Question 1

A string of mass $2.50\,\mathrm{kg}$ is under a tension of $200\,\mathrm{N}$. The length of the stretched string is $20.0\,\mathrm{m}$. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Accepted Answer

Given:
Mass of string, m = 2.50 kg
Tension, T = 200 N
Length, L = 20.0 m

Step 1: Calculate the linear mass density (mass per unit length) $ \mu = \frac{m}{L} = \frac{2.50}{20.0} = 0.125\, \mathrm{kg/m} $.

Step 2: Calculate the speed of the transverse wave on the string using $ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40\, \mathrm{m/s} $.

Step 3: Time taken for disturbance to travel length L is $ t = \frac{L}{v} = \frac{20.0}{40} = 0.5\, \mathrm{s} $.

Answer: The disturbance takes 0.5 seconds to reach the other end. — The speed of a transverse wave on a stretched string depends on the tension and linear mass density. Once speed is found, time is distance divided by speed.

Question 2

A stone dropped from the top of a tower of height $300\,\mathrm{m}$ 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 $340\,\mathrm{m\,s^{-1}}$? ($g = 9.8\,\mathrm{m\,s^{-2}}$)

Accepted Answer

Given:
Height, h = 300 m
Speed of sound, v = 340 m/s
Acceleration due to gravity, g = 9.8 m/s²

Step 1: Calculate time taken by stone to fall:
Using $ h = \frac{1}{2}gt^2 $,
$ t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 300}{9.8}} = \sqrt{61.22} \approx 7.82 \text{ s} $.

Step 2: Time taken by sound to travel up:
$ t_2 = \frac{h}{v} = \frac{300}{340} \approx 0.882 \text{ s} $.

Step 3: Total time when splash is heard:
$ t = t_1 + t_2 = 7.82 + 0.882 = 8.70 \text{ s} $.

Answer: The splash is heard at the top after approximately 8.7 seconds. — First find the time for the stone to hit the water using free fall formula, then add the time sound takes to travel back up.

Question 3

A steel wire has a length of $12.0\,\mathrm{m}$ and a mass of $2.10\,\mathrm{kg}$. 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 $20^{\circ}\mathrm{C} = 343\,\mathrm{m\,s^{-1}}$.

Accepted Answer

Given:
Length, L = 12.0 m
Mass, m = 2.10 kg
Speed of wave, v = 343 m/s

Step 1: Calculate linear mass density $ \mu = \frac{m}{L} = \frac{2.10}{12.0} = 0.175 \mathrm{kg/m} $.

Step 2: Use formula for wave speed on string:
$ v = \sqrt{\frac{T}{\mu}} \Rightarrow T = \mu v^2 = 0.175 \times (343)^2 $.

Step 3: Calculate tension:
$ T = 0.175 \times 117, 649 = 20,588.6 \mathrm{N} $.

Answer: The tension required is approximately 20,589 N. — The tension is found by rearranging the wave speed formula and substituting the given values.

Question 4

Use the formula $v = \sqrt{\frac{\gamma P}{\rho}}$ to explain why the speed of sound in air

(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.

Accepted Answer

Given formula for speed of sound in air:
$ v = \sqrt{\frac{\gamma P}{\rho}} $

(a) Independence of pressure:
Using ideal gas law, $ P = \rho R T / M $, so $ \rho = \frac{P M}{R T} $.
Substitute $ \rho $ in speed formula:
$ v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma P}{\frac{P M}{R T}}} = \sqrt{\frac{\gamma R T}{M}} $.
Pressure $ P $ cancels out, so speed is independent of pressure.

(b) Increase with temperature:
From above, $ v = \sqrt{\frac{\gamma R T}{M}} $, so speed $ v \propto \sqrt{T} $.
As temperature increases, speed increases.

(c) Increase with humidity:
Humid air has water vapor which has lower molar mass than dry air.
Since $ v = \sqrt{\frac{\gamma R T}{M}} $, decreasing molar mass $ M $ increases speed.
Therefore, speed increases with humidity. — Using ideal gas law to replace density, pressure cancels out. Temperature and molar mass affect speed as shown.

Question 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. $y = f(x \pm vt)$. Is the converse true? Examine if the following functions for $y$ can possibly represent a travelling wave:

(a) $(x - vt)^2$
(b) $\log [(x + vt) / x_0]$
(c) $1 / (x + vt)$

Accepted Answer

A travelling wave can be represented as $ y = f(x \pm vt) $, but the converse is not always true because the function must satisfy the wave equation.

(a) $ y = (x - vt)^2 $ is a function of $ x - vt $, but it does not satisfy the wave equation because the second derivatives do not satisfy the wave equation. Hence, it is not a travelling wave.

(b) $ y = \log \left( \frac{x + vt}{x_0} \right) $ depends on $ x + vt $, but logarithmic functions do not satisfy the linear wave equation. So, it is not a travelling wave.

(c) $ y = \frac{1}{x + vt} $ depends on $ x + vt $, but again this function does not satisfy the wave equation due to its non-linearity and singularity. Hence, not a travelling wave.

Conclusion: Although these functions depend on $ x \pm vt $, they do not satisfy the wave equation and thus do not represent travelling waves. — Only functions that satisfy the wave equation represent travelling waves. Merely depending on $ x \pm vt $ is not sufficient.

Question 6

A bat emits ultrasonic sound of frequency $1000\mathrm{kHz}$ 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 $340\mathrm{ms}^{-1}$ and in water $1486\mathrm{ms}^{-1}$.

Accepted Answer

Given:
Frequency, f = 1000 kHz = 1,000,000 Hz
Speed in air, v_air = 340 m/s
Speed in water, v_water = 1486 m/s

(a) Wavelength of reflected sound in air:
$ \lambda_{air} = \frac{v_{air}}{f} = \frac{340}{1,000,000} = 3.4 \times 10^{-4} \mathrm{m} = 0.34 \mathrm{mm} $.

(b) Wavelength of transmitted sound in water:
$ \lambda_{water} = \frac{v_{water}}{f} = \frac{1486}{1,000,000} = 1.486 \times 10^{-3} \mathrm{m} = 1.486 \mathrm{mm} $.

Answer:
(a) 0.34 mm
(b) 1.486 mm — Wavelength is speed divided by frequency. Frequency remains same across media.

Question 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 $1.7 \, \mathrm{km} \, \mathrm{s}^{-1}$? The operating frequency of the scanner is 4.2 MHz.

Accepted Answer

Given:
Speed of sound in tissue, v = 1.7 km/s = 1700 m/s
Frequency, f = 4.2 MHz = 4,200,000 Hz

Calculate wavelength:
$ \lambda = \frac{v}{f} = \frac{1700}{4,200,000} = 4.05 \times 10^{-4} \mathrm{m} = 0.405 \mathrm{mm} $.

Answer: The wavelength of sound in the tissue is approximately 0.405 mm. — Wavelength is speed divided by frequency.

Question 8

A transverse harmonic wave on a string is described by

$$
y (x, t) = 3. 0 \sin (3 6 t + 0. 0 1 8 x + \pi / 4)
$$

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?

Accepted Answer

Given:
Wave function: $ y(x,t) = 3.0 \sin(36 t + 0.018 x + \pi/4) $
Units: x and y in cm, t in s

(a) To check if travelling or stationary:
Form is $ y = A \sin(kx \pm \omega t + \phi) $.
Here, argument is $ 36 t + 0.018 x + \pi/4 = 0.018 x + 36 t + \pi/4 $.
Rewrite as $ 0.018 x + 36 t + \pi/4 = 0.018 (x + 2000 t) + \pi/4 $ since $ \frac{36}{0.018} = 2000 $.
Since the argument is $ k(x + vt) $, wave travels in negative x-direction with speed $ v = 2000 $ cm/s = 20 m/s.

Answer: It is a travelling wave moving in the negative x-direction at 20 m/s.

(b) Amplitude $ A = 3.0 $ cm.
Angular frequency $ \omega = 36 $ rad/s.
Frequency $ f = \frac{\omega}{2\pi} = \frac{36}{2\pi} \approx 5.73 $ Hz.

(c) Initial phase at origin (x=0, t=0):
$ \phi = \pi/4 $.

(d) Wavelength $ \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.018} \approx 349.07 $ cm = 3.49 m.
Least distance between two successive crests is wavelength $ \lambda = 3.49 $ m. — Identify wave form, find wave number and angular frequency, determine direction from sign, calculate frequency and wavelength.

Question 9

Exercise 14.8, plot the displacement $(y)$ versus $(t)$ graphs for $x = 0, 2$ and $4\mathrm{cm}$. 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?

Accepted Answer

To plot displacement y versus time t for x = 0, 2 cm, and 4 cm, use the given wave equation (not fully provided here). The graphs are sinusoidal oscillations with the same frequency and amplitude but different phases. The shape of the graphs is sinusoidal. The oscillatory motion differs from point to point in phase only; amplitude and frequency remain constant. — In a travelling wave, displacement at any point oscillates sinusoidally with time. The amplitude and frequency are constant throughout the medium, but the phase changes with position x. Hence, the graphs at different x values are sinusoidal with the same amplitude and frequency but shifted in phase.

Question 10

10 For the travelling harmonic wave

$$
y (x, t) = 2. 0 \cos 2 \pi (1 0 t - 0. 0 0 8 0 x + 0. 3 5)
$$

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) $4\mathrm{m}$
(b) $0.5\mathrm{m}$
(c) $\lambda /2$
(d) $3\lambda /4$

Accepted Answer

Given wave: y(x,t) = 2.0 cos 2π(10 t - 0.0080 x + 0.35)

Step 1: Identify wave number k and wavelength λ.
From the argument: 2π(10 t - 0.0080 x + 0.35), the spatial term is 2π × 0.0080 x = k x
So, k = 2π × 0.0080 = 0.016π cm⁻¹
Convert k to m⁻¹: 0.0080 cm⁻¹ = 0.8 m⁻¹ (since 1 cm = 0.01 m, 0.0080 cm⁻¹ = 0.0080 × 100 = 0.8 m⁻¹)

Wavelength λ = 2π / k = 2π / (2π × 0.0080) = 1 / 0.0080 cm = 125 cm = 1.25 m

Phase difference Δφ = k Δx

(a) For Δx = 4 m:
Δφ = k × Δx = 2π × 0.0080 × (4 × 100) cm = 2π × 0.0080 × 400 = 2π × 3.2 = 6.4π radians
But since k in m⁻¹ is 0.8, Δφ = 0.8 × 4 = 3.2 radians

(b) For Δx = 0.5 m:
Δφ = 0.8 × 0.5 = 0.4 radians

(c) For Δx = λ/2 = 1.25 / 2 = 0.625 m:
Δφ = 0.8 × 0.625 = 0.5 radians
But phase difference over half wavelength is π radians theoretically, so re-check:
Actually, phase difference Δφ = (2π / λ) × Δx
= (2π / 1.25) × (1.25 / 2) = 2π × 0.5 = π radians

(d) For Δx = 3λ/4 = 3 × 1.25 / 4 = 0.9375 m:
Δφ = (2π / 1.25) × 0.9375 = 2π × 0.75 = 3π/2 radians

Summary:
(a) 4 m → phase difference = 3.2 radians (or 6.4π if using cm units, but consistent units give 3.2 radians)
(b) 0.5 m → 0.4 radians
(c) λ/2 → π radians
(d) 3λ/4 → 3π/2 radians — The phase difference between two points separated by distance Δx is Δφ = k Δx, where k = 2π/λ. First find wavelength λ from the wave number k, then calculate phase difference for each Δx. Ensure consistent units (meters).

Question 11

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

$$
y (x, t) = 0. 0 6 \sin \left(\frac {2 \pi}{3} xight) \cos (1 2 0 \pi t)
$$

where $x$ and $y$ are in m and $t$ in s. The length of the string is $1.5\mathrm{m}$ and its mass is $3.0	imes 10^{-2}\mathrm{kg}$.

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.

Accepted Answer

(a) The given displacement is y(x,t) = 0.006 sin(2πx/3) cos(120π t). This is a product of a spatial sine function and a temporal cosine function, characteristic of a stationary wave (standing wave). So, it represents a stationary wave.

(b) A stationary wave can be seen as the superposition of two travelling waves moving in opposite directions:

Each travelling wave: y = A sin(kx - ωt) and y = A sin(kx + ωt)

Given k = 2π / 3 (from sin(2πx/3))

Wavelength λ = 2π / k = 3 m

Angular frequency ω = 120π rad/s

Frequency f = ω / (2π) = 120π / (2π) = 60 Hz

Wave speed v = f λ = 60 × 3 = 180 m/s

(c) To find tension T:

Mass m = 3.0 × 10⁻² kg, length L = 1.5 m

Linear mass density μ = m / L = 0.03 / 1.5 = 0.02 kg/m

Wave speed v = sqrt(T / μ) ⇒ T = μ v² = 0.02 × (180)² = 0.02 × 32400 = 648 N

Therefore, tension in the string is 648 N. — The given displacement is a product of sine and cosine functions, indicating a stationary wave. Wavelength and frequency are found from the wave number and angular frequency. Wave speed is frequency times wavelength. Tension is calculated using wave speed and linear mass density.

Question 12

(i) 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?

Accepted Answer

(i) (a) Frequency: Yes, all points on the string oscillate with the same frequency because the string vibrates as a whole at a particular frequency determined by the source or boundary conditions.

(b) Phase: No, all points do not oscillate with the same phase. The phase varies along the length of the string. For example, in a standing wave, nodes and antinodes have different phases.

(c) Amplitude: No, the amplitude varies from point to point. At nodes, amplitude is zero, and at antinodes, it is maximum.

(ii) To find the amplitude at a point 0.375 m from one end, we need the expression for amplitude variation along the string. For a string fixed at both ends of length L, the amplitude at a distance x from one end in the nth mode is given by:

A(x) = 2A_0

sin(nπx/L)

Assuming fundamental mode (n=1) and total length L = 0.75 m (since 0.375 m is half), the amplitude at x = 0.375 m is:

A(0.375) = 2A_0 sin(π * 0.375 / 0.75) = 2A_0 sin(π/2) = 2A_0 * 1 = 2A_0

Thus, the amplitude at 0.375 m is maximum (twice the amplitude constant A_0).

Note: The exact numerical amplitude depends on the maximum amplitude A_0 which must be given or measured. — Frequency is uniform because the entire string vibrates at the same frequency. Phase varies because points oscillate differently along the string length, especially in standing waves. Amplitude varies from zero at nodes to maximum at antinodes. Using the standing wave amplitude formula, amplitude at 0.375 m (midpoint) is maximum.

Question 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 \sqrt{x - vt}
(c) y = 3 sin (5x - 0.5t) + 4 cos (5x - 0.5t)
(d) y = cos x sin t + cos 2x sin 2t

Accepted Answer

(a) y = 2 cos (3x) sin (10t): This is a product of a spatial cosine and a temporal sine function. This form y = f(x)g(t) is characteristic of a stationary (standing) wave.

(b) y = 2 \sqrt{x - vt}: This is not a sinusoidal function and does not represent a wave solution of the wave equation. Hence, it is neither a travelling nor a stationary wave.

(c) y = 3 sin (5x - 0.5t) + 4 cos (5x - 0.5t): Using trigonometric identity, this can be combined into a single sinusoidal travelling wave:

Let y = R sin(5x - 0.5t + \phi), where R = \sqrt{3^2 + 4^2} = 5, and \phi = 	an^{-1}(4/3).

Thus, this represents a travelling wave.

(d) y = cos x sin t + cos 2x sin 2t: This is a sum of two terms each of the form spatial function times temporal function, characteristic of stationary waves (superposition of modes).

Summary:
(i) Travelling wave: (c)
(ii) Stationary wave: (a), (d)
(iii) None: (b) — Stationary waves are products of spatial and temporal functions separately. Travelling waves are functions of (x ± vt). Non-sinusoidal or non-wave-like functions do not represent waves.

Question 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?

Accepted Answer

(a) The fundamental frequency f = 45 Hz.

Length of wire, L is not given explicitly, but can be found from mass and linear density:

Mass m = 3.5 × 10⁻² kg

Linear mass density μ = 4.0 × 10⁻² kg/m

Length L = m / μ = (3.5 × 10⁻²) / (4.0 × 10⁻²) = 0.875 m

For fundamental mode, wavelength λ = 2L = 2 × 0.875 = 1.75 m

Wave speed v = f × λ = 45 × 1.75 = 78.75 m/s

(b) Tension T in the string is related to wave speed and linear density by:

v = sqrt(T / μ) => T = μ v²

T = 4.0 × 10⁻² × (78.75)² = 4.0 × 10⁻² × 6201.56 = 248.06 N

Therefore,

(a) Speed of transverse wave = 78.75 m/s
(b) Tension in the string = 248.06 N — First, find length from mass and linear density. Then use fundamental mode wavelength formula. Calculate wave speed from frequency and wavelength. Use wave speed and linear density to find tension.

Question 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.

Accepted Answer

Given:
Frequency f = 340 Hz
Tube lengths at resonance: L₁ = 25.5 cm = 0.255 m, L₂ = 79.3 cm = 0.793 m

Since the tube is open at one end and closed at the other, resonance occurs at odd multiples of quarter wavelengths:

L = (2n - 1) λ/4, where n = 1, 2, 3...

For first resonance (n=1):
L₁ = λ/4 => λ = 4L₁ = 4 × 0.255 = 1.02 m

For second resonance (n=2):
L₂ = 3λ/4 => λ = 4L₂/3 = 4 × 0.793 / 3 = 1.057 m

Average wavelength:
λ_avg = (1.02 + 1.057)/2 = 1.0385 m

Speed of sound v = f × λ_avg = 340 × 1.0385 = 353.1 m/s

Therefore, the speed of sound in air at the temperature of the experiment is approximately 353 m/s. — Use resonance condition for tube closed at one end: lengths correspond to odd multiples of quarter wavelengths. Calculate wavelength from both lengths and average. Multiply by frequency to get speed of sound.

WAVES

WAVES — Study Notes

14.1 Introduction

14.2 Transverse and longitudinal waves

14.3 Displacement relation in a progressive wave

Practice Questions — WAVES

All 7 Chapters in Physics Part-II