WBCHSE Class 11 Physics Superposition Of Waves Question and Answers

Superposition Of Waves Long Answer Type Questions

Question 1. When you sing in the shower why is the sound heard so good?
Answer:

It is because of superposition. The sound waves you produce in the air bounce off the walls and interfere with each other in a way that produces a sweet sound.

Question 2. Which from an of a string the tone clamped will at be both absent ends, in the when note it is emitted strode at a distance of one-third of its length from one end?
Answer:

The stretched string AB is struck at the point C, where AC = \(\frac{1}{3}\)AB. As a result, an antinode will be formed at C; a node can never be formed there. So the modes of vibration, as shown in the figure, will be absent. This means that the absent ones would correspond to the 3rd, 6th 9th,… harmonics.

Question 3. What would be the change in the fundamental frequency of a stringed instrument if

1. The length of the string is doubled,
2. The tension is doubled,
3. The diameter is doubled?

Answer:

1. The frequency is inversely proportional to the length of the string. So, if the length is doubled, the fundamental frequency would be reduced to half.
2. Frequency is proportional to the square root of the tension. So, if the tension is doubled, the fundamental frequency would be √2. or 1.414 times of its initial value.
3. If the diameter is doubled, the area of the cross-section becomes 22 or 4 times the initial value. If the materials of the string is the same, the mass per unit length also becomes 4 times its original value. The frequency is inversely proportional to the square root of the mass per unit length. So, in this case, the frequency would be \(\frac{1}{\sqrt{4}} \text { or } \frac{1}{2}\) times, i.e., half the initial value.

Question 4. Compare the fundamental frequencies of the tones emitted by two pipes of equal length, one open at both ends and the other closed at one end.
Answer:

Let V be the velocity of sound in air and l, be the length of each of the pipes.

∴ The frequency of the fundamental tone emitted by the pipe open at both ends, \(n_1=\frac{V}{2 l}\)

and the frequency of the fundamental tone emitted by the pipe closed at one end, \( n_2=\frac {V}{4 l}.\)

∴ \(\frac{n_1}{n_2}=\frac{\frac{V}{2 l}}{\frac{V}{4 l}}=2 \quad \text { or, } n_1=2 n_2\)

So, the fundamental frequency for a pipe open at both ends is twice that of a pipe of the same length but closed at one end.

Example 5. A closed pipe and an open pipe emit fundamental tones of the same frequency. Find out the ratio of their lengths.
Answer:

Let the length of the closed pipe and the open pipe be l₁ and l₂, respectively. V is the velocity of sound in air.

For the closed pipe, the fundamental frequency is \(n_1=\frac{V}{4 l_1}\); for the open pipe, it is \(n_2=\frac{V}{2 l_2}\)

In the problem, \(n_1=n_2 \text { or, } \frac{V}{4 l_1}=\frac{V}{2 l_2} \text { or, } \frac{l_1}{l_2}=\frac{1}{2}\)

Therefore, the ratio of their lengths is  1: 2

Question 6. Would the frequencies of the tones emitted from a closed or an open pipe change If the temperature of the air column in the pipe increases?
Answer:

The frequencies of the tones are proportional to the velocity of sound in air, which increases with the increase of temperature. So, as the temperature increases, the frequencies of the tones emitted from the air column in the pipe would also increase.

Question 7. How would the fundamental frequency emitted from an organ pipe, open at both ends, change, If

1. An open end is suddenly closed,
2. The length of the pipe is increased,
3. The diameter of the pipe is increased?
4. What would happen if the air is blown heavily through an open end?

Answer:

1. If an open end is suddenly closed, the pipe open at both ends becomes a closed pipe of equal length. In this case, the fundamental frequency would be halved.
2. The frequency of the emitted fundamental tone is inversely proportional to the length of the pipe. So, the frequency would decrease with the increase in the length of the pipe.
3. The end error of a pipe increases with the increase in diameter. As a result, the effective length of the pipe increases. The fundamental frequency is inversely proportional to the length of the pipe; so the frequency would decrease due to this increase in effective length.
4. A heavier blow would increase the loudness of the emitted tones. Moreover, the hither (Le, 2nd, 3rd, 4th,…) harmonics would be. formed more easily.

Question 8. Why is the musical sound emitted from an open pipe more pleasant than that emitted from a closed pipe?
Answer:

A dosed pipe can emit the fundamental rone and its odd harmonics. But an open pipe can emit both the odd and the even harmonics as well as the fundamental tone. So, a note emitted from, an open pipe contains a larger number of constituent tones.

The quality of a note depends on the number of overtones present in it. Store the number of overtones present in a note, the more pleasant it sounds. This results in a higher quality of sound emitted from an open pipe.

Question 9. A stretched vibrating string is touched at a distance of 1/3 rd of its length from one end. What would happen to the musical sound emitted?
Answer:

The point of touch would be a nodal point. So only the 3rd, 6th. 9th…. harmonics, which have a node at 1/3 would be emitted. No other harmonics have a node at 1/3. So, all of them would be suppressed.

Question 10. Two tuning forks, vibrating simultaneously, produce 6 beats per second. The first of them has a frequency of 312 Hz. Some amount of wax is added to one arm of the second tuning fork; the number of beats per second reduces to 3. Find out the frequency of this second tuning fork. Is it possible to increase the beat frequency to 6 per second by adding some more wax to die second tuning fork?
Answer:

Initially, the number of beats per second is 6. So the frequency of the 2nd fork is either 312-*-6 = 318 Hz or 312 – 6 = 306 Hz. After putting the wax, the beat frequency is 3 per second.

• So the changed frequency of the 2nd fork is, either 312÷3=315Hz or 312 – 3 = 309 Hz. But the addition of wax reduces the frequency of the 2nd fork; so it cannot change from 306 Hz to 309 Hz; the actual change is from 318 Hz to 315 Hz. This means that the actual frequency of the 2nd tuning fork is 318 Hz.
• By adding a sufficient amount of wax on the 2nd fork, its frequency can be reduced from 318 Hz to 306 Hz. Then again the beat frequency would be 312 – 306 = 6 per second.

Question 11. Two Identical wires of equal lengths are stretched In such a way that their simultaneous vibrations produce 6 beats per second. The tension In one of the wires is changed slightly and it is observed that the beat frequency remains the same. How is it possible?
Answer:

The frequency of a stretched string is proportional to the square root of the tension. If the fundamental frequency in the 1st wire is more than that in the 2nd wire, the tension in the 1st ware (T₁) > the tension in the 2nd wire (T₂).

Then, n₁ = n₂ ÷ 6, where n₁ and n₂ are the fundamental frequencies of the two wires respectively. Now, the tension T₂ is increased gradually until the fundamental frequency of the 2nd wire changes to n₂‘ = n₂ ÷ 12. Under these circumstances, the frequency difference becomes, \(n_2^{\prime}-n_1=\left(n_2+12\right)-\left(n_2+6\right)=6\)

This means that 6 beats would be heard again per second.

Question 12. Three tuning forks of frequencies n-x,n, and n + x are vibrated simultaneously. If the amplitudes of vibration are equal, show that the forks would form beats.
Answer:

Let A be the amplitude of the vibration of each tuning fork.

Then, the equations of the waves are y₁ = Asin2π(n- x)t, y₂ = Asin2πnt if and y₃ = Asin2π(n + x)t.

∴ The equation of the resultant wave is

y = \(y_1+y_2+y_3\)

= \(A[\sin 2 \pi(n-x) t+\sin 2 \pi n t+\sin 2 \pi(n+x) t]\)

= \(A[\{\sin 2 \pi(n+x) t+\sin 2 \pi(n-x) t\}+\sin 2 \pi n t]\)

= \(A[2 \sin 2 \pi n t \cos 2 \pi x t+\sin 2 \pi n t]\)

= \(A(1+2 \cos 2 \pi x t) \sin 2 \pi n t\)

∴ The amplitude of the resultant wave = A(1 + 2 cos2πxt)

This amplitude clearly depends on time. As a result, the intensity of the emitted sound would also increase and decrease periodically with time [intensity ∝ (amplitude)²]. Thus beats are produced. So, beats can be produced not only due to the superposition of two waves but also due to the superposition of more than two waves.

Question 13. Three sources emitting sound waves of the same amplitude have frequencies 400 Hz, 401 Hz, and 402 Hz, respectively. Find out the number of beats heard per second.
Answer:

⇒ \(y_1=A \sin (2 \pi \cdot 400 t) ; y_2=A \sin (2 \pi \cdot 401 t)\)

⇒ \(y_3=A \sin (2 \pi \cdot 402 t)\)

∴ y = \(y_1+y_2+y_3\)

= \(A[\sin (2 \pi \cdot 400 t)+\sin (2 \pi \cdot 401 t)+\sin (2 \pi \cdot 402 t)\} \)

= \(A\left[\sin (2 \pi \cdot 401 t)+2 \sin \left\{2 \pi \frac{402+400}{2} t\right\}\right.\)

. \(\left.\cdot \cos \left\{2 \pi \frac{402-400}{2} t\right\}\right]\)

= \(A[\sin (2 \pi \cdot 401 t)+2 \sin (2 \pi \cdot 401 t) \cos 2 \pi t]\)

= \(A(1+2 \cos 2 \pi t) \sin (2 \pi \cdot 401 t)\)

Here, the amplitude of the resultant wave, \(A^{\prime}=A(1+\cos 2 \pi t).\)

This is maximum when \(\cos 2 \pi t=1=\cos 2 n \pi[n=0,1,2,3, \cdots]\)

or, t = \(n=0,1,2, \cdots\)

Then, the time interval between two consecutive maxima is Is; so the beat frequency = 1 per second.

Question 14. A 100 cm long stretched string is struck at a point 25 cm from one of its ends. Which of the overtones would be absent in the emitted note?
Answer:

At the striking point, an antinode would be formed. So, the overtones having a node at the point of 25 cm would not be formed. As 25 cm = \(\frac{100 \mathrm{~cm}}{4}=\frac{1}{4}\)(l = length of the stretched string), the absent overtones would be the 4th, 8th, 12th, …. harmonics.

Question 15. In a particular vibrating mode of a stretched string of length l clamped at both ends, n nodes are formed. What is the wavelength of the stationary wave formed in this mode?
Answer:

Both endpoints are nodes. So, the number of loops formed between the n nodes = n-1. Every loop has a length = \(\frac{\lambda}{2} \text {. So, } \frac{l}{n-1}=\frac{\lambda}{2} \text { or, } \lambda=\frac{2 l}{n-1} \text {. }\)

Question 16. The equation of two progressive waves superposing on a string is y₁ = Asin[k(x- ct)] and y₂ = Asin[k(x+ ct)]. What is the distance between two consecutive nodes?
Answer:

Here, \(y_1 \text { or } y_2=A \sin [k(x \mp c t)]\)

The general form of the two waves, y = \(A \sin \frac{2 \pi}{\lambda}(x \mp V t)\)

So, k = \(\frac{2 \pi}{\lambda} \text { or, } \lambda=\frac{2 \pi}{k}\)

∴ The distance between two consecutive nodes = \(\frac{\lambda}{2}=\frac{\pi}{k}\).

Question 17. Two waves represented as y₁ = A₁ sin ωt and y₂ = A₂ cosωt superpose at a point in space. Find out the amplitude of the resultant wave at that point.
Answer:

⇒ \(y_1=A_1 \sin \omega t ; y_2=A_2 \cos \omega t=A_2 \sin \left(\omega t+\frac{\pi}{2}\right) \text {. }\)

So, the phase difference between the two waves = \(\frac{\pi}{2}\).

Then, the amplitude of the resultant wave is A = \(\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \frac{\pi}{2}}=\sqrt{A_1^2+A_2^2} .\)

Question 18. The length of a stretched string between two rigid supports is 40 cm. What would be the maximum length of a stationary wave that can be formed in the string?
Answer:

Let the maximum wavelength that can be formed for the given string be λ. For the stationary wave of maximum length, only two nodes are present and they are at the two ends of the string. There is only one antinode between them. So the string vibrates in a single loop. Then the length of this loop = \(\frac{\lambda}{2}\) = 40 cm; so, λ = 80 cm.

Question 19. The equation of a transverse progressive wave is y = 0.02sin(x+40t) m. Find out the tension In a wire of linear density 10^-4 kg · m^-1, if the wave travels along it.
Answer:

Comparing the given equation with the standard form y = A sin(ωt+ kx), we have k = 1 and ω = 40.

So, the wave velocity, V = \(\frac{\omega}{k}=\frac{40}{1}=40 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

If T = tension and m = linear density of the wire,

V = \(\sqrt{\frac{T}{m}} \text { or, } T=V^2 m=(40)^2 \times 10^{-4}=0.16 \mathrm{~N}\)

Question 20. The superposition of two progressive waves produces a stationary wave represented as y = Acos(0.01x)sin(100t) m. What is the velocity I of the two-component waves?
Answer:

Two progressive waves represented as y₁ = A sin(ωt- kx) and y₂ = A sin(ωt+ kx) superpose to form a stationary wave y = y₁ + y₂ = 2A cos kx sin ωt.

Comparing the given equation with this standard form, we have k = 0.01 m^-1 and ω = 100 Hz. So the wave velocity,

V = \(\frac{\omega}{k}=\frac{100}{0.01}=10^4 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

Example 21. A uniform wire has length I and area of cross-section α and the density of its material is ρ. If the wire is stretched with a tension T, what would be the velocity of a transverse wave traveling along It?
Answer:

Volume of unit length of the wire = unit length x area of cross-section = 1 x α = α; then linear density = mass per unit length, m = ρα.

∴ Velocity of a transverse wave, V = \(\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\rho \alpha}}\)