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Page No 389: - Chapter 15 Waves class 11 ncert solutions Physics - SaraNextGen [2024]


Question 15.11:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m3b8096db.gif

Where and are in m and in s. The length of the string is 1.5 m and its mass is 3.0 ×10–2 kg.

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.

Answer:

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

y (xt) = 2a sin kx cos ωt

This equation is similar to the given equation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m406f1ee5.gif

Hence, the given function represents a stationary wave.

(b) A wave travelling along the positive x-direction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m55435c4d.gif

The wave travelling along the negative x-direction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_47037b14.gif

The superposition of these two waves yields:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m485d3d8c.gif

The transverse displacement of the string is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_4c3100b8.gif

Comparing equations (i) and (ii), we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m10e2942d.gif

∴Wavelength, λ = 3 m

It is given that:

120π = 2πν

Frequency, ν = 60 Hz

Wave speed, = νλ

= 60 × 3 = 180 m/s

(c) The velocity of a transverse wave travelling in a string is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_1f380174.gif

Where,

Velocity of the transverse wave, v = 180 m/s

Mass of the string, m = 3.0 × 10–2 kg

Length of the string, = 1.5 m

Mass per unit length of the string, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_4d176e9a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_11258213.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4726/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m127ba69d.gif

Tension in the string = T

From equation (i), tension can be obtained as:

T = v2μ

= (180)2 × 2 × 10–2

= 648 N

Question 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?

Answer:

Answer:

(i)

(a) Yes, except at the nodes

(b) Yes, except at the nodes

(c) No

(ii) 0.042 m

Explanation:

(i)

(a) All the points on the string oscillate with the same frequency, except at the nodes which have zero frequency.

(b) All the points in any vibrating loop have the same phase, except at the nodes.

(c) All the points in any vibrating loop have different amplitudes of vibration.

(ii) The given equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4727/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_5b9e987a.gif

For = 0.375 m and t = 0

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4727/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m3fa36ba3.gif

Question 15.13:

Given below are some functions of and to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a traveling wave, (ii) a stationary wave or (iii) none at all:

(a) = 2 cos (3x) sin (10t)

(b) https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4728/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m242c379b.gif

(c) = 3 sin (5– 0.5t) + 4 cos (5– 0.5t)

(d) = cos sin + cos 2sin 2t

Answer:

(a) The given equation represents a stationary wave because the harmonic terms kx and ωt appear separately in the equation.

(b) The given equation does not contain any harmonic term. Therefore, it does not represent either a travelling wave or a stationary wave.

(c) The given equation represents a travelling wave as the harmonic terms kx and ωt are in the combination of kx – ωt.

(d) The given equation represents a stationary wave because the harmonic terms kx and ωt appear separately in the equation. This equation actually represents the superposition of two stationary waves.

Question 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 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Answer:

(a) Mass of the wire, m = 3.5 × 10–2 kg

Linear mass density,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4729/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m326a82bf.gif

Frequency of vibration, ν = 45 Hz

∴Length of the wire,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4729/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m7faef762.gif

The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4729/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_3ab61161.gif

For fundamental node, n = 1:

λ = 2l

λ = 2 × 0.875 = 1.75 m

The speed of the transverse wave in the string is given as:

= νλ= 45 × 1.75 = 78.75 m/s

(b) The tension produced in the string is given by the relation:

v2µ

= (78.75)2 × 4.0 × 10–2 = 248.06 N

Question 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.

Answer:

Frequency of the turning fork, ν = 340 Hz

Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4730/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m63a9531f.jpg

Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4730/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_7d424435.gif

Where,

Length of the pipe, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4730/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_26caea1.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4730/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m48413ae5.gif

The speed of sound is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4730/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m46d5bf55.gif = 340 × 1.02 = 346.8 m/s

Question 15.16:

A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?

Answer:

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

Fundamental frequency of vibration, ν = 2.53 kHz = 2.53 × 10Hz

When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends, as shown in the given figure.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4731/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_347c6472.jpg

The distance between two successive nodes ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4731/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_a8e2120.gif .

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4731/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_2b3c4e98.gif

The speed of sound in steel is given by the relation:

v = νλ

= 2.53 × 103 × 2

= 5.06 × 103 m/s

= 5.06 km/s

Question 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).

Answer:

Answer: First (Fundamental); No

Length of the pipe, = 20 cm = 0.2 m

Source frequency = nth normal mode of frequency, νn = 430 Hz

Speed of sound, v = 340 m/s

In a closed pipe, the nth normal mode of frequency is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4732/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_589a5a64.gif

Hence, the first mode of vibration frequency is resonantly excited by the given source.

In a pipe open at both ends, the nth mode of vibration frequency is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4732/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_2ea323d.gif

Since the number of the mode of vibration (n) has to be an integer, the given source does not produce a resonant vibration in an open pipe.

Question 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?

Answer:

Frequency of string A, fA = 324 Hz

Frequency of string B = fB

Beat’s frequency, n = 6 Hz

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4733/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m76d3d4fc.gif

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/193/4733/NS_8-10-28_Ravinder_11_Physics_15_27_NRJ_LVN_html_m782911b7.gif

Question 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.

Answer:

(a) A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an antinode is a point where the amplitude of vibration is the maximum and pressure is the minimum.

Therefore, a displacement node is nothing but a pressure antinode, and vice versa.

(b) Bats emit very high-frequency ultrasonic sound waves. These waves get reflected back toward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses.

(c) The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

(d) Solids have shear modulus. They can sustain shearing stress. Since fluids do not have any definite shape, they yield to shearing stress. The propagation of a transverse wave is such that it produces shearing stress in a medium. The propagation of such a wave is possible only in solids, and not in gases.

Both solids and fluids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal waves can propagate through solids and fluids.

(e) A pulse is actually is a combination of waves having different wavelengths. These waves travel in a dispersive medium with different velocities, depending on the nature of the medium. This results in the distortion of the shape of a wave pulse.

Also Read : Page-No-390:-Chapter-15-Waves-class-11-ncert-solutions-Physics

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