GSEB Class 11 Physics Solutions Chapter 15 Waves

Question 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?
Answer: Here, the tension \( T = 200 \, \text{N} \). The string's length \( l = 20.0 \, \text{m} \). The mass of the string \( M = 2.50 \, \text{kg} \).
First, we find the mass per unit length of the string, \( m = \frac{M}{l} = \frac{2.50}{20.0} = 0.125 \, \text{kg m}^{-1} \).
We know that the velocity of a transverse wave is given by:
\( v = \sqrt{\frac{T}{m}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40 \, \text{ms}^{-1} \)
Let \( t \) be the time taken for the transverse jerk to travel from one end to the other end of the string.
Then, \( t = \frac{\text{length of string}}{\text{velocity of string}} = \frac{l}{v} = \frac{20}{40} = 0.5 \, \text{s} \).
In simple words: We calculated the string's thickness per unit length, then used the tension and thickness to find the wave's speed. Finally, we divided the string's length by this speed to determine how long the disturbance would take to travel.

Exam Tip: Remember to calculate the linear mass density (\( m \)) first if it's not given directly. Ensure units are consistent throughout the calculation.

Question 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 ms\(^{-1}\)? (g = 9.8 ms\(^{-2}\))
Answer: Here, the initial velocity of the stone, \( u = 0 \). The acceleration due to gravity \( a = g = 9.8 \, \text{ms}^{-2} \). The tower's height \( h = 300 \, \text{m} \). The velocity of sound in air \( v = 340 \, \text{ms}^{-1} \).
Let \( t_1 \) be the time it takes for the stone to fall down.
Let \( t_2 \) be the time it takes for the sound to travel from the bottom to the top of the tower.
If \( t \) is the total time after which the splash is heard at the top, then \( t = t_1 + t_2 \) (i)
To find \( t_1 \), we use the equation of motion:
\( s = ut + \frac{1}{2}at^2 \)
Here, \( s = h \), \( u = 0 \), \( a = g \), \( t = t_1 \).
\( h = 0 + \frac{1}{2}gt_1^2 \)
\( t_1^2 = \frac{2h}{g} \)
\( t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 300}{9.8}} = \sqrt{\frac{600}{9.8}} \approx \sqrt{61.22} \approx 7.82 \, \text{s} \) (ii)
To find \( t_2 \), we use the formula:
\( t_2 = \frac{\text{Distance}}{\text{Velocity}} = \frac{\text{Height of tower}}{\text{Velocity of sound}} = \frac{h}{v} \)
\( t_2 = \frac{300}{340} = \frac{15}{17} \approx 0.88 \, \text{s} \) (iii)
From (i), (ii), and (iii), we get the total time:
\( t = t_1 + t_2 = 7.82 + 0.88 = 8.7 \, \text{s} \).
In simple words: First, we calculated the time it took for the stone to fall. Then, we found how long it took for the sound of the splash to reach the top. Adding these two times gave us the total time until the splash was heard.

Exam Tip: This problem involves two distinct parts: the stone's fall (kinematics) and the sound's travel (constant speed). Make sure to calculate both times separately before summing them up.

Question 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°C = 343
Answer: Here, the length of the steel wire \( l = 12.0 \, \text{m} \). The mass of the steel wire \( M = 2.10 \, \text{kg} \).
Let \( T \) be the tension needed in the wire.
If \( m \) is the mass per unit length of the wire, then \( m = \frac{M}{l} = \frac{2.10}{12.0} = 0.175 \, \text{kg m}^{-1} \).
The speed of transverse waves on the wire \( V \) should be equal to the speed of sound in dry air at 20°C, which is \( 343 \, \text{ms}^{-1} \) (given).
We know that the speed of a transverse wave in a wire is given by:
\( v = \sqrt{\frac{T}{m}} \)
Squaring both sides, \( v^2 = \frac{T}{m} \)
So, \( T = mv^2 \)
Substituting the values: \( T = 0.175 \times (343)^2 \)
\( T = 0.175 \times 117649 \)
\( T = 20588.575 \, \text{N} \)
Rounding to two significant figures (as in the given mass density):
\( T \approx 2.06 \times 10^4 \, \text{N} \).
In simple words: We first found the wire's mass per meter. Then, using the formula for wave speed on a string, we calculated the tension required to make the wave travel as fast as sound in air.

Exam Tip: Pay close attention to unit consistency. The given speed of sound is in m/s, so ensure all other quantities like mass and length are in standard SI units (kg and m) before calculation.

Question 4. Use the formula \( v = \sqrt{\frac{\gamma \mathbf{P}}{\rho}} \) to explain why the speed of sound in air:
(a) is independent of pressure.
(b) increases with temperature.
(c) inicreases with humidity.
Answer:
(a) The formula for the velocity of sound in air is \( v = \sqrt{\frac{\gamma \mathbf{P}}{\rho}} \), where \( \gamma \) is a constant for a given gas (like air), \( \mathbf{P} \) is the pressure, and \( \rho \) is the density of the gas.
According to the ideal gas equation, we know that \( \mathbf{P}V = RT \), where \( V \) is volume, \( R \) is the gas constant, and \( T \) is the temperature. The density \( \rho = \frac{M}{V} \), where \( M \) is the mass of the gas.
So, \( V = \frac{M}{\rho} \). Substituting this into the ideal gas equation:
\( \mathbf{P} \frac{M}{\rho} = RT \implies \frac{\mathbf{P}}{\rho} = \frac{RT}{M} \)
Thus, the velocity formula becomes \( v = \sqrt{\frac{\gamma RT}{M}} \).
For a specific gas, \( \gamma \), \( R \), and \( M \) (molecular weight) are constants. So, if the temperature \( T \) remains constant, the term \( \frac{\mathbf{P}}{\rho} \) also remains constant. This means the speed of sound \( v \) is independent of the pressure of air (gas) if the temperature stays unchanged.

(b) From the derived formula \( v = \sqrt{\frac{\gamma RT}{M}} \), since \( \gamma \), \( R \), and \( M \) are constants for a specific gas, we can see that \( v \propto \sqrt{T} \).
This indicates that the velocity of sound in a gas is directly proportional to the square root of its absolute temperature. Consequently, we can conclude that the speed of sound in air increases as the temperature rises.

(c) Humidity affects the density of air. Moist air is a mixture of dry air and water vapor. The molecular mass of water (\( \text{H}_2\text{O} \approx 18 \, \text{g/mol} \)) is less than the average molecular mass of dry air (which is approximately \( 29 \, \text{g/mol} \), composed mainly of \( \text{N}_2 \approx 28 \, \text{g/mol} \) and \( \text{O}_2 \approx 32 \, \text{g/mol} \)).
Therefore, as humidity increases, the proportion of lighter water vapor molecules in the air mixture rises, causing the overall density \( \rho \) of the moist air to decrease. Since \( v \propto \frac{1}{\sqrt{\rho}} \), a decrease in density leads to an increase in the speed of sound. This is why sound travels faster on a rainy day (higher humidity) than on a dry day.
In simple words: (a) The speed of sound doesn't change with pressure if the temperature stays the same because pressure and density change in a way that keeps their ratio constant. (b) Sound travels faster when the air is hotter because its speed is linked to the square root of the temperature. (c) When air is more humid, it becomes lighter because water vapor molecules are lighter than air molecules, making sound travel faster.

Exam Tip: When explaining the relationship between sound speed and gas properties, always refer back to the key formula \( v = \sqrt{\frac{\gamma \mathbf{P}}{\rho}} \) and then use the ideal gas law to show how \( \frac{\mathbf{P}}{\rho} \) or \( \frac{T}{M} \) varies or stays constant with changes in pressure, temperature, or humidity. For humidity, remember to mention the molecular mass of water being less than air.

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, 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) \( \log \left[\frac{x+v t}{x_{0}}\right] \)
(c) \( \frac{1}{(x+vt)} \)
Answer: No, the converse is not always true.
The basic requirement for a wave function to represent a travelling wave is that for all values of \( x \) and \( t \), the wave function must have a finite value. This means the function should not become infinitely large at any point.
Let's check the given functions:
(a) \( y = (x - vt)^2 \): This function always has a finite value for all \( x \) and \( t \). It correctly represents a travelling wave because \( x \) and \( t \) appear in the combination \( (x - vt) \).
(b) \( y = \log \left[\frac{x+v t}{x_{0}}\right] \): This function represents a travelling wave, as \( x \) and \( t \) appear in the combination \( (x + vt) \). However, it will become undefined or negatively infinite if \( \frac{x+vt}{x_0} \le 0 \). For a valid wave function, it must be finite and well-defined everywhere. Assuming \( x_0 \) is a positive constant and \( x+vt > 0 \), it can represent a wave.
(c) \( y = \frac{1}{(x+vt)} \): This function does represent a travelling wave as \( x \) and \( t \) appear in the combination \( (x + vt) \). However, the function becomes infinitely large (undefined) when \( (x+vt) = 0 \). A true wave function must be finite for all values of \( x \) and \( t \). Therefore, this function typically does not represent a physical travelling wave.
Out of the given functions for \( y \), only (a) always satisfies the condition of having a finite value for all \( x \) and \( t \). (b) and (c) have conditions where they might become undefined or infinite.
In simple words: No, just having \( x \) and \( t \) together as \( (x \pm vt) \) isn't enough for a function to be a wave. A wave function must always have a normal, measurable value. Function (a) works because it's always normal. Function (b) is usually fine but could have issues if the inside of the log becomes zero or negative. Function (c) doesn't work because it becomes huge and breaks down when the bottom part is zero.

Exam Tip: When examining if a function represents a travelling wave, first check if the variables \( x \) and \( t \) appear as a linear combination like \( (ax \pm bt) \). Second, and crucially, check if the function remains finite and well-defined for all possible values of \( x \) and \( t \). Logarithms and fractions are common places where functions might become undefined.

Question 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\(^{-1}\).
Answer: Here, the frequency of the ultrasonic sound \( \nu = 1000 \, \text{kHz} = 1000 \times 10^3 \, \text{Hz} = 10^6 \, \text{s}^{-1} \).
The speed of sound in air \( v_a = 340 \, \text{ms}^{-1} \).
The speed of sound in water \( v_w = 1486 \, \text{ms}^{-1} \).

(a) For reflected sound:
After reflection, the ultrasonic sound travels back in air. The frequency of the wave remains unchanged during reflection. Let \( \lambda_a \) be the wavelength of the reflected sound.
Using the relationship \( v_a = \nu \lambda_a \), we get:
\( \lambda_a = \frac{v_a}{\nu} = \frac{340 \, \text{ms}^{-1}}{10^6 \, \text{s}^{-1}} \)
\( \lambda_a = 3.4 \times 10^{-4} \, \text{m} \).

(b) For transmitted sound:
The transmitted sound travels through water. The frequency of the wave remains unchanged as it passes from air to water. Let \( \lambda_w \) be the wavelength of the transmitted sound.
Using the relationship \( v_w = \nu \lambda_w \), we get:
\( \lambda_w = \frac{v_w}{\nu} = \frac{1486 \, \text{ms}^{-1}}{10^6 \, \text{s}^{-1}} \)
\( \lambda_w = 14.86 \times 10^{-4} \, \text{m} \)
\( \lambda_w = 1.49 \times 10^{-3} \, \text{m} \) (approximately).
In simple words: For the reflected sound, we used the speed of sound in air and the frequency to find its wavelength. For the sound that went into the water, we used the speed of sound in water and the same frequency to find its wavelength. The frequency never changes when sound moves between materials.

Exam Tip: Remember that the frequency of a wave (like sound) does not change when it passes from one medium to another or when it reflects. Only its speed and wavelength change according to the properties of the new medium.

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 kms\(^{-1}\)? The operating frequency of the scanner is 4.2 MHz.
Answer: Here, the speed of sound in the tissue \( v = 1.7 \, \text{kms}^{-1} = 1.7 \times 10^3 \, \text{ms}^{-1} \).
The frequency of the scanner \( \nu = 4.2 \, \text{MHz} = 4.2 \times 10^6 \, \text{Hz} \).
Let \( \lambda \) be the wavelength of sound in the tissue.
From the relationship \( v = \nu \lambda \), we can find the wavelength:
\( \lambda = \frac{v}{\nu} \)
Substituting the given values:
\( \lambda = \frac{1.7 \times 10^3 \, \text{ms}^{-1}}{4.2 \times 10^6 \, \text{Hz}} \)
\( \lambda = \frac{1.7}{4.2} \times 10^{-3} \, \text{m} \)
\( \lambda \approx 0.40476 \times 10^{-3} \, \text{m} \)
\( \lambda \approx 4.05 \times 10^{-4} \, \text{m} \).
Rounding to two significant figures, \( \lambda \approx 4.1 \times 10^{-4} \, \text{m} \).
In simple words: To find the sound's wavelength inside the tissue, we divide its speed in the tissue by the scanner's operating frequency. This gives us how long each wave is.

Exam Tip: Always convert all quantities to their base SI units (meters, seconds, Hertz) before performing calculations. Pay attention to prefixes like 'kilo' (10\(^3\)) and 'mega' (10\(^6\)).

Question 8. A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36t + 0.018x + \( \frac{\pi}{4} \)) 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?
Answer: The equation of the given transverse harmonic wave is:
\( y(x, t) = 3.0 \sin (36t + 0.018x + \frac{\pi}{4}) \) (i)
The standard equation for a travelling wave is:
\( y(x, t) = A \sin \left[ \frac{2\pi}{T}t \pm \frac{2\pi}{\lambda}x + \phi_0 \right] \) or \( y(x, t) = A \sin (\omega t \pm kx + \phi_0) \) (ii)

(a) Comparing equation (i) with the standard form (ii):
This is a travelling wave because \( x \) and \( t \) appear together in a linear combination inside the sine function. The plus sign \( (+0.018x) \) indicates that the wave is travelling in the negative \( x \) direction (from right to left).
From comparison, we have: \( \omega = 36 \) rad/s and \( k = 0.018 \) rad/cm.
The speed of the wave is \( v = \frac{\omega}{k} = \frac{36}{0.018} = 2000 \, \text{cm/s} = 20 \, \text{m/s} \).
So, the speed is 20 m/s and the direction is from right to left (negative \( x \) direction).

(b) From comparison with (ii):
Amplitude \( A = 3.0 \, \text{cm} = 3.0 \times 10^{-2} \, \text{m} \).
The angular frequency \( \omega = 36 \, \text{rad/s} \).
We know \( \omega = 2\pi\nu \), where \( \nu \) is the frequency.
So, \( \nu = \frac{\omega}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \approx \frac{18}{3.14} \approx 5.73 \, \text{Hz} \).

(c) The initial phase at the origin (when \( x = 0 \)) is \( \phi_0 = \frac{\pi}{4} \, \text{rad} \).

(d) The least distance between two successive crests in the wave is the wavelength \( \lambda \).
From comparison, \( k = \frac{2\pi}{\lambda} = 0.018 \, \text{cm}^{-1} \).
So, \( \lambda = \frac{2\pi}{0.018} = \frac{2 \times 3.14159}{0.018} \approx \frac{6.28318}{0.018} \approx 349.06 \, \text{cm} \).
\( \lambda \approx 3.49 \, \text{m} \).
In simple words: (a) This is a travelling wave because of its form, and the plus sign means it moves to the left at a speed of 20 m/s. (b) Its greatest height (amplitude) is 3 cm, and it vibrates about 5.73 times per second (frequency). (c) The starting point of the wave at the origin is \( \frac{\pi}{4} \) radians. (d) The distance from one wave peak to the next (wavelength) is about 3.49 meters.

Exam Tip: For travelling waves, a \( (kx - \omega t) \) or \( (\omega t - kx) \) form means propagation in the positive x-direction, while \( (kx + \omega t) \) or \( (\omega t + kx) \) means propagation in the negative x-direction. Pay careful attention to the signs in the argument of the sine or cosine function.

Question 9. For the wave described in exercise 15.8 plot the displacement (y) versus time (t) graphs for x = 0,2 and 4 cm. What are the shape of these graphs? In which aspects does oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Answer: Here, the given wave is: \( y(x, t) = 3.0 \sin (36t + 0.018x + \frac{\pi}{4}) \)
Let \( y_1, y_2 \) and \( y_3 \) be the displacements of the wave for \( x = 0, 2 \) and \( 4 \, \text{cm} \).
For \( x = 0 \, \text{cm} \): \( y_1(0, t) = 3.0 \sin (36t + \frac{\pi}{4}) \)
For \( x = 2 \, \text{cm} \): \( y_2(2, t) = 3.0 \sin (36t + 0.018 \times 2 + \frac{\pi}{4}) = 3.0 \sin (36t + 0.036 + \frac{\pi}{4}) \)
For \( x = 4 \, \text{cm} \): \( y_3(4, t) = 3.0 \sin (36t + 0.018 \times 4 + \frac{\pi}{4}) = 3.0 \sin (36t + 0.072 + \frac{\pi}{4}) \)
The shape of these graphs (y vs. t) for different fixed \( x \) values will be sinusoidal (like a sine wave). This is because the functional form is a sine function of time.
In a travelling wave, the oscillatory motion differs from one point to another only in terms of its **phase**. The amplitude and frequency remain constant across all points in a travelling wave. The phase difference between points is determined by their spatial separation.
For the points \( x=0, 2 \), and \( 4 \, \text{cm} \), the phases are \( \frac{\pi}{4} \), \( (\frac{\pi}{4} + 0.036) \), and \( (\frac{\pi}{4} + 0.072) \) respectively. These are all different, showing a phase shift.
The amplitude for all these waves is 3.0 cm, and the angular frequency is 36 rad/s, meaning the frequency \( \nu = \frac{36}{2\pi} \, \text{s}^{-1} \), which is constant.
In simple words: When you draw graphs of how the wave moves over time at different spots, they all look like smooth, curving sine waves. In a travelling wave, the way things jiggle (amplitude) and how often they jiggle (frequency) are the same everywhere. But the timing of the jiggling (phase) is different at different places along the wave.

Exam Tip: For travelling waves, remember that amplitude and frequency are characteristic properties of the wave itself and remain constant throughout the medium. Phase, however, varies with position, leading to phase differences between different points.

Question 10. For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t – 0.0080x + 035) 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 cm, (b) 0.5 m (c) \( \frac{\lambda}{2} \) (d) \( \frac{3\lambda}{4} \).
Answer: The given equation of a travelling harmonic wave is:
\( y(x, t) = 2.0 \cos 2\pi (10t – 0.0080x + 0.35) \) (1)
The standard equation of a travelling harmonic wave is:
\( y(x,t) = A \cos \left[ 2\pi \left(\frac{t}{T} - \frac{x}{\lambda}\right) + \phi_0 \right] \) (2)
Comparing equation (1) and (2), we get:
\( \frac{1}{T} = 10 \implies T = 0.1 \, \text{s} \)
\( \frac{1}{\lambda} = 0.0080 \, \text{cm}^{-1} \)
\( \phi_0 = 0.35 \times 2\pi \, \text{rad} \)
So, \( \frac{2\pi}{\lambda} = 2\pi \times 0.0080 = 0.016\pi \, \text{rad/cm} \) (3)
We know that the phase difference \( \Delta\phi \) between two points separated by a path difference \( \Delta x \) is given by: \( \Delta\phi = \frac{2\pi}{\lambda} \times \Delta x \) (4)

(a) When path difference \( \Delta x = 4 \, \text{cm} \):
Using (3) and (4):
\( \Delta\phi = (2\pi \times 0.0080) \times 4 \)
\( \Delta\phi = 0.016\pi \times 4 = 0.064\pi \, \text{rad} \).

(b) When path difference \( \Delta x = 0.5 \, \text{m} = 50 \, \text{cm} \):
Using (3) and (4):
\( \Delta\phi = (2\pi \times 0.0080) \times 50 \)
\( \Delta\phi = 0.016\pi \times 50 = 0.8\pi \, \text{rad} \).

(c) When path difference \( \Delta x = \frac{\lambda}{2} \):
\( \Delta\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi \, \text{rad} \).

(d) When path difference \( \Delta x = \frac{3\lambda}{4} \):
\( \Delta\phi = \frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = \frac{3\pi}{2} \, \text{rad} \).
In simple words: First, we matched the given wave equation to a standard form to find important values like the wavelength. Then, for each specified distance between two points, we used a formula to calculate how much their phases would differ.

Exam Tip: The key to solving phase difference problems is using the formula \( \Delta\phi = \frac{2\pi}{\lambda} \times \Delta x \). Make sure to correctly extract \( \frac{2\pi}{\lambda} \) (the wave number \( k \)) from the given wave equation and ensure consistent units for \( \Delta x \) and \( \lambda \).

Question 11. The transverse displacement of a string (damped at its both ends) is given by y(x, t) = 0.06 sin (\( \frac{2\pi x}{3} \)) 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\(^{-2}\) kg. Answer the following:
(a) Does the function represent a travelling or a stationary wave?
(b) Interpret wave as a superposition of two waves travelling in opposite directions. What are the wavelength, frequency and speed of each wave?
(c) Determine the tension in the string?
Answer: The given function for transverse displacement is:
\( y(x, t) = 0.06 \sin \left(\frac{2\pi x}{3}\right) \cos (120\pi t) \) (i)

(a) This function separates the \( x \) and \( t \) variables into distinct sine and cosine terms. This specific form, \( y(x,t) = A \sin(kx) \cos(\omega t) \), is characteristic of a **stationary wave** (or standing wave), not a travelling wave.

(b) A stationary wave can be interpreted as the superposition of two travelling waves of the same amplitude, frequency, and speed, moving in opposite directions. The standard representation for such a stationary wave is \( y(x, t) = 2A \sin(kx) \cos(\omega t) \).
Comparing the given equation (i) with \( y(x, t) = A' \sin(kx) \cos(\omega t) \):
We have \( k = \frac{2\pi}{3} \, \text{m}^{-1} \) and \( \omega = 120\pi \, \text{rad/s} \).
For each of the constituent travelling waves:
Wavelength \( \lambda \): \( k = \frac{2\pi}{\lambda} \implies \frac{2\pi}{3} = \frac{2\pi}{\lambda} \implies \lambda = 3 \, \text{m} \).
Frequency \( \nu \): \( \omega = 2\pi\nu \implies 120\pi = 2\pi\nu \implies \nu = \frac{120\pi}{2\pi} = 60 \, \text{Hz} \).
Speed \( v \): \( v = \nu\lambda = 60 \, \text{Hz} \times 3 \, \text{m} = 180 \, \text{m/s} \).
So, each constituent wave has a wavelength of 3 m, a frequency of 60 Hz, and a speed of 180 m/s.

(c) To determine the tension in the string, we use the formula for the speed of a transverse wave on a string: \( v = \sqrt{\frac{T}{m}} \), where \( T \) is tension and \( m \) is the linear mass density.
First, calculate the linear mass density \( m \). The string's length \( L = 1.5 \, \text{m} \) and its mass \( M = 3.0 \times 10^{-2} \, \text{kg} \).
\( m = \frac{M}{L} = \frac{3.0 \times 10^{-2} \, \text{kg}}{1.5 \, \text{m}} = 2.0 \times 10^{-2} \, \text{kg m}^{-1} \).
Now, rearrange the speed formula to find tension: \( T = mv^2 \).
Using the speed found in part (b), \( v = 180 \, \text{m/s} \):
\( T = (2.0 \times 10^{-2} \, \text{kg m}^{-1}) \times (180 \, \text{m/s})^2 \)
\( T = (2.0 \times 10^{-2}) \times (32400) \)
\( T = 648 \, \text{N} \).
In simple words: (a) The given wave is a stationary wave because its formula shows separate parts for position and time. (b) It's like two waves moving against each other, each with a 3-meter wavelength, 60 vibrations per second, and travelling at 180 meters per second. (c) We found the string's thickness per meter, then used the wave's speed to calculate that the string needs to be under 648 Newtons of tension.

Exam Tip: Differentiate clearly between travelling waves and stationary waves based on their mathematical forms. Remember that the speed of a wave on a string depends on both the tension and the linear mass density, so be prepared to calculate both.

Question 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:
(i) The wave described in Exercise 15.11 is a stationary wave, given by \( y(x, t) = 0.06 \sin \left(\frac{2\pi x}{3}\right) \cos (120\pi t) \).
Let's analyze the properties:
(a) **Frequency:** All points on the string (except at nodes) oscillate with the **same frequency**. This is because the time-dependent part \( \cos (120\pi t) \) is common for all points, meaning all particles vibrate with the same angular frequency \( \omega = 120\pi \, \text{rad/s} \), hence the same frequency \( \nu = 60 \, \text{Hz} \).
(b) **Phase:** All points on the string between two consecutive nodes oscillate in the **same phase**. However, points in adjacent segments (separated by a node) oscillate in opposite phases (a phase difference of \( \pi \) radians). For instance, if one segment moves up, the next moves down. So, not all points have the exact same phase across the entire string.
(c) **Amplitude:** No, all points on the string do **not** oscillate with the same amplitude. In a stationary wave, the amplitude of oscillation for a point at position \( x \) is given by \( A(x) = 0.06 \sin \left(\frac{2\pi x}{3}\right) \). This amplitude varies with \( x \). It is zero at the nodes (where \( \sin(\frac{2\pi x}{3}) = 0 \)) and maximum at the antinodes (where \( |\sin(\frac{2\pi x}{3})| = 1 \)).

(ii) We need to find the amplitude of a point 0.375 m away from one end. Let \( x = 0.375 \, \text{m} \).
The amplitude function is \( A(x) = 0.06 \sin \left(\frac{2\pi x}{3}\right) \).
Substitute \( x = 0.375 \):
\( A(0.375) = 0.06 \sin \left(\frac{2\pi \times 0.375}{3}\right) \)
\( A(0.375) = 0.06 \sin \left(\frac{0.75\pi}{3}\right) \)
\( A(0.375) = 0.06 \sin \left(\frac{\pi}{4}\right) \)
We know that \( \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \approx 0.707 \).
\( A(0.375) = 0.06 \times \frac{1}{\sqrt{2}} = \frac{0.06}{\sqrt{2}} \)
\( A(0.375) = \frac{0.06 \times \sqrt{2}}{2} = 0.03 \sqrt{2} \)
\( A(0.375) \approx 0.03 \times 1.414 \approx 0.04242 \, \text{m} \).
Rounding to two significant figures, the amplitude is approximately \( 0.042 \, \text{m} \).
In simple words: (i) In a stationary wave, all parts of the string wiggle at the same rate (frequency). However, their wiggling might be in step or exactly opposite (phase), and how much they wiggle (amplitude) is different depending on where they are on the string, being zero at fixed points and largest in the middle. (ii) For a spot 0.375 meters from an end, its maximum wiggle size (amplitude) is about 0.042 meters.

Exam Tip: For stationary waves, remember that particles oscillate with the same frequency (except nodes), but their amplitude varies with position, and phase can be in-phase or out-of-phase by \( \pi \) depending on the segment.

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:
1. \( y = 2\cos (3x) \sin (10t) \)
2. \( y = 2\sqrt{x-vt} \)
3. \( y = 3 \sin (5x – 0.5t) + 4 \cos (5x – 0.5t) \)
4. \( y = \cos x \sin t + \cos 2x \sin 2t \).
Answer:
1. The function \( y = 2\cos (3x) \sin (10t) \) has separate harmonic functions for \( x \) and \( t \). This form, \( f(x)g(t) \), is characteristic of a **stationary wave**.
2. The function \( y = 2\sqrt{x-vt} \) does not contain any harmonic (sine or cosine) functions. While it has \( (x-vt) \) in combination, the square root means it's not oscillatory. Therefore, this function represents **none at all** (it's not a wave function in the usual sense).
3. The function \( y = 3 \sin (5x – 0.5t) + 4 \cos (5x – 0.5t) \) combines sine and cosine functions, where \( x \) and \( t \) appear in the combination \( (5x - 0.5t) \). This is a general form of a **travelling harmonic wave**. (It can be rewritten as a single harmonic function \( A \sin(kx - \omega t + \phi) \)).
4. The function \( y = \cos x \sin t + \cos 2x \sin 2t \) is a sum of two terms, each of which has separate harmonic functions for \( x \) and \( t \). Each term \( (\cos x \sin t) \) and \( (\cos 2x \sin 2t) \) represents a stationary wave. The superposition of stationary waves also results in a **stationary wave**.
In simple words: 1. This is a stationary wave because its parts for position and time are separated. 2. This is not a wave at all because it lacks the repeating, wavelike pattern of sine or cosine. 3. This is a travelling wave because both terms involve position and time together in the same specific way. 4. This is also a stationary wave, as it's made up of two simpler stationary waves added together.

Exam Tip: To classify wave functions: look for combinations of \( (x \pm vt) \) inside sine/cosine for travelling waves. For stationary waves, look for separate \( f(x) \) and \( g(t) \) harmonic terms multiplied together (e.g., \( \sin(kx) \cos(\omega t) \)). If there are no harmonic functions, it's generally not a wave.

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 density is 4.0 × 10\(^{-2}\) kgm\(^{-1}\). What is (a) the speed of a transverse wave in the string and (b) the tension in the string?
Answer: Here, the frequency of vibration \( \nu = 45 \, \text{Hz} \). The mass of the wire \( M = 3.5 \times 10^{-2} \, \text{kg} \). The linear density (mass per unit length) \( m = 4.0 \times 10^{-2} \, \text{kgm}^{-1} \).

To find the length of the wire \( L \): \( m = \frac{M}{L} \implies L = \frac{M}{m} = \frac{3.5 \times 10^{-2} \, \text{kg}}{4.0 \times 10^{-2} \, \text{kgm}^{-1}} = 0.875 \, \text{m} \).

(a) For a string vibrating in its fundamental mode, the wavelength \( \lambda = 2L \).
\( \lambda = 2 \times 0.875 \, \text{m} = 1.75 \, \text{m} \).
The speed of a transverse wave \( v \) in the string is given by \( v = \nu\lambda \).
\( v = 45 \, \text{Hz} \times 1.75 \, \text{m} = 78.75 \, \text{ms}^{-1} \).
So, the speed of the transverse wave is \( 78.75 \, \text{ms}^{-1} \).

(b) The tension \( T \) in the string can be found using the formula for the speed of a transverse wave: \( v = \sqrt{\frac{T}{m}} \).
Squaring both sides, \( v^2 = \frac{T}{m} \).
So, \( T = mv^2 \).
Substitute the values of \( m \) and \( v \):
\( T = (4.0 \times 10^{-2} \, \text{kgm}^{-1}) \times (78.75 \, \text{ms}^{-1})^2 \)
\( T = (4.0 \times 10^{-2}) \times (6201.5625) \)
\( T = 248.0625 \, \text{N} \).
Rounding to a reasonable number of significant figures (e.g., three, based on input data): \( T \approx 248 \, \text{N} \).
In simple words: First, we used the mass and linear density to figure out the string's length. (a) Then, knowing it vibrates in its simplest way, we found the wavelength and multiplied it by the frequency to get the wave's speed. (b) Lastly, using that wave speed and the linear density, we calculated the tension required in the string.

Exam Tip: In fundamental mode, the length of the string \( L \) is half the wavelength \( \lambda \), i.e., \( \lambda = 2L \). Always derive \( L \) if not given directly, and ensure all units are consistent (SI units are preferred) before applying formulas.

Question 15. A metre-long tube open at one end, with a movable poston 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.
Answer: The tube, being open at one end and closed by a piston at the other, functions as a closed organ pipe. The fixed frequency source is a tuning fork with \( \nu = 340 \, \text{Hz} \).
The resonance lengths for a closed pipe are given as \( l_1 = 25.5 \, \text{cm} = 0.255 \, \text{m} \) and \( l_2 = 79.3 \, \text{cm} = 0.793 \, \text{m} \).
For a closed organ pipe, the resonance frequencies are given by \( \nu_n = (2n-1)\frac{v}{4L} \), where \( n = 1, 2, 3, \ldots \) corresponds to the 1st, 3rd, 5th harmonics, etc.
Let \( l_1 \) correspond to the \( n^{th} \) mode and \( l_2 \) correspond to the \( (n+1)^{th} \) mode.
So, for \( l_1 \): \( \nu = (2n-1)\frac{v}{4l_1} \) (1)
And for \( l_2 \): \( \nu = (2(n+1)-1)\frac{v}{4l_2} = (2n+1)\frac{v}{4l_2} \) (2)
Since the frequency \( \nu \) is the same (340 Hz) for both resonances:
\( (2n-1)\frac{v}{4l_1} = (2n+1)\frac{v}{4l_2} \)
\( \frac{2n-1}{l_1} = \frac{2n+1}{l_2} \)
Rearranging to solve for \( n \):
\( (2n-1)l_2 = (2n+1)l_1 \)
\( 2nl_2 - l_2 = 2nl_1 + l_1 \)
\( 2n(l_2 - l_1) = l_1 + l_2 \)
\( n = \frac{l_1 + l_2}{2(l_2 - l_1)} \)
Substitute the lengths in meters:
\( n = \frac{0.255 + 0.793}{2(0.793 - 0.255)} = \frac{1.048}{2(0.538)} = \frac{1.048}{1.076} \approx 0.974 \).
Since \( n \) must be an integer, this suggests that the initial interpretation of \( n \) and \( n+1 \) might need adjustment or that these are the lowest modes, possibly 1st and 3rd harmonics. Let's reconsider this, or use the concept of successive resonances for a closed pipe where \( l_2 - l_1 = \frac{\lambda}{2} \).
For a closed organ pipe, consecutive resonance lengths for a given frequency are typically \( L_n = \frac{(2n-1)\lambda}{4} \). The difference between two successive resonances is \( \frac{\lambda}{2} \).
So, \( l_2 - l_1 = \frac{\lambda}{2} \).
\( 0.793 \, \text{m} - 0.255 \, \text{m} = 0.538 \, \text{m} \).
Therefore, \( \frac{\lambda}{2} = 0.538 \, \text{m} \implies \lambda = 2 \times 0.538 \, \text{m} = 1.076 \, \text{m} \).
Now, using the wave equation \( v = \nu\lambda \):
\( v = 340 \, \text{Hz} \times 1.076 \, \text{m} \)
\( v = 365.84 \, \text{m/s} \).
Let's verify with the original formula for \( n \). If we assume the resonance lengths are \( \frac{\lambda}{4} \) and \( \frac{3\lambda}{4} \), then \( l_1 = \frac{\lambda}{4} \) and \( l_2 = \frac{3\lambda}{4} \).
Then \( \lambda = 4 l_1 = 4 \times 0.255 \, \text{m} = 1.02 \, \text{m} \).
And \( v = \nu \lambda = 340 \, \text{Hz} \times 1.02 \, \text{m} = 346.8 \, \text{m/s} \).
Also, using \( l_2 = \frac{3\lambda}{4} \implies \lambda = \frac{4l_2}{3} = \frac{4 \times 0.793 \, \text{m}}{3} = \frac{3.172}{3} \approx 1.057 \, \text{m} \).
Then \( v = \nu \lambda = 340 \, \text{Hz} \times 1.057 \, \text{m} \approx 359.38 \, \text{m/s} \).
The method \( l_2 - l_1 = \frac{\lambda}{2} \) accounts for *any* two successive harmonics, so it is generally more robust.
So, the estimated speed of sound in air is \( 365.84 \, \text{m/s} \).
The solution provided in the OCR calculates \( n=1 \) using the initial equations, let's follow that. \( n = \frac{0.255 + 0.793}{2(0.793 - 0.255)} = \frac{1.048}{1.076} \approx 0.974 \). This suggests \( n \) is very close to 1. Let's assume \( n=1 \) for the first resonance \( l_1 \). Then \( \nu = (2(1)-1)\frac{v}{4l_1} = \frac{v}{4l_1} \). \( v = 4\nu l_1 = 4 \times 340 \, \text{Hz} \times 0.255 \, \text{m} = 346.8 \, \text{m/s} \). Using \( n=2 \) for the second resonance \( l_2 \). \( \nu = (2(2)-1)\frac{v}{4l_2} = \frac{3v}{4l_2} \). \( v = \frac{4\nu l_2}{3} = \frac{4 \times 340 \, \text{Hz} \times 0.793 \, \text{m}}{3} \approx \frac{1078.5}{3} \approx 359.5 \, \text{m/s} \). The discrepancy arises from the finite precision of the given lengths. The OCR solution uses \( n=1 \) and then calculates \( v \) using \( l_1 \). So let's use \( n=1 \) with \( l_1 \). \( v = 4 \nu l_1 = 4 \times 340 \, \text{Hz} \times 0.255 \, \text{m} = 346.8 \, \text{m/s} \). This is approximately \( 347 \, \text{m/s} \).
In simple words: We treated the tube as a closed organ pipe. By using the frequency of the tuning fork and the two lengths where resonance occurred, we could calculate the wavelength. Then, multiplying the wavelength by the frequency gave us an estimate for the speed of sound in the air inside the tube.

Exam Tip: For a closed organ pipe, remember that only odd harmonics are possible. The resonance lengths correspond to \( \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \ldots \). The difference between any two successive resonance lengths for the same frequency is always \( \frac{\lambda}{2} \). This relationship is very useful for accurately calculating the wavelength and then the speed of sound.

Question 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?
Answer: A rod clamped in the middle has antinodes (A) at its ends and a node (N) at the clamping point. In the fundamental mode, the length of the rod is half the wavelength.

Here, length of rod, \( l = 100 \) cm
Frequency, \( \nu = 2.53 \) kHz \( = 2.53 \times 10^3 \) Hz
We know that the length of the rod is \( l = \frac{1}{2}\lambda \) or \( \lambda = 2l \)
So, wavelength \( \lambda = 2 \times 100 = 200 \) cm
If \( v \) is the speed of sound in steel, then \( v = \nu\lambda \)
\( v = (2.53 \times 10^3) \times 200 \)
\( v = 506 \times 10^3 \) cms\(^{-1} \)
\( v = 5.06 \times 10^3 \) ms\(^{-1} \)
Therefore, \( v = 5.06 \) kms\(^{-1} \).In simple words: When a steel rod is held in the middle, and it vibrates, the sound travels really fast through it. The speed depends on how long the rod is and how quickly it vibrates.

Exam Tip: For fundamental modes in rods, remember that the clamped points are nodes and free ends are antinodes. The wavelength is directly related to the rod's length based on these conditions.

Question 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 ms⁻¹).
Answer: Here, length of pipe, \( l = 20 \) cm \( = 0.20 \) m
Speed of sound, \( v = 340 \) ms\(^{-1} \)
Frequency of exciting source, \( \nu_n = 430 \) Hz
We know that the frequency of the nth mode of vibration of a closed pipe is given by:
\( \nu_n = (2n-1) \frac{v}{4l} \)
Substituting the given values:
\( 430 = (2n-1) \frac{340}{4 \times 0.20} \)
\( 430 = (2n-1) \frac{340}{0.80} \)
\( 430 = (2n-1) \times 425 \)
So, \( (2n-1) = \frac{430}{425} \approx 1.01 \)
\( 2n - 1 \approx 1 \)
\( 2n \approx 2 \)
\( n \approx 1 \)
This means the organ pipe is in the first harmonic or fundamental mode of vibration.

Now, for an open pipe, the frequency of the nth mode of vibration is given by:
\( \nu_n = n \frac{v}{2l} \)
If the same source \( \nu_n = 430 \) Hz is used:
\( 430 = n \frac{340}{2 \times 0.2} \)
\( 430 = n \frac{340}{0.4} \)
\( 430 = n \times 850 \)
So, \( n = \frac{430}{850} \approx 0.5 \)
Since \( n \) must be an integer (it represents a mode number), \( n = 0.5 \) is not a valid mode. Therefore, the same source cannot be in resonance with the open pipe.

Alternatively, for a closed-end pipe:
Fundamental frequency \( \nu_1 = \frac{v}{4l} = \frac{340}{4 \times 0.2} = \frac{340}{0.8} = 425 \) Hz
This is the first harmonic. Since \( 430 \) Hz is close to \( 425 \) Hz, it excites the first harmonic.
For an open-end pipe:
Fundamental frequency \( \nu_1 = \frac{v}{2l} = \frac{340}{2 \times 0.2} = \frac{340}{0.4} = 850 \) Hz
The source frequency \( 430 \) Hz is not a harmonic of \( 850 \) Hz (like \( 1 \times 850, 2 \times 850, \dots \)). Hence, the same source cannot be in resonance with the open pipe.
In simple words: For a pipe closed at one end, the sound source makes it vibrate in its basic, first mode. But if you open both ends of the pipe, the same sound source won't make it resonate because the numbers don't match up for the way an open pipe vibrates.

Exam Tip: Remember the frequency formulas for open and closed pipes: \( \nu_n = (2n-1) \frac{v}{4l} \) for closed and \( \nu_n = n \frac{v}{2l} \) for open. Always check if 'n' results in an integer for resonance.

Question 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: We know that frequency \( \nu \) is proportional to the square root of tension \( \sqrt{T} \). Therefore, a decrease in the tension of a string reduces its frequency.

Let the original frequency of A be \( \nu_A \) and B be \( \nu_B \).
Original beat frequency = \( |\nu_A - \nu_B| = 6 \) Hz (given)
Given \( \nu_A = 324 \) Hz
So, \( |324 - \nu_B| = 6 \)
This means either \( 324 - \nu_B = 6 \) or \( 324 - \nu_B = -6 \).
Case 1: \( 324 - \nu_B = 6 \implies \nu_B = 324 - 6 = 318 \) Hz
Case 2: \( 324 - \nu_B = -6 \implies \nu_B = 324 + 6 = 330 \) Hz

When the tension in string A is slightly reduced, its frequency \( \nu_A \) decreases.
The new beat frequency is 3 Hz.
If \( \nu_B = 330 \) Hz:
When \( \nu_A \) decreases, the difference \( |\nu_A - 330| \) would initially increase if \( \nu_A \) was \( 324 \) Hz (meaning \( \nu_A < \nu_B \)). For example, if \( \nu_A \) becomes \( 320 \) Hz, then \( |320 - 330| = 10 \) Hz, which is an increase from 6 Hz. But the problem states the beat frequency reduces to 3 Hz. So this case is incorrect.

Therefore, we must consider \( \nu_B = 318 \) Hz:
When \( \nu_A = 324 \) Hz, then \( |\nu_A - \nu_B| = |324 - 318| = 6 \) Hz.
If \( \nu_A \) decreases (e.g., to \( \nu'_A \)), the new beat frequency is \( |\nu'_A - 318| = 3 \) Hz.
For the beat frequency to decrease, \( \nu'_A \) must be closer to \( 318 \) Hz than \( 324 \) Hz was. This happens if \( \nu_A \) was initially higher than \( \nu_B \).
So, \( \nu_A \) decreases from 324 Hz. Let's say it becomes \( 321 \) Hz. Then \( |321 - 318| = 3 \) Hz, which matches the reduced beat frequency.
Thus, the frequency of string B is \( 318 \) Hz.
In simple words: When two sitar strings are slightly off-tune, they make "beats". If you loosen one string (string A) and the beats slow down, it means string A's original sound was higher than string B's. By checking the numbers, string B must have a frequency of 318 Hz.

Exam Tip: For beat frequency problems, always consider both possible values for the unknown frequency, then use the information about how a change (like reducing tension) affects the beat frequency to determine the correct value.

Question 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 wave but only longitudinal waves can propagate in gases, and
(e) of pulse gets distorted during propagation in a dispersive medium.
Answer:
(a) In a sound wave, a displacement node is a point where the amplitude of oscillation (i.e., displacement) is zero. Here, a compression and a rarefaction meet, resulting in maximum pressure, so it is called a pressure antinode. Conversely, a displacement antinode is a point where displacement is maximum, but the pressure is minimum, which makes it a pressure node. Hence, a displacement node coincides with a pressure antinode, and a displacement antinode coincides with a pressure node.
In simple words: In sound waves, where particles don't move (displacement node), the pressure is actually at its highest (pressure antinode). And where particles move the most (displacement antinode), the pressure is lowest (pressure node). They are opposites.

Exam Tip: Clearly differentiate between displacement and pressure properties in sound waves; a node for one is an antinode for the other.

Question 19. (b) bats can ascertain distances, directions, nature and sizes of the obstacles without any'eyes',
Answer: Bats emit ultrasonic waves, which are high-frequency waves with small wavelengths, when they fly. These ultrasonic waves bounce back after reflecting from obstacles. Their ears are incredibly sensitive and well-trained, allowing them to gather information not only about the distance of the obstacle but also its nature and the reflecting surface.
In simple words: Bats use special high-pitched sounds that bounce off things. Their super-sensitive ears listen to these echoes, which tell them how far away things are, what they are made of, and even their shape, without needing to see.

Exam Tip: Focus on the properties of ultrasonic waves (high frequency, small wavelength) and the concept of echo-location to explain how bats navigate.

Question 19. (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
Answer: The quality or timbre of the sound produced by a musical instrument relies on the number and intensity of overtones (harmonics) present. Since the number and relative strengths of overtones differ between the sounds generated by a violin and a sitar, we can easily tell apart the notes they produce, even if they have the same fundamental frequency.
In simple words: Even if a violin and a sitar play the same note (same frequency), they sound different because of their "timbre." This timbre comes from the different extra sounds, called overtones, that each instrument makes along with the main note.

Exam Tip: Explain that timbre, determined by the presence and intensity of overtones, is the key factor allowing us to distinguish sounds from different instruments with the same fundamental frequency.

Question 19. (d) solids can support both longitudinal and transverse wave but only longitudinal waves can propagate in gases, and
Answer: Solids possess both volume elasticity (resistance to change in volume) and shear elasticity (resistance to change in shape). Because of these two types of elasticity, both longitudinal (compressional) and transverse (shear) waves can travel through them. On the other hand, gases only have volume elasticity and lack shear elasticity. Consequently, only longitudinal waves, which involve compressions and rarefactions, can propagate through gases.
In simple words: Solids can handle both pushing-and-pulling waves (longitudinal) and side-to-side waves (transverse) because they can spring back from both squeezing and twisting. But gases can only handle pushing-and-pulling waves, not side-to-side, because they don't resist twisting or changing shape.

Exam Tip: Connect the type of wave propagation (longitudinal/transverse) directly to the elastic properties (volume/shear) of the medium (solid/gas).

Question 19. (e) of pulse gets distorted during propagation in a dispersive medium.
Answer: A sound pulse is essentially a combination of waves with various wavelengths. In a dispersive medium, waves of different wavelengths travel at different speeds and sometimes in different directions. This variation in speed causes the components of the pulse to spread out or change their relative positions, which distorts the original shape of the pulse. Therefore, a plane wavefront does not remain a plane wavefront in a dispersive medium.
In simple words: When a sound "pulse" (a short burst of sound) travels through certain materials (dispersive mediums), it gets messed up. This happens because the different parts of the sound pulse, which have different wavelengths, travel at different speeds and can spread out, changing the pulse's original shape.

Exam Tip: Define a dispersive medium as one where wave speed depends on wavelength, and explain how this leads to pulse distortion due to different components traveling at different speeds.

Question 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 ms⁻¹, (b) recedes from the platform with a speed of 10 ms⁻¹.
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms⁻¹.
Answer: Here, frequency of source, \( \nu = 400 \) Hz
Speed of sound in still air, \( v = 340 \) ms\(^{-1} \)
Speed of source, \( v_s = 10 \) ms\(^{-1} \)

(i) (a) When the train approaches the platform (observer is stationary):
The apparent frequency heard by the observer is given by:
\( \nu' = \nu \left( \frac{v}{v - v_s} \right) \)
\( \nu' = 400 \left( \frac{340}{340 - 10} \right) \)
\( \nu' = 400 \left( \frac{340}{330} \right) \)
\( \nu' \approx 412.1 \) Hz
In simple words: When the train comes closer, the sound waves get squished together, making the whistle sound higher pitched to someone on the platform. The new frequency is about 412.1 Hz.

Exam Tip: For approaching sources, the denominator in the Doppler effect formula uses \( v - v_s \), which leads to an increased apparent frequency. Remember to always multiply by the source frequency \( \nu \).

Question 20. (i) (b) recedes from the platform with a speed of 10 ms⁻¹.
Answer: (b) When the train recedes from the platform (observer is stationary):
The apparent frequency heard by the observer is given by:
\( \nu' = \nu \left( \frac{v}{v + v_s} \right) \)
\( \nu' = 400 \left( \frac{340}{340 + 10} \right) \)
\( \nu' = 400 \left( \frac{340}{350} \right) \)
\( \nu' \approx 388.6 \) Hz \( \approx 389 \) Hz
In simple words: When the train moves away, the sound waves spread out, making the whistle sound lower pitched to someone on the platform. The new frequency is about 389 Hz.

Exam Tip: For receding sources, the denominator in the Doppler effect formula uses \( v + v_s \), leading to a decreased apparent frequency. Make sure to apply the correct sign in the denominator.

Question 20. (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms⁻¹.
Answer: (ii) The speed of sound in the medium (air) remains unchanged in both cases, which is \( 340 \) ms\(^{-1} \). The Doppler effect only alters the perceived frequency due to relative motion, not the actual speed of sound waves in the medium.
In simple words: The actual speed of sound in the air doesn't change whether the train is coming or going. It stays at 340 meters per second. The Doppler effect only makes the pitch sound different, not how fast the sound is traveling.

Exam Tip: Understand that the speed of sound is a property of the medium and does not change due to the relative motion of the source or observer. Only the observed frequency and wavelength change.

Question 21. A train standing in a station yard blows a whistle of frequency 400 Hz in still air. A wind starts blowing in the direction from the yard to the station with a speed of 10 ms⁻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 case when the air is still and the observer runs towards the yard at a speed of 10 ms⁻¹? The speed of sound in still air can be taken as 340 ms⁻¹.
Answer: Here, source frequency, \( \nu = 400 \) Hz
Speed of sound in still air, \( v = 340 \) ms\(^{-1} \)
Speed of wind, \( w = 10 \) ms\(^{-1} \)

(a) When wind blows towards the listener at rest:
For a stationary observer on the platform, the frequency does not change because there is no relative motion between the source (train) and the observer. So, frequency received by observer \( \nu' = \nu = 400 \) Hz.
Since the wind is blowing in the same direction as the sound waves are moving, the apparent velocity of sound \( v' \) is:
\( v' = v + w \)
\( v' = 340 + 10 = 350 \) ms\(^{-1} \)
The wavelength of sound for the stationary listener is:
\( \lambda' = \frac{v'}{\nu} = \frac{350}{400} = 0.875 \) m
In simple words: When wind blows along with the sound, the listener still hears the same whistle pitch because they aren't moving relative to the train. But the sound travels faster because of the wind, so its wavelength changes. The speed of sound becomes 350 m/s, and the wavelength is 0.875 m.

Exam Tip: Remember that wind affects the effective speed of sound but not the frequency observed by a stationary listener relative to the source. The wavelength will adjust according to \( \lambda = v/\nu \).

Question 21. (b) Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms⁻¹?
Answer: (b) When the air is still and the listener moves towards the stationary source (train) with speed \( v_o = 10 \) ms\(^{-1} \):
The apparent frequency \( \nu'' \) for the observer is given by:
\( \nu'' = \nu \left( \frac{v + v_o}{v} \right) \)
\( \nu'' = 400 \left( \frac{340 + 10}{340} \right) \)
\( \nu'' = 400 \left( \frac{350}{340} \right) \)
\( \nu'' \approx 411.76 \) Hz \( \approx 412 \) Hz
In this case, the wavelength of sound remains unchanged, as the medium is still. Only the relative speed between the sound waves and the observer changes.
Since the observed frequency (412 Hz) is different from the previous case (400 Hz), and the wavelength also changes in case (a) but not in case (b), the two situations are not exactly identical.
In simple words: If the air is still but the listener runs towards the train, they hear a higher pitch (about 412 Hz) because they are moving into the sound waves. This is different from the wind blowing case because the observed frequency and wavelength are not the same in both scenarios.

Exam Tip: Distinguish between the effect of wind on sound speed (and thus wavelength) and the Doppler effect due to observer/source motion (which changes frequency and relative speed). These are not always identical situations.