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Class- XI-CBSE-Physics Waves Practice more on Waves Page - 1 www.embibe.com CBSE NCERT Solutions for Class 11 Physics Chapter 15 Back of Chapter Questions 15.1. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? Solution: Given: Mass of the string, = 2.50 kg Tension in the string, = 200 N Length of the string, = 20.0 m Then, mass per unit length (linear mass density) of the string, = = 2.50 20 = 0.125 kg m −1 The transverse wave speed () in the string is determined by the relation: =√ =√ 200 0.125 = √1600 = 40 m s −1 Hence, time taken by the disturbance to reach the other end, = = 20 40 = 0.5 s 15.2. A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s −1 ? ( = 9.8 m s −2 ) Solution: Given: Height of the tower, ℎ = 300 m Initial velocity of the stone, =0 Acceleration, = = 9.8 m s −2 Speed of sound in air = = 340 m s −1 Let 1 be the time taken by stone to reach the pond’s surface.
Transcript
Page 1: Class- XI-CBSE-Physics Waves...Class- XI-CBSE-Physics Waves Practice more on Waves Page - 1 CBSE NCERT Solutions for Class 11 Physics Chapter 15 Back of Chapter Questions 15.1. A string

Class- XI-CBSE-Physics Waves

Practice more on Waves Page - 1 www.embibe.com

CBSE NCERT Solutions for Class 11 Physics Chapter 15

Back of Chapter Questions

15.1. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched

string is 20.0 m. If the transverse jerk is struck at one end of the string, how long

does the disturbance take to reach the other end?

Solution:

Given:

Mass of the string, 𝑀 = 2.50 kg

Tension in the string, 𝑇 = 200 N

Length of the string, 𝑙 = 20.0 m

Then, mass per unit length (linear mass density) of the string, 𝜇 =𝑀

𝑙=

2.50

20=

0.125 kg m−1

The transverse wave speed (𝑣) in the string is determined by the relation:

𝑣 = √𝑇

𝜇

= √200

0.125= √1600 = 40 m s−1

Hence, time taken by the disturbance to reach the other end, 𝑡 =𝑙

𝑣=

20

40= 0.5 s

15.2. A stone dropped from the top of a tower of height 300 m high splashes into the

water of a pond near the base of the tower. When is the splash heard at the top given

that the speed of sound in air is 340 m s−1? (𝑔 = 9.8 m s−2)

Solution:

Given:

Height of the tower, ℎ = 300 m

Initial velocity of the stone, 𝑢 = 0

Acceleration, 𝑎 = 𝑔 = 9.8 m s−2

Speed of sound in air = 𝑣 = 340 m s−1

Let 𝑡1 be the time taken by stone to reach the pond’s surface.

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Then we use, 𝑠 = 𝑢𝑡 +1

2𝑎𝑡2 to calculate 𝑡1 by substituting 𝑠 = ℎ, 𝑡 = 𝑡1, 𝑎 =

𝑔 and 𝑢 = 0

ℎ = 0 × 𝑡1 +1

2𝑔𝑡1

2

Therefore, 𝑡1 = √2ℎ

𝑔= √

2×300

9.8= 7.82 s

Also, let 𝑡2 be the time taken by sound to reach top of the tower of height 300 m.

Then, 𝑡2 =ℎ

𝑣=

300

340= 0.88 s

Hence, the total time after which the sound of splash is heard = 𝑡1 + 𝑡2 = 7.82 +

0.88 = 8.7 s

15.3. A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the

tension in the wire so that speed of a transverse wave on the wire equals the speed

of sound in dry air at 20 oC = 343 m s−1.

Solution:

Given:

Mass of the steel wire, 𝑀 = 2.10 kg

Speed of the transverse wave, 𝑣 = 343 ms−1

Length of the steel wire, 𝑙 = 12.0 m

Then, mass per unit length (linear mass density) of the steel wire, 𝜇 =𝑀

𝑙=

2.10

12.0=

0.175 kg m−1

The transverse wave speed (𝑣) in the steel wire is determined by the relation,

𝑣 = √𝑇

𝜇, where 𝑇 is the tension in the wire

Squaring on both the sides, we get

𝑣2 =𝑇

𝜇⇒ 𝑇 = 𝜇𝑣2

Hence, 𝑇 = 0.175 × 3432 = 20,588.575 ≈ 2.06 × 104 N

15.4. Use the formula 𝑣 = √𝛾𝑃

𝜌 to explain why the speed of sound in air

(a) is independent of pressure,

(b) increases with temperature,

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(c) increases with humidity.

Solution:

(a) Given: 𝑣 = √𝛾𝑃

𝜌 … (i)

Where, 𝜌 = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =𝑀𝑎𝑠𝑠

𝑉𝑜𝑙𝑢𝑚𝑒=

𝑀

𝑉

𝑀 = Molecular weight of gas

𝑉 = Volume of gas

So, equation (i) reduces to:

𝑣 = √𝛾𝑃𝑉

𝑀 … (iib

Using the ideal gas equation for 𝑛 = 1:

𝑃𝑉 = 𝑅𝑇

For constant 𝑇, 𝑃𝑉 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Since both 𝑀 and 𝛾 are constants, 𝑣 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Hence, the speed of sound in air is not dependent on pressure at a constant

temperature.

(b) Given: 𝑣 = √𝛾𝑃

𝜌 … (i)

Where, 𝜌 = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =𝑀𝑎𝑠𝑠

𝑉𝑜𝑙𝑢𝑚𝑒=

𝑀

𝑉 ⇒𝜌𝑉 = 𝑀 … (ii)

𝑀 = Molecular weight of gas

𝑉 = Volume of gas

So, equation (i) reduces to:

𝑣 = √𝛾𝑃𝑉

𝑀

Using the ideal gas equation for 𝑛 = 1:

𝑃𝑉 = 𝑅𝑇

𝑃 =𝑅𝑇

𝑉 … (iii)

Putting (iii) in (i), we get

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𝑣 = √𝛾𝑅𝑇

𝜌𝑉= √

𝛾𝑅𝑇

𝑀 … (iv)

Since 𝑀, 𝛾 and 𝑅 are all constants, 𝑣 ∝ √𝑇.

Hence, the speed of sound in air increases with increase in temperature.

(c) Let 𝑣𝑑 and 𝑣𝑚 be the speeds of sound in dry air and moist air respectively.

Let 𝜌𝑑 and 𝜌𝑚 be the densities of dry air and moist air respectively.

We know the formula:

𝑣 = √𝛾𝑃

𝜌

Therefore, the speed of sound in moist air is:

𝑣𝑚 = √𝛾𝑃

𝜌𝑚 …… (i)

And the speed of sound in dry air is:

𝑣𝑑 = √𝛾𝑃

𝜌𝑑 …… (ii)

On dividing equations (i) and (ii) we get:

𝑣𝑚

𝑣𝑑= √

𝛾𝑃

𝜌𝑚×

𝜌𝑑

𝛾𝑃 ……. (ii)

However, the presence of water vapour reduces the density of air, i.e.,

𝜌𝑑 < 𝜌𝑚

∴ 𝑣𝑚 > 𝑣𝑑

Hence, the speed of sound in moist air is greater than it is in dry air. Thus,

in a gaseous medium, the speed of sound increases with humidity.

15.5. You have learnt that a travelling wave in one dimension is represented by a function

𝑦 = 𝑓(𝑥, 𝑡) where 𝑥 and 𝑡 must appear in the combination 𝑥 − 𝑣𝑡 or 𝑥 + 𝑣𝑡, i.e.,

𝑦 = 𝑓(𝑥 ± 𝑣𝑡). Is the converse true? Examine if the following functions for 𝑦 can

possibly represent a travelling wave:

(a) (𝑥 − 𝑣𝑡)2

(b) 𝑙𝑜𝑔 [𝑥+𝑣𝑡

𝑥𝑜]

(c) 1

(𝑥+𝑣𝑡)

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Solution:

The converse is not true. An obvious requirement for an acceptable function for a

travelling wave is that it should be finite everywhere and at all times.

(a) No, the function does not represent a travelling wave as it is not having

finite value every where and at all times.

Example: For 𝑥 = ∞ and 𝑡 = ∞, function has infinite value.

(b) By substituting 𝑥 = 0 and 𝑡 = 0, the given function tends to infinity. So,

the function cannot possibly represent a travelling wave.

(c) By substituting 𝑥 = 0 and 𝑡 = 0, the given function tends to infinity. So,

the function cannot possibly represent a travelling wave.

15.6. A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a

water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted

sound? Speed of sound in air is 340 ms−1 and in water 1486 ms−1.

Solution:

(a) Given:

Frequency of the ultrasonic sound, 𝜈 = 1000 kHz = 106 Hz

Speed of sound in air, 𝑣𝑎 = 340 ms−1

The wavelength (𝜆𝑟) of the reflected sound is given by:

𝜆𝑟 =𝑣𝑎

𝜈=

340

106= 3.4 × 10−4 m

(b) Given:

Frequency of the ultrasonic sound, 𝜈 = 1000 kHz = 106 Hz

Speed of sound in water, 𝑣𝑤 = 1486 ms−1

The wavelength (𝜆𝑡) of the transmitted sound is given by:

𝜆𝑡 =𝑣𝑤

𝜈=

1486

106= 1.49 × 10−3 m

15.7. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the

wavelength of sound in the tissue in which the speed of sound is 1.7 km s−1? The

operating frequency of the scanner is 4.2 MHz.

Solution:

Given: Speed of sound in the tissue, 𝑣 = 1.7 km s−1 = 1.7 × 103 m s−1

Operating frequency of an ultrasonic scanner, 𝜈 = 4.2 MHz = 4.2 × 106Hz

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The wavelength of sound in the tissue= 𝜆 =𝑣

𝜈=

1.7×103

4.2×106= 4.1 × 10−4 m

15.8. A transverse harmonic wave on a string is described by

𝑦(𝑥, 𝑡) = 3.0 sin (36𝑡 + 0.018𝑥 +𝜋

4)

where 𝑥 and 𝑦 are in cm and 𝑡 in s. The positive direction of 𝑥 is from left to right.

(a) Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

(b) What are its amplitude and frequency?

(c) What is the initial phase at the origin?

(d) What is the least distance between two successive crests in the wave?

Solution:

The equation of a progressive wave travelling from right to left is given by the

displacement equation:

𝑦(𝑥, 𝑡) = 𝑎 sin (𝜔𝑡 + 𝑘𝑥 + 𝜙) … (i)

The given equation is:

𝑦(𝑥, 𝑡) = 3.0 sin (36𝑡 + 0.018𝑥 +𝜋

4) … (ii)

By comparing equations (i) and (ii), we observe that equation (ii) represents a

travelling wave, propagating from right to left.

So, 𝑎 = 3.0 cm, 𝜔 = 36 rad s−1, 𝑘 = 0.018 cm−1 and 𝜙 =𝜋

4 .

We know that,

𝜈 =𝜔

2𝜋 and 𝜆 =

2𝜋

𝑘

But, 𝑣 = 𝜈𝜆

Therefore, 𝑣 = (𝜔

2𝜋) × (

2𝜋

𝑘) =

𝜔

𝑘=

36

0.018= 2000 cm s−1 = 20 m s−1

(a) Thus, the wave is a travelling wave with speed equal to 20 m s−1 and

travelling from right to left.

(b) Amplitude of wave, 𝑎 = 3.0 cm = 0.03 m

Frequency of wave, 𝜈 =𝜔

2𝜋=

36

2𝜋= 5.7 Hz

(c) On comparing equations (i) and (ii), we get initial phase angle, 𝜙 =𝜋

4

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(d) The wavelength is the distance between two consecutive troughs or crests.

So, 𝜆 =2𝜋

𝑘=

2𝜋

0.018= 348.89 cm = 349 cm = 3.5 m

15.9. For the wave described in Exercise 15.8, plot the displacement (𝑦) versus (𝑡)

graphs for 𝑥 = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects

does the oscillatory motion in travelling wave differ from one point to another:

amplitude, frequency or phase?

Solution:

For the given wave,

𝑦(𝑥, 𝑡) = 3.0 sin (36𝑡 + 0.018𝑥 +𝜋

4)

For 𝑥 = 0, 𝑦(0, 𝑡) = 3.0 sin (36𝑡 +𝜋

4) … (i)

⇒ Amplitude, 𝑎 = 3 cm

⇒ Initial phase angle, 𝜙 =𝜋

4

⇒ Angular frequency, 𝜔 = 36 rad s−1

⇒ Time period, 𝑇 =2𝜋

𝜔=

2𝜋

36=

𝜋

18 s

For 𝑥 = 2, 𝑦(2, 𝑡) = 3.0 sin (36𝑡 + 0.036 +𝜋

4)

For 𝑥 = 4, 𝑦(4, 𝑡) = 3.0 sin (36𝑡 + 0.072 +𝜋

4)

For 𝑥 = 0, substituting various values of ‘𝑡’ in equation (i), we get

𝑡(s) 0 𝑇

8

2𝑇

8

3𝑇

8

4𝑇

8

5𝑇

8

6𝑇

8

7𝑇

8

𝑇

𝑦(cm) 3√2

2

3 3√2

2

0 −3√2

2

−3 −3√2

2

0 3√2

2

Which can be plotted as below:

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Similarly, the graphs for 𝑥 = 2 and 𝑥 = 4 can be plotted.

These graphs have same amplitude and frequency, but they differ in phase given by

𝜋

4,

𝜋

4+ 0.036 and

𝜋

4+ 0.072.

15.10. For the travelling harmonic wave

𝑦(𝑥, 𝑡) = 2.0 cos 2𝜋(10𝑡 − 0.0080𝑥 + 0.35)

where 𝑥 and 𝑦 are in cm and 𝑡 in s. Calculate the phase difference between

oscillatory motion of two points separated by a distance of

(a) 4 𝑚,

(b) 0.5 m,

(c) 𝜆

2,

(d) 3𝜆

4

Solution:

The given equation can be rewritten as follows:

𝑦(𝑥, 𝑡) = 2.0 cos (20𝜋𝑡 − 0.016𝜋𝑥 + 0.70𝜋)

Where,

Amplitude, 𝑎 = 2.0 cm

Angular frequency, 𝜔 = 20𝜋 rad s−1

Angular wave number (𝑘) = 0.016𝜋 cm−1

In an oscillatory motion, the phase difference between two points separated by a

distance of ∆𝑥 is given by,

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∆𝜙 =2𝜋

𝜆∆𝑥

∆𝜙 = 𝑘∆𝑥

For ∆𝑥 = 4 m = 400 cm

∆𝜙 = 0.016𝜋 × 400 = 6.4𝜋 rad

For ∆𝑥 = 0.5 m = 50 cm

∆𝜙 = 0.016𝜋 × 50 = 0.8𝜋 rad

(a) For ∆𝑥 =𝜆

2

∆𝜙 =2𝜋

𝜆∆𝑥

∆𝜙 =2𝜋

𝜆×

𝜆

2= 𝜋 rad

For ∆𝑥 =3𝜆

4

∆𝜙 =2𝜋

𝜆∆𝑥

∆𝜙 =2𝜋

𝜆×

3𝜆

4=

3𝜋

2 rad

15.11. The transverse displacement of a string (clamped at its both ends) is given by

𝑦(𝑥, 𝑡) = 0.06 sin (2𝜋𝑥

3) cos(120𝜋𝑡)

where 𝑥 and 𝑦 are in m and 𝑡 in s. The length of the string is 1.5 m and its mass is

3.0 × 10−2kg.

Answer the following:

(a) Does the function represent a travelling wave or a stationary wave?

(b) Interpret the wave as a superposition of two waves travelling in opposite

directions. What is the wavelength, frequency, and speed of each wave?

(c) Determine the tension in the string.

Solution:

(a) The general equation representing a stationary wave is given by:

𝑦(𝑥, 𝑡) = 2𝑎 sin (𝑘𝑥)cos (𝑤𝑡) … (i)

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Given equation:

𝑦(𝑥, 𝑡) = 0.06 sin (2𝜋𝑥

3) cos(120𝜋𝑡) … (ii)

Equations (i) and (ii) are similar to each other.

So, the given function represents a stationary wave.

(b) A wave pulse travelling in positive direction of x-axis is given by,

𝑦1(𝑥, 𝑡) = 𝑎 sin (𝑘𝑥 − 𝜔𝑡)

Similarly, a wave pulse travelling in negative direction of x-axis is given

by,

𝑦2(𝑥, 𝑡) = 𝑎 sin (𝑘𝑥 + 𝜔𝑡)

The superposition of these two waves gives:

𝑦 = 𝑦1 + 𝑦2 = 𝑎 sin (𝑘𝑥 − 𝜔𝑡) + 𝑎 sin (𝑘𝑥 + 𝜔𝑡)

𝑦 = 2𝑎 sin (𝑘𝑥)cos (𝜔𝑡) {𝑆𝑖𝑛𝑐𝑒, sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵) = 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵}

𝑦 = 2𝑎 sin (2𝜋

𝜆𝑥)cos (2𝜋𝜈𝑡) …(iii)

The transverse displacement is given by:

𝑦(𝑥, 𝑡) = 0.06 sin (2𝜋𝑥

3) cos(120𝜋𝑡) … (ii)

Comparing equations (ii) and (iii), we get

2𝜋

𝜆=

2𝜋

3

Therefore, the wavelength is 3 m.

By comparing, we also get

120𝜋 = 2𝜋𝜈

So, the frequency of the wave is 60 Hz.

Speed of the wave, 𝑣 = 𝜈𝜆

𝑣 = 𝜈𝜆 = 60 × 3 = 180 m s−1

(c) Given:

Mass of the string, 𝑀 = 3.0 × 10−2 kg

Speed of the transverse wave, 𝑣 = 180 ms−1

Length of the string, 𝑙 = 1.5 m

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Then, mass per unit length (linear mass density) of the string, 𝜇 =𝑀

𝑙=

3.0×10−2

1.5= 2 × 10−2 kg m−1

The transverse wave speed (𝑣) in the string is determined by the relation,

𝑣 = √𝑇

𝜇, where 𝑇 is the tension in the string

Squaring on both the sides, we get

𝑣2 =𝑇

𝜇⇒ 𝑇 = 𝜇𝑣2

⇒ 𝑇 = 2 × 10−2 × 1802 = 648 N

15.12. (i) For the wave on a string described in Exercise 15.11, do all the points on

the string oscillate with the same (a) frequency, (b) phase, (c) amplitude?

Explain your answers.

(ii) What is the amplitude of a point 0.375 m away from one end?

Solution:

Given:

(i) The transverse displacement of the string is given as:

𝑦(𝑥, 𝑡) = 0.06 sin (2𝜋𝑥

3) cos(120𝜋𝑡)

(a) Yes. The frequency of oscillation of all points on the string is same. This is

because, frequency is represented by the time dependent harmonic function

cos(120𝜋𝑡) of the stationary wave. It is clear from this function that

frequency is independent of 𝑥.

(b) Yes. The phase of all points on the string is same. This is because, phase is

represented by the time dependent harmonic function cos(120𝜋𝑡) of the

stationary wave. It is clear from this function that phase is independent of

𝑥.

(c) No. All the points on the string have different amplitudes at different points.

This is because, the amplitude of a stationary wave is given as:

𝑎 = 0.06 sin (2𝜋𝑥

3)

Here, as 𝑎 depends on 𝑥, amplitude of all the points on the string is not

same.

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(ii) Now, amplitude at a point 0.375 m away from one end is given by,

𝑎 = 0.06 sin (2𝜋 × 0.375

3) = 0.06 sin (

𝜋

4) = 0.06 × 0.707 = 0.042 m

15.13. Given below are some functions of 𝑥 and 𝑡 to represent the displacement (transverse

or longitudinal) of an elastic wave. State which of these represent (i) a travelling

wave, (ii) a stationary wave or (iii) none at all:

(a) 𝑦 = 2 cos(3𝑥) sin (10𝑡)

(b) 𝑦 = 2√𝑥 − 𝑣𝑡

(c) 𝑦 = 3 sin (5𝑥 − 0.5𝑡) + 4 cos(5𝑥 − 0.5𝑡)

(d) 𝑦 = cos 𝑥 sin 𝑡 + cos 2𝑥 sin 2𝑡

Solution:

(a) The general equation representing a stationary wave is given by:

𝑦(𝑥, 𝑡) = 2𝑎 sin (𝑘𝑥)cos (𝑤𝑡)

Basically, equation of a stationary wave is product of harmonic functions of

𝑥 and 𝑡 separately. Hence, the given function represents a stationary wave.

(b) It does not represent any wave.

(c) It represents a travelling harmonic wave.

(d) Here, the equation is sum of two functions, each representing a stationary

wave. Hence, it represents superposition of two stationary waves.

15.14. A wire stretched between two rigid supports vibrates in its fundamental mode with

a frequency of 45 Hz. The mass of the wire is 3.5 × 10−2 kg and its linear mass

density is 4.0 × 10−2kg m−1. What is (a) the speed of a transverse wave on the

string, and (b) the tension in the string?

Solution:

Given:

Wire is vibrating with a frequency = 45 Hz

Mass of the wire, 𝑀 = 3.5 × 10−2 kg

Linear mass density, 𝜇 =𝑀

𝑙= 4.0 × 10−2kg m−1

So, the length of the wire, 𝑙 =𝑀

𝜇=

3.5×10−2

4.0×10−2 =7

8= 0.875 m

Standing waves formed on a string of length 𝑙 have wavelength given by,

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𝜆 =2𝑙

𝑛, where 𝑛 = Number of nodes in the wire

i.e., 𝑛 = 1, 2, 3, … 𝑒𝑡𝑐.

But given that the wire vibrates in its fundamental node, so 𝑛 = 1

Then, 𝜆 = 2𝑙 = 2 × 0.875 = 1.75 m

The speed of a transverse wave on the string is given by,

𝑣 = 𝜈𝜆 = 45 × 1.75 = 78.75 m s−1 ≈ 79 m s−1

The transverse wave speed (𝑣) in the string is determined by the relation,

𝑣 = √𝑇

𝜇, where 𝑇 is the tension in the string

Squaring on both the sides, we get

𝑣2 =𝑇

𝜇⇒ 𝑇 = 𝜇𝑣2 = 4.0 × 10−2 × (78.75)2 = 248.06 N ≈ 248 N

15.15. A metre-long tube open at one end, with a movable piston at the other end, shows

resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when

the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the

temperature of the experiment. The edge effects may be neglected.

Solution:

Given:

Frequency of the tuning fork, 𝜈 = 340 Hz

One end of the tube is open and the other end is closed by the piston, so it behaves

as a closed organ pipe, which produces only odd harmonics.

Therefore, the pipe is in resonance with the fundamental mode and the third

harmonic (Since, 79.3 ≈ 3 × 25.5)

Below is the figure for a fundamental mode formed due to pipe open at one end and

closed at other end.

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𝑙1 =𝜆

4, where length of the pipe, 𝑙1 = 25.5 cm = 0.255 m

Therefore, 𝜆 = 4𝑙1 = 4 × 0.255 = 1.02 m

Speed of sound in air = 𝜈𝜆 = 340 × 1.02 = 347 m s−1

15.16. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of

longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of

sound in steel?

Solution:

Given:

Length of the steel rod, 𝑙 = 100 cm = 1 m

The fundamental frequency of vibrations of the rod, 𝜈 = 2.53 kHz = 2.53 ×

103Hz

When the rod is clamped at its middle, then in its fundamental mode of vibration,

antinodes (A) are formed at its two ends, and a node (N) is formed at the middle,

as shown in the given figure

It is obvious that, 𝐿 =𝜆

4+

𝜆

4=

𝜆

2

𝜆 = 2𝐿 = 2 m

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We know that, 𝑣 = 𝜈𝜆 = 2.53 × 103 × 2 = 5.06 × 103 m s−1

15.17. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is

resonantly excited by a 430 Hz source? Will the same source be in resonance with

the pipe if both ends are open? (speed of sound in air is 340 m s−1).

Solution:

Given:

Length of the pipe, 𝑙 = 20 cm = 0.2 m

Frequency of the source = 𝑛𝑡ℎ normal mode of frequency, 𝜈𝑛 = 430 Hz

Speed of sound in air = 340 m s−1

For a pipe closed at one end:

𝜈𝑛 = (2𝑛 − 1)𝑣

4𝑙, for 𝑛 = 1, 2, 3, …

430 = (2𝑛 − 1)340

4 × 0.2

2𝑛 − 1 =430 × 4 × 0.2

340= 1.01

𝑛 ≈ 1

Thus, the first mode of frequency of vibration is resonantly excited by a 430 Hz

source. For a pipe open at both ends:

𝜈𝑛 =𝑛𝑣

2𝑙, for 𝑛 = 1, 2, 3, …

𝑛 =2𝑙𝜈𝑛

𝑣=

2 × 0.2 × 430

340= 0.506

As the number of the mode of vibration (𝑛) has to be an integer, the same frequency

source of 430 Hz does not produce a resonant vibration in the pipe, whose both

ends are open.

15.18. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce

beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat

frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what

is the frequency of B?

Solution:

Given:

Original frequency of string A, 𝜈𝐴 = 324 Hz

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Let the original frequency of string B = 𝜈𝐵

Decrease in the tension of a string decreases its frequency.

If the original frequency of B (𝜈𝐵) were greater than that of A (𝜈𝐴), a further

increase in 𝜈𝐵 should have resulted in an increase in the beat frequency.

But, the beat frequency is found to decrease from 6 Hz to 3 Hz.

This shows that 𝜈𝐵 < 𝜈𝐴.

Since 𝜈𝐴 − 𝜈𝐵 = 6 Hz, and 𝜈𝐴 = 324 Hz, we get

324 − 𝜈𝐵 = 6

⇒𝜈𝐵 = 318 Hz

Hence, the original frequency of string B is 318 Hz.

15.19. Explain why (or how):

(a) in a sound wave, a displacement node is a pressure antinode and vice versa,

(b) bats can ascertain distances, directions, nature, and sizes of the obstacles

without any “eyes”,

(c) a violin note and sitar note may have the same frequency, yet we can

distinguish between the two notes,

(d) solids can support both longitudinal and transverse waves, but only

longitudinal waves can propagate in gases, and

(e) the shape of a pulse gets distorted during propagation in a dispersive

medium.

Solution:

(a) Since the displacement node is a point where displacement is zero, the

variation of pressure is maximum at this point. So, it is a pressure antinode.

(b) Bats can generate and detect ultrasonic waves (𝜈 > 20 kHz) so that they

can detect the object's distance by the interval between the waves they

generate and the echo they receive. They identify the nature and size of the

object from the intensity of the echo, and they can determine the direction

of the object from the tiny interval between getting the echo from their two

ears.

(c) The sound quality generated relies on the instruments ' overtones. However,

the Sitar and Violin may have the same fundamental frequency, but they

generate distinct overtones that are integral multiples of fundamental

frequencies and therefore we can distinguish between the two sounds.

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(d) Only longitudinal waves can travel through gases as they do not have shear

elasticity. But solids can support both transverse and longitudinal waves as

they have both shear and volume elasticity.

(e) The property of a dispersive medium tells that waves of various

wavelengths travel with different velocities or at distinct speeds in distinct

directions. Thus, when a sound pulse passes through it, which is a

combination of waves of various wavelengths, it becomes distorted.

15.20. A train, standing at the outer signal of a railway station blows a whistle of frequency

400 Hz in still air. (i) What is the frequency of the whistle for a platform observer

when the train (a) approaches the platform with a speed of 10 m s−1, (b) recedes

from the platform with a speed of 10 m s−1? (ii) What is the speed of sound in each

case? The speed of sound in still air can be taken as 340 m s−1.

Solution:

(a) Given:

Frequency of the whistle, 𝜈 = 400 Hz

Speed of sound in still air, 𝑣 = 340 m s−1

Speed of the train, 𝑣𝑠 = 10 m s−1

When the train moves towards the platform, the apparent frequency (𝑣′) of

the whistle is given by:

𝑣′ = (𝑣

𝑣 − 𝑣𝑠) 𝜈

= (340

340 − 10) × 400 = 412 Hz

(b) When the train recedes the platform, the apparent frequency (𝑣′′) of the

whistle is given by:

𝑣′′ = (𝑣

𝑣 + 𝑣𝑠) 𝜈

= (340

340 + 10) × 400 = 389 Hz

(ii) The speed of sound is not dependent on relative motion between source and

observer. Hence, the speed of sound in air in both cases remains the same.

15.21. A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air.

The wind starts blowing in the direction from the yard to the station with at a speed

of 10 m s−1. What are the frequency, wavelength, and speed of sound for an

observer standing on the station’s platform? Is the situation exactly identical to the

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case when the air is still and the observer runs towards the yard at a speed of

10 m s−1? The speed of sound in still air can be taken as 340 m s−1.

Solution:

Case 1:

For the stationary observer:

Source frequency in still air, 𝜈 = 400 Hz

Speed of sound, 𝑣 = 340 m s−1

Speed of the wind, 𝑣𝑜 = 10 m s−1

When the wind starts blowing in the direction from the yard to the station

with a speed of 10 m s−1.

Therefore, for an observer standing on the station’s platform, the effective

speed of sound, 𝑣𝑒𝑓𝑓 = 340 + 10 = 350 m s−1

As observer and source both are at rest, so the frequency remains the same,

i.e., 400 Hz.

The wavelength (𝜆) of the sound heard by the stationary observer is given

by,

𝜆 =𝑣𝑒𝑓𝑓

𝜈=

350

400= 0.875 m

Case 2:

For the running observer:

Velocity of the observer, 𝑣𝑜 = 10 m s−1

Since there is a relative motion between the sound source (train) and the

observer, when the observer is moving towards the source, the frequency of

the sound heard by the observer will not be equal to that of the source due

to Doppler effect.

So, the observed frequency is given by,

𝑣′ = (𝑣 + 𝑣𝑜

𝑣) 𝜈

= (340 + 10

340) × 400

= 411.8 Hz

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Also, as wavelength of sound waves is not affected by motion of the

observer, it remains unchanged.

Thus, the first situation is not identical to the case when the air is still and

observer runs towards the station-yard at a speed of 10 m s−1. In the latter

situation, medium is at rest. So, the effective speed of sound = 340 m s−1

Additional Exercises

15.22. A travelling harmonic wave on a string is described by

𝑦(𝑥, 𝑡) = 7.5 sin  (0.0050𝑥 + 12𝑡 +𝜋

4)

(a) what are the displacement and velocity of oscillation of a point at 𝑥 = 1 cm,

and 𝑡 = 1 s? Is this velocity equal to the velocity of wave propagation?

(b) Locate the points of the string which have the same transverse

displacements and velocity as the 𝑥 = 1 cm point at 𝑡 = 2 s, 5 s and 11 s.

Solution:

(a) The given harmonic wave is:

𝑦(𝑥, 𝑡) = 7.5 sin  (0.0050𝑥 + 12𝑡 +𝜋

4)

For x = 1 cm and t = 1 s,

𝑦(1,1) = 7.5 sin  (0.0050 + 12 +𝜋

4)

7.5 sin  (12.0050 +𝜋

4)

= 7.5 sin𝜃

Where,

𝜃 = 12.0050 +𝜋

4= 12.0050 +

3.14

4= 12.78 rad

=180

3.14× 12.79 = 732.81°

∴ 𝑦(1,1) = 7.5 sin (732.81°)

= 7.5 sin(90 × 80 + 12.81°) = 7.5sin12.81°

= 7.5 × 0.2217

= 1.6629 ≈ 1.663 cm

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The velocity of the oscillation at a given point and time is given as:

𝑣 =𝑑

𝑑𝑡𝑦(𝑥, 𝑡) = (

𝑑

𝑑𝑡) [7.5 sin (0.0050𝑥 + 12𝑡 +

𝜋

4)]

= 7.5 × 12 cos (0.0050𝑥 + 12𝑡 +𝜋

4)

At 𝑥 = 1 cm  and 𝑡 = 1 s

𝑣 =𝑑

𝑑𝑡𝑦(1,1) = 90cos (12.005 +

𝜋

4)

= 90cos(732.81°) = 90cos(90 × 8 + 12.81°)

= 90cos(12.81°)

= 90 × 0.975 = 87.75 cm/s

Now, the equation of propagating wave is given by:

𝑦(𝑥, 𝑡) = 𝑎sin(𝑘𝑥 + 𝑤𝑡 + 𝜙)

Where,

𝑘 =2𝜋

𝜆

∴ 𝜆 =2𝜋

𝑘

And 𝜔 = 2𝜋𝜈

∴ 𝜈 =𝜔

2𝜋

Speed, 𝑣 = 𝜈𝜆 =𝜔

𝑘

Where,

𝜔 = 12 rad/s

𝑘 = 0.0050 m−1

∴ 𝑣 =12

0.0050= 2400 cm/s

Hence, the velocity of the wave oscillation at 𝑥 = 1 cm and 𝑡 = 1 s is not

equal to the velocity of the wave propagation.

(b) The relation between wavelength and propagation constant is given by,

𝑘 =2𝜋

𝜆

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Therefore, 𝜆 =2𝜋

𝑘=

2×3.14

0.005= 1256 cm = 12.56 m

All the points situated at distance 𝑛𝜆, (where 𝑛 is an integer) from the point

𝑥 = 1 cm have the same transverse velocity and displacement.

15.23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a

medium.

(a) Does the pulse have a definite

(i) frequency, (ii) wavelength, (iii) speed of propagation?

(b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split

of second after every 20 s), is the frequency of the note produced by the

whistle equal to 1

20 or 0.05 Hz?

Solution:

(a) The narrow sound pulse such as a short pip by a whistle does not have a

definite frequency or wavelength. But, the speed of the sound pulse remains

the same, which is the same as that of the speed of sound in that medium.

(b) No, the frequency of the note produced by the whistle is not equal to 1

20 or

0.05 Hz. As it is only the frequency at which the pulse gets repeated.

15.24. One end of a long string of linear mass density 8.0 × 10−3 kg m−1 is connected to

an electrically driven tuning fork of frequency 256 Hz. The other end passes over

a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all

the incoming energy so that reflected waves at this end have negligible amplitude.

At 𝑡 = 0, the left end (fork end) of the string 𝑥 = 0 has zero transverse

displacement (𝑦 = 0) and is moving along positive 𝑦-direction. The amplitude of

the wave is 5.0 cm. Write down the transverse displacement 𝑦 as function of 𝑥 and

𝑡 that describes the wave on the string.

Solution:

Linear mass density, 𝜇 = 8.0 × 10−3 kg m−1

Frequency of the tuning fork, 𝜈 = 256 Hz

Mass of the pan, 𝑚 = 90 kg

Amplitude of the wave, 𝑎 = 5.0 cm = 0.05 m … (i)

Tension in the string, 𝑇 = 𝑚𝑔 = 90 × 9.8 = 882 N

The transverse wave speed (𝑣) in the string is determined by the relation,

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𝑣 = √𝑇

𝜇, where 𝑇 is the tension in the string

= √882

8.0 × 10−3 = 332 m s−1

Angular frequency, 𝜔 = 2𝜋𝜈

= 2 × 3.14 × 256

= 1608.5 = 1.61 × 103 rad s−1 … (ii)

Wavelength, 𝜆 =𝑣

𝜈=

332

256 m

Therefore, propagation constant, 𝑘 =2𝜋

𝜆

=2×3.14

332

256

= 4.84 m−1 … (iii)

The equation of a travelling wave is given by the displacement function:

𝑦(𝑥, 𝑡) = 𝑎 sin (𝜔𝑡 − 𝑘𝑥) … (iv)

Substituting the values from equations (i), (ii), and (iii) in equation (iv), we get the

displacement function:

𝑦(𝑥, 𝑡) = 0.05 sin(1.61 × 103 𝑡 − 4.84 𝑥) m, where 𝑥, 𝑦 are in m and 𝑡 in s.

15.25. A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An

enemy submarine moves towards the SONAR with a speed of 360 km/h. What is

the frequency of sound reflected by the submarine? Take the speed of sound in

water to be 1450 m/s.

Solution:

Given:

Frequency of the SONAR system, 𝜈 = 40.0 kHz = 40 × 103Hz

Speed of sound in water, 𝑣 = 1450 m/s

Speed of the enemy submarine (observer), 𝑣𝑜 = 360 km/h = 360 ×5

18=

100 m/s

As the source (SONAR) is at rest and the enemy submarine (observer) moves it,

Therefore,

𝑣′ = (𝑣 + 𝑣𝑜

𝑣) 𝜈

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= (1450 + 100

1450) × 40 × 103 = 42.76 kHz

This frequency (𝑣′) is reflected by the enemy submarine and is observed by the

SONAR (which now acts as an observer).

Thus, in this case, 𝑣s = 360 km/h = 360 ×5

18= 100 m/s

Therefore, the frequency of sound reflected by the submarine is given by:

𝑣′′ = (𝑣

𝑣 − 𝑣s) 𝜈′ = (

1450

1450 − 100) × 42.76 kHz = 45.9 kHz

15.26. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can

experience both transverse (𝑆) and longitudinal (𝑃) sound waves. Typically the

speed of 𝑆 wave is about 4.0 km s−1, and that of 𝑃 wave is 8.0 km s−1. A

seismograph records 𝑃 and 𝑆 waves from an earthquake. The first 𝑃 wave arrives

4 min before the first 𝑆 wave. Assuming the waves travel in straight line, at what

distance does the earthquake occur?

Solution:

Given:

Speed of 𝑆 wave, 𝑣s = 4.0 km s−1

Speed of 𝑃 wave, 𝑣p = 8.0 km s−1

The time gap between 𝑃 and 𝑆 waves reaching the seismograph, 𝑡 = 4 min =

240 s

Let the point of earthquake occur at a distance of 𝐷 km from the seismograph.

Also, let 𝑡𝑠 and 𝑡𝑝 are the respective times taken by 𝑆 and 𝑃 waves to reach the

seismograph from the location of the earthquake.

So, 𝑡 = 𝑡𝑠 − 𝑡𝑝 =𝐷

𝑣s−

𝐷

𝑣p=

𝐷

4−

𝐷

8=

𝐷

8= 240 s

⇒ 𝐷 = 8 × 240 = 1920 km

Therefore, the earthquake occurs at a distance of 1920 km from the seismograph.

15.27. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the

sound emission frequency of the bat is 40 kHz. During one fast swoop directly

toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air.

What frequency does the bat hear reflected off the wall?

Solution:

Given:

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Frequency of sound emitted by the bat, 𝜈 = 40 kHz

Let the velocity of sound in the air be 𝑣.

So, the velocity of the bat, 𝑣b = 0.03𝑣

The apparent frequency striking the wall is given by,

𝑣′ = (𝑣

𝑣 − 𝑣b) 𝜈

= (𝑣

𝑣 − 0.03𝑣) × 40 kHz

=40

0.97 kHz

This frequency (𝑣′) is reflected by the wall and is received by the bat moving

towards the wall.

So, 𝑣s = 0

𝑣′′ = (𝑣 + 𝑣𝑏

𝑣) 𝜈′

= (𝑣 + 0.03𝑣

𝑣) ×

40

0.97

= (1.03 × 40

0.97) = 42.47 kHz

⧫ ⧫ ⧫


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