WAVES-Exercises

Q1. 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

To determine the time taken by the disturbance to travel along the string, we first need the speed of transverse waves on a stretched string. This speed depends on the tension in the string and its mass per unit length.

The mass per unit length of the string is obtained by dividing the total mass by the total length of the stretched string. Using the given values, the linear mass density is calculated as

\[ \mu = \frac{m}{L} = \frac{2.50}{20.0} = 0.125 \ \text{kg m}^{-1} \]

The speed of a transverse wave travelling along a stretched string under tension is given by

\[ v = \sqrt{\frac{T}{\mu}} \]

Substituting the known values of tension and linear mass density, we obtain

\[ \begin{aligned} v &= \sqrt{\frac{200}{0.125}} \\ &= \sqrt{1600} \\ &= 40 \ \text{m s}^{-1} \end{aligned} \]

The time taken by the disturbance to travel the full length of the string is the ratio of distance to wave speed. Therefore,

\[ \begin{aligned} t &= \frac{L}{v} \\ &= \frac{20.0}{40} \\ &= 0.50 \ \text{s} \end{aligned} \]

Hence, the transverse disturbance takes 0.50 seconds to reach the other end of the string.

Q2. A stone dropped from the top of a tower of height 300 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 = \(\mathrm{9.8\ m\ s^{–2}}\))

Solution

When the stone is released from rest at the top of the tower, it first takes some time to fall freely through air before splashing into the water. This time of fall is determined using the equation of motion for uniformly accelerated motion under gravity.

\[ \begin{aligned} h &= \frac{1}{2} g t_1^2 \\ 300 &= \frac{1}{2} \times 9.8 \times t_1^2 \end{aligned} \]

Solving for the time of fall,

\[ \begin{aligned} t_1^2 &= \frac{600}{9.8} \\ t_1 &\approx 7.83 \ \text{s} \end{aligned} \]

After the stone hits the water, the sound of the splash travels upward through air to the top of the tower. The time taken by sound to travel this distance is obtained using the relation between speed, distance, and time.

\[ \begin{aligned} t_2 &= \frac{\text{distance}}{\text{speed of sound}} \\ &= \frac{300}{340} \\ &\approx 0.88 \ \text{s} \end{aligned} \]

The total time after the stone is dropped at which the splash is heard at the top is the sum of the time of fall and the time taken by sound to travel upward.

\[ \begin{aligned} t &= t_1 + t_2 \\ &= 7.83 + 0.88 \\ &\approx 8.71 \ \text{s} \end{aligned} \]

Thus, the splash is heard at the top of the tower approximately 8.7 seconds after the stone is released.

Q3. 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\(\mathrm{ m\ s^{–1}}\).

Solution

The speed of a transverse wave on a stretched wire depends on the tension in the wire and its mass per unit length. To make this speed equal to the speed of sound in air, we first determine the linear mass density of the steel wire using its given mass and length.

\[ \mu = \frac{m}{L} = \frac{2.10}{12.0} = 0.175 \ \text{kg m}^{-1} \]

The speed of a transverse wave on a stretched wire is given by the relation

\[ v = \sqrt{\frac{T}{\mu}} \]

Since the wave speed is required to be equal to the speed of sound in air at 20 °C, we substitute \(v = 343 \ \text{m s}^{-1}\) and rearrange the expression to obtain the tension in the wire.

\[ \begin{aligned} T &= \mu v^2 \\ &= 0.175 \times (343)^2 \end{aligned} \]

Evaluating the expression,

\[ \begin{aligned} T &= 0.175 \times 117\,649 \\ &\approx 2.06 \times 10^{4} \ \text{N} \end{aligned} \]

Therefore, the tension required in the steel wire so that the transverse wave travels with the speed of sound in air is approximately \(2.06 \times 10^{4}\ \text{N}\).

Q4. Use the formula \(v = \sqrt{\dfrac{γρ}{P}}\) to explain why the speed of sound in air
(a)is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.

Solution

The speed of sound in a gas is given by the relation \( v = \sqrt{\dfrac{\gamma P}{\rho}} \), where \( \gamma \) is the ratio of specific heats, \( P \) is the pressure of the gas, and \( \rho \) is its density. This expression helps us understand how different physical factors affect the propagation of sound in air.

To explain the effect of pressure, we recall that for an ideal gas at constant temperature, the density is directly proportional to pressure. Using the gas law, the density of air can be written as

\[ \rho = \frac{P M}{R T} \]

Substituting this expression for density into the formula for speed of sound, we obtain

\[ \begin{aligned} v &= \sqrt{\frac{\gamma P}{\rho}} \\ &= \sqrt{\frac{\gamma P}{\dfrac{P M}{R T}}} \\ &= \sqrt{\frac{\gamma R T}{M}} \end{aligned} \]

Since pressure no longer appears in the final expression, the speed of sound in air is independent of pressure, provided the temperature remains constant.

The above result also shows that the speed of sound is directly proportional to the square root of absolute temperature. As the temperature increases, the average kinetic energy of air molecules increases, enabling sound waves to travel faster. Hence, the speed of sound in air increases with temperature.

Humidity refers to the presence of water vapour in air. When humidity increases, lighter water vapour molecules partially replace heavier oxygen and nitrogen molecules, reducing the average molecular mass of air. From the relation

\[ v = \sqrt{\frac{\gamma R T}{M}} \]

it is clear that a decrease in the effective molecular mass \( M \) results in an increase in the speed of sound. Therefore, sound travels faster in humid air than in dry air.

Thus, the given formula explains why the speed of sound in air is independent of pressure, increases with temperature, and increases with humidity.

Q5. 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 \(x – v t\) or \(x + v t\), i.e. \(y = f (x ± v t)\). 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) \(1/(x + vt)\)

Solution

A one-dimensional travelling wave is any disturbance whose shape moves without change along the x-axis with a constant speed. Mathematically, this requirement is satisfied if the displacement can be written as a function of the single combined variable \(x - vt\) or \(x + vt\). Therefore, any expression of the form \(y = f(x \pm vt)\), where \(f\) is an arbitrary function, represents a travelling wave. Hence, the converse statement is indeed true.

To examine the given functions, we check whether the displacement depends only on a single combination of position and time.

For the function \(y = (x - vt)^2\), the displacement depends entirely on the variable \((x - vt)\). Although the shape of the wave is parabolic rather than sinusoidal, the entire profile shifts unchanged in the positive x-direction with speed \(v\). Therefore, this function does represent a travelling wave.

In the case \(y = \log\!\left(\frac{x + vt}{x_0}\right)\), the displacement again depends only on the single variable \((x + vt)\), with \(x_0\) being a constant introduced for dimensional consistency. Since the form of the function remains unchanged as it moves in the negative x-direction with speed \(v\), this function can also represent a travelling wave.

For the function \(y = \dfrac{1}{x + vt}\), the dependence is solely on the combined variable \((x + vt)\). This implies that the entire waveform moves rigidly along the x-axis without distortion, travelling in the negative direction with speed \(v\). Hence, this function too represents a travelling wave.

Thus, all the given functions satisfy the condition of depending only on \(x \pm vt\) and can therefore represent travelling waves, confirming that the converse statement is true.

Q6. 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 \(\mathrm{m\ s^{–1}}\) and in water 1486 \(\mathrm{m\ s^{–1}}\)

Solution

The bat emits ultrasonic sound waves of frequency \(1000\ \text{kHz}\), which is equal to \(1.0 \times 10^{6}\ \text{Hz}\). When these waves strike the water surface, part of the sound is reflected back into air and part is transmitted into water. It is important to note that the frequency of sound remains unchanged during reflection and transmission, while the wavelength changes according to the speed of sound in the medium.

For the reflected sound, the wave continues to travel in air. The wavelength of sound in a medium is given by the relation

\[ \lambda = \frac{v}{f} \]

Substituting the given speed of sound in air and the frequency of the ultrasonic wave, we obtain

\[ \begin{aligned} \lambda_{\text{air}} &= \frac{340}{1.0 \times 10^{6}} \\ &= 3.4 \times 10^{-4}\ \text{m} \end{aligned} \]

Thus, the wavelength of the reflected ultrasonic sound in air is \(3.4 \times 10^{-4}\ \text{m}\).

For the transmitted sound, the wave travels through water. Although the frequency remains the same, the speed of sound in water is much higher, leading to a different wavelength. Using the same relation,

\[ \begin{aligned} \lambda_{\text{water}} &= \frac{1486}{1.0 \times 10^{6}} \\ &= 1.486 \times 10^{-3}\ \text{m} \end{aligned} \]

Hence, the wavelength of the reflected sound in air is \(3.4 \times 10^{-4}\ \text{m}\), while the wavelength of the transmitted sound in water is \(1.486 \times 10^{-3}\ \text{m}\).

Q7. 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 \(\mathrm{km\ s^{–1}}\) ? The operating frequency of the scanner is 4.2 MHz.

Solution

An ultrasonic scanner operates by transmitting high-frequency sound waves into body tissue and analysing the reflected waves. The wavelength of these waves inside the tissue depends on the speed of sound in the tissue and the operating frequency of the scanner.

The given speed of sound in the tissue is \(1.7\ \text{km s}^{-1}\), which is first converted into SI units for consistency.

\[ v = 1.7 \times 10^{3}\ \text{m s}^{-1} \]

The operating frequency of the ultrasonic scanner is \(4.2\ \text{MHz}\), which can be written as

\[ f = 4.2 \times 10^{6}\ \text{Hz} \]

The wavelength of sound is related to its speed and frequency by the expression

\[ \lambda = \frac{v}{f} \]

Substituting the given values, we obtain

\[ \begin{aligned} \lambda &= \frac{1.7 \times 10^{3}}{4.2 \times 10^{6}} \\ &\approx 4.05 \times 10^{-4}\ \text{m} \end{aligned} \]

Therefore, the wavelength of the ultrasonic sound in the tissue is approximately \(4.05 \times 10^{-4}\ \text{m}\), which is suitable for resolving small structures such as tumours.

Q8. A transverse harmonic wave on a string is described by \(y(x, t) = 3.0 \sin (36 t + 0.018 x + π/4)\) 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 given transverse wave is described by the equation \( y(x,t) = 3.0 \sin(36t + 0.018x + \pi/4) \), where the displacement depends on both position and time through a combined argument. Since the argument of the sine function contains the term \(36t + 0.018x\), which can be written in the form \((\omega t + kx)\), the wave represents a travelling wave rather than a stationary wave.

A travelling wave of the form \(y = A \sin(\omega t + kx + \phi)\) propagates in the negative x-direction. Comparing the given equation with the standard form, we identify the angular frequency \(\omega = 36\ \text{rad s}^{-1}\) and the wave number \(k = 0.018\ \text{cm}^{-1}\). The speed of propagation of the wave is given by the ratio \(\omega/k\).

\[ \begin{aligned} v &= \frac{\omega}{k} \\ &= \frac{36}{0.018} \\ &= 2000\ \text{cm s}^{-1} \end{aligned} \]

Thus, the wave travels with a speed of \(2000\ \text{cm s}^{-1}\) or \(20\ \text{m s}^{-1}\) in the negative x-direction, that is, from right to left.

The amplitude of the wave is the coefficient of the sine function. Hence, the amplitude is \(3.0\ \text{cm}\). The frequency of the wave is related to the angular frequency by \(f = \omega / 2\pi\).

\[ \begin{aligned} f &= \frac{36}{2\pi} \\ &\approx 5.73\ \text{Hz} \end{aligned} \]

The initial phase of the wave at the origin is obtained by setting \(x = 0\) and \(t = 0\) in the wave equation. This gives the phase equal to \(\pi/4\). Therefore, the initial phase at the origin is \(\pi/4\ \text{rad}\).

The least distance between two successive crests corresponds to the wavelength of the wave. The wavelength is related to the wave number by \(\lambda = 2\pi/k\).

\[ \begin{aligned} \lambda &= \frac{2\pi}{0.018} \\ &\approx 349\ \text{cm} \end{aligned} \]

Hence, the wave is a travelling wave moving from right to left with a speed of \(20\ \text{m s}^{-1}\), an amplitude of \(3.0\ \text{cm}\), a frequency of about \(5.73\ \text{Hz}\), an initial phase of \(\pi/4\), and a wavelength of approximately \(3.49\ \text{m}\).

Q9. For the wave described in Exercise 14.8, 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

The transverse wave considered here is the same as in Exercise 14.8 and is given by the equation \( y(x,t) = 3.0 \sin(36t + 0.018x + \pi/4) \), where displacement \(y\) is in cm, position \(x\) is in cm, and time \(t\) is in seconds. To study how the displacement varies with time at fixed positions, we substitute specific values of \(x\) into the wave equation.

At the position \(x = 0\), the displacement becomes

\[ \begin{aligned} y(0,t) &= 3.0 \sin(36t + \pi/4) \end{aligned} \]

This represents a simple harmonic variation of displacement with time. The graph of \(y\) versus \(t\) at \(x = 0\) is therefore a sine curve with amplitude 3.0 cm and angular frequency 36 rad s\(^{-1}\).

At the position \(x = 2\ \text{cm}\), the displacement is given by

\[ \begin{aligned} y(2,t) &= 3.0 \sin(36t + 0.036 + \pi/4) \end{aligned} \]

This equation also represents simple harmonic motion with time. The shape of the \(y\) versus \(t\) graph remains sinusoidal, having the same amplitude and frequency as at \(x = 0\), but the phase is shifted by an additional amount due to the term \(0.018x\).

Similarly, at the position \(x = 4\ \text{cm}\), the displacement becomes

\[ \begin{aligned} y(4,t) &= 3.0 \sin(36t + 0.072 + \pi/4) \end{aligned} \]

Once again, the displacement varies sinusoidally with time. The graph of \(y\) versus \(t\) has the same amplitude and frequency as the earlier cases, but it is further shifted along the time axis because of the increased phase term.

Thus, for \(x = 0\), \(2\ \text{cm}\), and \(4\ \text{cm}\), all displacement–time graphs are sine curves of identical shape. The oscillatory motion at different points of a travelling wave does not differ in amplitude or frequency. The only difference from one point to another is the phase of oscillation, which changes continuously with position along the direction of wave propagation.

Hence, every particle of the medium executes simple harmonic motion with the same amplitude and frequency, while the phase varies from point to point, giving rise to the travelling nature of the wave.

Q10. For the travelling harmonic wave \(y(x, t) = 2.0 \cos 2π (10t – 0.0080 x + 0.35)\) 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 travelling harmonic wave is described by the equation \( y(x,t) = 2.0 \cos 2\pi(10t - 0.0080x + 0.35) \). From this expression, it is clear that the phase of the wave is the quantity inside the cosine multiplied by \(2\pi\). The term involving position, \(-0.0080x\), determines how the phase changes from one point to another along the string.

The phase difference between two points separated by a distance \(\Delta x\) is obtained from the spatial part of the phase. Hence, the phase difference is

\[ \begin{aligned} \Delta \phi &= 2\pi (0.0080\, \Delta x) \end{aligned} \]

Here, \(\Delta x\) must be expressed in centimetres, since \(x\) is given in cm. The wavelength of the wave can also be obtained from the wave equation by noting that one complete phase change of \(2\pi\) corresponds to one wavelength.

\[ \begin{aligned} 0.0080\, \lambda &= 1 \\ \lambda &= 125\ \text{cm} = 1.25\ \text{m} \end{aligned} \]

For a separation of \(4\ \text{m}\), the distance in centimetres is \(400\ \text{cm}\). Substituting into the phase difference expression,

\[ \begin{aligned} \Delta \phi &= 2\pi (0.0080 \times 400) \\ &= 2\pi \times 3.2 \\ &= 6.4\pi\ \text{rad} \end{aligned} \]

For a separation of \(0.5\ \text{m}\), which corresponds to \(50\ \text{cm}\),

\[ \begin{aligned} \Delta \phi &= 2\pi (0.0080 \times 50) \\ &= 2\pi \times 0.4 \\ &= 0.8\pi\ \text{rad} \end{aligned} \]

If the separation between the two points is \(\lambda/2\), then \(\Delta x = \lambda/2 = 62.5\ \text{cm}\). The phase difference becomes

\[ \begin{aligned} \Delta \phi &= 2\pi (0.0080 \times 62.5) \\ &= 2\pi \times 0.5 \\ &= \pi\ \text{rad} \end{aligned} \]

For a separation of \(3\lambda/4\), the distance is \(93.75\ \text{cm}\). Substituting this value,

\[ \begin{aligned} \Delta \phi &= 2\pi (0.0080 \times 93.75) \\ &= 2\pi \times 0.75 \\ &= \frac{3\pi}{2}\ \text{rad} \end{aligned} \]

Thus, the phase differences between the two points are \(6.4\pi\) rad for 4 m, \(0.8\pi\) rad for 0.5 m, \(\pi\) rad for a separation of \(\lambda/2\), and \(3\pi/2\) rad for a separation of \(3\lambda/4\).

Q11. The transverse displacement of a string (clamped at its both ends) is given by \(y(x, t) = 0.06 sin \frac{2\pi}{3}x cos (120 \pi t)\) where \(x\) and \(y\) are in \(m\) and \(t\) in s. The length of the string is 1.5 m and its mass is \(\mathrm{3.0 ×10^{–2}\ kg}\). 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.

Solution

The transverse displacement of the string is given by \( y(x,t) = 0.06 \sin\!\left(\frac{2\pi}{3}x\right)\cos(120\pi t) \). The displacement is expressed as a product of a function of position and a function of time. Such a form is characteristic of a stationary wave rather than a travelling wave, because the spatial and temporal dependences are separated and fixed nodes are present along the string.

A stationary wave can be interpreted as the superposition of two identical travelling waves moving in opposite directions. Using the trigonometric identity

\[ \sin kx \cos \omega t = \frac{1}{2}\big[\sin(kx-\omega t)+\sin(kx+\omega t)\big], \]

the given expression may be rewritten as

\[ \begin{aligned} y(x,t) &= 0.06 \sin\!\left(\frac{2\pi}{3}x\right)\cos(120\pi t) \\ &= 0.03 \Big[\sin\!\left(\frac{2\pi}{3}x - 120\pi t\right) + \sin\!\left(\frac{2\pi}{3}x + 120\pi t\right)\Big]. \end{aligned} \]

This shows that the wave is formed by two travelling waves of equal amplitude \(0.03\ \text{m}\), one moving in the positive x-direction and the other in the negative x-direction. From this expression, the wave number is \(k = \frac{2\pi}{3}\ \text{rad m}^{-1}\) and the angular frequency is \(\omega = 120\pi\ \text{rad s}^{-1}\).

The wavelength of each travelling wave is obtained from \( \lambda = \frac{2\pi}{k} \), which gives

\[ \begin{aligned} \lambda &= \frac{2\pi}{2\pi/3} \\ &= 3\ \text{m}. \end{aligned} \]

The frequency of the wave is related to the angular frequency by \( f = \frac{\omega}{2\pi} \). Hence,

\[ \begin{aligned} f &= \frac{120\pi}{2\pi} \\ &= 60\ \text{Hz}. \end{aligned} \]

The speed of each travelling wave is then given by \( v = f\lambda \), which yields

\[ \begin{aligned} v &= 60 \times 3 \\ &= 180\ \text{m s}^{-1}. \end{aligned} \]

To determine the tension in the string, we first calculate its linear mass density. The mass of the string is \(3.0 \times 10^{-2}\ \text{kg}\) and its length is \(1.5\ \text{m}\), so

\[ \begin{aligned} \mu &= \frac{m}{L} \\ &= \frac{3.0 \times 10^{-2}}{1.5} \\ &= 2.0 \times 10^{-2}\ \text{kg m}^{-1}. \end{aligned} \]

The speed of a transverse wave on a stretched string is related to the tension \(T\) by \( v = \sqrt{\frac{T}{\mu}} \). Solving for the tension,

\[ \begin{aligned} T &= \mu v^2 \\ &= (2.0 \times 10^{-2})(180)^2 \\ &= 648\ \text{N}. \end{aligned} \]

Thus, the given expression represents a stationary wave formed by two oppositely travelling waves of wavelength 3 m, frequency 60 Hz, and speed 180 m s\(^{-1}\), and the tension in the string is 648 N.

Q12. (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

The wave referred to in Exercise 15.11 is described by the stationary wave equation \( y(x,t) = 0.06 \sin\!\left(\frac{2\pi}{3}x\right)\cos(120\pi t) \). In a stationary wave, every point on the string executes simple harmonic motion with the same time-dependent factor. Since the cosine term \( \cos(120\pi t) \) is common to all points, all particles oscillate with the same angular frequency and hence the same frequency.

However, the phase of oscillation is not the same for all points. The displacement depends on the position through the factor \( \sin\!\left(\frac{2\pi}{3}x\right) \), which can be positive or negative. Points lying between two adjacent nodes on the same loop oscillate in the same phase, whereas points on opposite sides of a node oscillate in opposite phases.

The amplitude of oscillation also varies from point to point. In a stationary wave, the amplitude is zero at the nodes and maximum at the antinodes. From the given equation, the amplitude at a position \(x\) is

\[ A(x) = 0.06 \left| \sin\!\left(\frac{2\pi}{3}x\right) \right| \]

To find the amplitude of a point located \(0.375\ \text{m}\) away from one end, we substitute \(x = 0.375\ \text{m}\) into the above expression:

\[ \begin{aligned} A &= 0.06 \sin\!\left(\frac{2\pi}{3} \times 0.375\right) \\ &= 0.06 \sin\!\left(\frac{\pi}{4}\right) \\ &= 0.06 \times \frac{1}{\sqrt{2}} \\ &\approx 4.24 \times 10^{-2}\ \text{m} \end{aligned} \]

Thus, all points on the string oscillate with the same frequency, their phases depend on position, their amplitudes vary along the string, and the amplitude at a point 0.375 m from one end is approximately \(4.24 \times 10^{-2}\ \text{m}\).

Q13. 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) \(y = cos x sin t + cos 2x sin 2t\)

Solution

An elastic wave represents a physical disturbance that must be finite, single-valued, and well defined for all relevant values of position and time. A travelling wave is characterised by a dependence on the combined variable \(x \pm vt\), whereas a stationary wave typically appears as a product of a function of position and a function of time, indicating fixed nodes and antinodes.

For the function \(y = 2\cos(3x)\sin(10t)\), the displacement is written as a product of a purely spatial term and a purely temporal term. This is the standard mathematical form of a stationary wave, showing that different points oscillate with the same frequency but different amplitudes depending on position. Hence, this function represents a stationary wave.

In the expression \(y = 2\sqrt{x - vt}\), although the variables appear in the combination \((x - vt)\), the square-root dependence causes the displacement to be undefined for \(x - vt < 0\). Since a physical wave must be well defined and finite over the entire region of propagation, this function cannot represent a valid elastic wave. Therefore, it represents none at all.

The function \(y = 3\sin(5x - 0.5t) + 4\cos(5x - 0.5t)\) contains both terms depending on the same combined variable \((5x - 0.5t)\). These terms can be combined into a single harmonic function of the form \(A\sin(5x - 0.5t + \phi)\). Since the displacement depends only on \((x - vt)\), this expression represents a travelling wave.

For the function \(y = \cos x \sin t + \cos 2x \sin 2t\), each term individually represents a stationary wave, but they have different angular frequencies and different spatial dependences. Their superposition does not produce a single stationary wave pattern with fixed nodes, nor does it reduce to a function of \((x \pm vt)\). Hence, this expression does not represent a pure travelling wave or a stationary wave.

Thus, the first function represents a stationary wave, the second represents no physical wave, the third represents a travelling wave, and the fourth represents none at all.

Q14. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is \(\mathrm{3.5 × 10^{–2}\ kg}\) and its linear mass density is \(\mathrm{4.0× 10^{–2}\ kg\ m^{–1}}\). What is
(a) the speed of a transverse wave on the string, and
(b) the tension in the string?

Solution

The wire is stretched between two rigid supports and vibrates in its fundamental mode. In this mode, the length of the wire is equal to half the wavelength of the stationary wave formed on it. The given mass of the wire and its linear mass density allow us to first determine the length of the wire.

The length of the wire is obtained using the relation between mass and linear mass density:

\[ \begin{aligned} L &= \frac{m}{\mu} \\ &= \frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}} \\ &= 0.875\ \text{m}. \end{aligned} \]

In the fundamental mode of vibration, the wavelength is related to the length of the string by \( \lambda = 2L \). Therefore, the wavelength of the wave on the wire is

\[ \begin{aligned} \lambda &= 2 \times 0.875 \\ &= 1.75\ \text{m}. \end{aligned} \]

The speed of a transverse wave on the string is given by the product of frequency and wavelength. Using the given frequency of 45 Hz,

\[ \begin{aligned} v &= f\lambda \\ &= 45 \times 1.75 \\ &= 78.75\ \text{m s}^{-1}. \end{aligned} \]

Thus, the speed of the transverse wave on the string is \(78.75\ \text{m s}^{-1}\).

The tension in the string is related to the wave speed and linear mass density by the relation \( v = \sqrt{\frac{T}{\mu}} \). Solving this for the tension,

\[ \begin{aligned} T &= \mu v^2 \\ &= (4.0 \times 10^{-2})(78.75)^2 \\ &\approx 2.48 \times 10^{2}\ \text{N}. \end{aligned} \]

Hence, the speed of the transverse wave on the wire is approximately \(78.8\ \text{m s}^{-1}\), and the corresponding tension in the string is about \(2.5 \times 10^{2}\ \text{N}\).

Q15. 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

The tube is open at one end and closed at the other by a movable piston, so it behaves as a closed organ pipe. Resonance occurs when the length of the air column corresponds to an odd multiple of one-quarter of the wavelength of sound.

If successive resonant lengths are observed for the same frequency, the difference between these lengths corresponds to half the wavelength of sound in air. The two given resonant lengths are \(L_1 = 25.5\ \text{cm}\) and \(L_2 = 79.3\ \text{cm}\).

The difference between these lengths is

\[ \begin{aligned} \Delta L &= L_2 - L_1 \\ &= 79.3 - 25.5 \\ &= 53.8\ \text{cm}. \end{aligned} \]

For a closed tube, the difference between two successive resonant lengths is equal to \(\lambda/2\). Hence, the wavelength of sound in air is

\[ \begin{aligned} \lambda &= 2 \Delta L \\ &= 2 \times 53.8 \\ &= 107.6\ \text{cm} \\ &= 1.076\ \text{m}. \end{aligned} \]

The speed of sound in air is obtained using the relation \(v = f\lambda\), where the frequency of the tuning fork is 340 Hz.

\[ \begin{aligned} v &= 340 \times 1.076 \\ &\approx 3.66 \times 10^{2}\ \text{m s}^{-1}. \end{aligned} \]

Thus, the speed of sound in air at the temperature of the experiment is approximately \(3.66 \times 10^{2}\ \text{m s}^{-1}\).

Q16. 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

The steel rod has a total length of 100 cm and is clamped at its middle. When a rod is clamped at the centre, longitudinal vibrations form with a node at the middle and antinodes at both free ends. In the fundamental mode of longitudinal vibration, the length of the rod corresponds to half the wavelength of the sound wave travelling through the material.

Thus, the wavelength of the longitudinal wave in the rod is

\[ \begin{aligned} \lambda &= 2L \\ &= 2 \times 1.0 \\ &= 2.0\ \text{m}. \end{aligned} \]

The fundamental frequency of longitudinal vibrations of the rod is given as \(2.53\ \text{kHz}\), which is expressed in SI units as

\[ f = 2.53 \times 10^{3}\ \text{Hz}. \]

The speed of sound in the steel rod is related to the frequency and wavelength by the relation \(v = f\lambda\). Substituting the known values,

\[ \begin{aligned} v &= f\lambda \\ &= (2.53 \times 10^{3}) \times 2.0 \\ &= 5.06 \times 10^{3}\ \text{m s}^{-1}. \end{aligned} \]

Therefore, the speed of sound in steel is approximately \(5.06 \times 10^{3}\ \text{m s}^{-1}\).

Q17. 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 \(\mathrm{340\ m\ s^{–1}}\)).

Solution

The pipe has a length of 20 cm, which is equal to 0.20 m, and is initially closed at one end. Such a pipe supports only odd harmonics. For a pipe closed at one end, the resonant frequencies are given by

\[ f_n = \frac{(2n-1)v}{4L}, \quad n = 1,2,3,\ldots \]

Using the given speed of sound \(v = 340\ \text{m s}^{-1}\) and the length \(L = 0.20\ \text{m}\), the fundamental frequency of the pipe is

\[ \begin{aligned} f_1 &= \frac{340}{4 \times 0.20} \\ &= \frac{340}{0.80} \\ &= 425\ \text{Hz}. \end{aligned} \]

The source frequency is 430 Hz, which is very close to the fundamental frequency of the pipe. Hence, the pipe closed at one end is resonantly excited in its fundamental mode, which corresponds to the first harmonic.

Now consider the same pipe when both ends are open. An open pipe supports all harmonics, and its resonant frequencies are given by

\[ f_n = \frac{n v}{2L}, \quad n = 1,2,3,\ldots \]

The fundamental frequency of the open pipe is therefore

\[ \begin{aligned} f_1 &= \frac{340}{2 \times 0.20} \\ &= \frac{340}{0.40} \\ &= 850\ \text{Hz}. \end{aligned} \]

Since 430 Hz is neither equal to the fundamental frequency nor to any integral multiple of it, the given source frequency does not match any resonant mode of the pipe when both ends are open.

Thus, the 430 Hz source excites the fundamental mode of the pipe when one end is closed, but it will not produce resonance when the pipe is open at both ends.

Q18. 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

When two strings of nearly equal frequencies are played together, beats are produced whose frequency is equal to the absolute difference between the two frequencies. Let the original frequency of string A be \(f_A = 324\ \text{Hz}\) and the frequency of string B be \(f_B\).

Initially, the beat frequency is given as 6 Hz. Therefore,

\[ \begin{aligned} |f_A - f_B| &= 6 \\ |324 - f_B| &= 6. \end{aligned} \]

This gives two possible values for \(f_B\),

\[ f_B = 330\ \text{Hz} \quad \text{or} \quad f_B = 318\ \text{Hz}. \]

Now, the tension in string A is slightly reduced. Since the frequency of a stretched string is proportional to the square root of its tension, a reduction in tension causes the frequency of string A to decrease. As a result, the new frequency of string A becomes less than 324 Hz.

After reducing the tension, the beat frequency is observed to be 3 Hz. This means the difference between the new frequency of string A and the frequency of string B has decreased. Such a reduction in beat frequency is possible only if the frequency of string B is lower than the original frequency of string A. Hence, string B must originally have had a frequency less than 324 Hz.

From the two possible values obtained earlier, the physically consistent value is

\[ f_B = 318\ \text{Hz}. \]

Therefore, the frequency of sitar string B is \(318\ \text{Hz}\).

Q19. 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

In a sound wave, the particles of the medium oscillate back and forth along the direction of wave propagation, producing regions of compression and rarefaction. At a displacement node, the particles do not move from their mean positions, but the surrounding particles crowd together and spread apart most strongly. This produces maximum pressure variation at that point, making it a pressure antinode. Conversely, at a displacement antinode, particles oscillate with maximum amplitude, so their instantaneous positions are spread out in such a way that pressure variations cancel out, resulting in a pressure node.

Bats are able to determine distances, directions, shapes, and even the nature of obstacles through a process known as echolocation. They emit short bursts of ultrasonic sound and listen carefully to the echoes reflected from surrounding objects. By measuring the time delay between emission and reception of the echo, a bat can estimate the distance of an object. Differences in intensity and frequency of the reflected sound, caused by the motion, size, and texture of the object, help the bat infer direction, shape, and surface characteristics without relying on vision.

Two musical notes, such as those from a violin and a sitar, may have the same fundamental frequency and hence the same pitch, but they can still sound different because of their timbre or quality. This difference arises from the presence of different overtones and harmonics and their relative intensities. Each instrument produces a unique combination of harmonics along with the fundamental frequency, giving rise to a characteristic waveform that the human ear can distinguish.

Solids can support both longitudinal and transverse waves because they possess elastic properties that allow them to resist both compression and shear deformation. In contrast, gases can sustain only longitudinal waves because they have negligible resistance to shear stress. As a result, transverse disturbances cannot propagate through gases, while compressional waves such as sound can travel efficiently.

In a dispersive medium, the speed of wave propagation depends on the frequency or wavelength of the wave. A pulse is made up of many frequency components, and as it travels through such a medium, different components move with different speeds. This causes the initially compact pulse to spread out and change shape over time, leading to distortion during propagation.

These explanations illustrate how fundamental wave principles account for a wide range of physical phenomena observed in sound, music, and wave motion.