)
Since c ∝ √T, the velocity of sound increases with the increase in temperature of the air.
Again with the increase in humidity, the amount of water vapour in air also increases. The density of water vapor Is less than the density of air. So the density ρ Is less in the rainy season and the velocity of sound increases.
Hence, the velocity of sound in air is higher in the rainy season In comparison to that in winter.
Question 17. At what temperature the velocity of sound In air Is twice its value at 0°C?
Answer:
The velocity of sound (c) in the air is proportional to the square root of its absolute temp (T), i.e., c α √T.
So the velocity of sound is doubled if the absolute temperature becomes four times. Since the initial temperature is 0°C or 273 K,
The required final temperature = 4x 273 K = 1092 K
= (1092 – 273)°C = 819°C.
Question 18. The velocity of sound is generally greater in solids than in gases why?
Answer:
The velocity of sound in a solid is given by, \(V_s=\sqrt{\frac{Y}{\rho}} ;\)
Y = Young’s modulus of the solid, ρ = density of the solid
The velocity of sound in a gas is given by, \(V_g=\sqrt{\frac{k}{\rho^{\prime}}}\)
k = bulk modulus of the gas, ρ’ = density of the gas.
Although the density of a solid medium is more than that of a gaseous medium, the modulus of elasticity of a solid (Y) is many times greater than that of a gas (k).
Hence, \(\frac{Y}{\rho}>\frac{k}{\rho^{\prime}}\)
Therefore, Vs > Vg, i.e., the velocity of sound is greater in solids than in gases.
Question 19. How will the velocity of sound be affected at high altitudes?
Answer:
The temperature decreases with increasing altitude. Furthermore, the humidity of air is generally less in hill areas. So, the velocity of sound will decrease due to a decrease in both temperature and humidity.
Question 20. The speed of sound in water and in air are 1500 m • s-1 and 300 m • s-1 respectively. A source of sound of frequency 600 Hz is placed inside water. If the emitted sound enters air from the source, what will be its frequency?
Answer:
The frequency of the emitted sound will remain the same, i.e., 600 Hz, because the frequency of a wave does not change due to refraction.
Question 21. Determine the relationship between the rms velocity. f of the molecule of a gas and the velocity of sound in the gas.
Answer:
From the kinetic theory of gases, we have,
rms velocity of the gas molecules = \(\sqrt{\frac{3 p}{\rho}}\)
Again from the calculations of Newton and Laplace we have, the velocity of sound in a gaseous medium = \(\sqrt{\frac{\gamma p}{\rho}}\)
[γ = ratio of the two specific heats of the gas]
So, the ratio of the rms velocity and the velocity of sound in a gaseous medium = \(\sqrt{\frac{3}{\gamma}}\)
Question 22. If the velocity of sound in oxygen at STP is v, what will be the velocity of sound in helium under the same conditions?
Answer:
Velocity of sound, c = \(\sqrt{\frac{\gamma p}{\rho}}\),
i.e., if the pressure remains the same, for two different gases, \(\frac{c_1}{c_2}=\sqrt{\frac{\gamma_1}{\gamma_2}} \cdot \sqrt{\frac{\rho_2}{\rho_1}}\)
The ratio of the densities of oxygen and helium = \(\frac{16}{2}\) = 8
Again, for oxygen \(\gamma_1=\frac{7}{5}\) (oxygen is diatomic)
And for helium \(\gamma_2=\frac{5}{3}\) (helium is monatomic)
So, \(\frac{\gamma_1}{\gamma_2}=\frac{7 / 5}{5 / 3}=\frac{21}{25}\)
∴ \(\frac{c_1}{c_2}=\sqrt{\frac{21}{25}} \times \sqrt{\frac{1}{8}}\)
As \(c_1=v, c_2=v \sqrt{\frac{25 \times 8}{21}}=3.1 v\).