The Physics of Waves

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,

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

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.

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

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,

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

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 (molar mass ≈ 18 g mol⁻¹) partially replace heavier oxygen and nitrogen molecules (effective molar mass of dry air ≈ 29 g mol⁻¹), 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.

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.

The bat emits ultrasonic sound waves of frequency 1000 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. 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 \(\lambda = v/f\). Substituting the given speed of sound in air and the frequency of the ultrasonic wave,

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

For the transmitted sound, the wave travels through water. The frequency remains the same, but the speed of sound in water is much higher, so the wavelength is greater. Using the same relation,

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

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 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 MHz, which is expressed as

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

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

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

Substituting the values,

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

The given transverse wave is described by \( y(x,t) = 3.0 \sin(36t + 0.018x + \pi/4) \). The displacement depends on both position and time through a combined argument \((\omega t + kx)\). Since the spatial and temporal parts are mixed together — not separated as a product — this is a travelling wave, not a stationary wave.

A travelling wave of the form \(y = A \sin(\omega t + kx + \phi)\) propagates in the negative x-direction (i.e., from right to left), because the phase \((\omega t + kx)\) is constant along a wave crest only when \(x\) decreases as \(t\) increases. Comparing with the standard form:

\[\omega = 36\ \text{rad s}^{-1}, \qquad k = 0.018\ \text{cm}^{-1}\]

The speed of propagation is

\[\begin{aligned}v &= \frac{\omega}{k} = \frac{36}{0.018} = 2000\ \text{cm s}^{-1} = 20\ \text{m s}^{-1}\end{aligned}\]

The wave travels at 20 m s⁻¹ in the negative x-direction — from right to left.

The amplitude is the coefficient of the sine function:

\[A = 3.0\ \text{cm}\]

The frequency is related to the angular frequency by \(f = \omega/2\pi\):

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

Setting \(x = 0\) and \(t = 0\) in the wave equation, the phase becomes the constant term:

\[\phi_0 = \frac{\pi}{4}\ \text{rad}\]

The least distance between two successive crests is the wavelength \(\lambda = 2\pi/k\):

\[\begin{aligned}\lambda &= \frac{2\pi}{0.018} \approx 349\ \text{cm} \approx 3.49\ \text{m}\end{aligned}\]

The wave equation from Exercise 14.8 is \( y(x,t) = 3.0 \sin(36t + 0.018x + \pi/4) \), where \(y\) and \(x\) are in cm and \(t\) in seconds. To plot y–t graphs at fixed positions, we substitute each value of \(x\).

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

This is a sinusoidal oscillation with amplitude 3.0 cm, angular frequency 36 rad s⁻¹, and initial phase \(\pi/4\). The y–t graph is a sine curve starting at \(y = 3.0\sin(\pi/4) = 3.0/\sqrt{2} \approx 2.12\ \text{cm}\) at \(t = 0\).

\[\begin{aligned}y(2,t) &= 3.0 \sin\!\left(36t + 0.018 \times 2 + \frac{\pi}{4}\right) \\&= 3.0 \sin\!\left(36t + 0.036 + \frac{\pi}{4}\right)\end{aligned}\]

The shape remains sinusoidal with the same amplitude (3.0 cm) and the same angular frequency (36 rad s⁻¹), but the phase is shifted forward by \(0.036\) rad compared to \(x = 0\).

\[\begin{aligned}y(4,t) &= 3.0 \sin\!\left(36t + 0.018 \times 4 + \frac{\pi}{4}\right) \\&= 3.0 \sin\!\left(36t + 0.072 + \frac{\pi}{4}\right)\end{aligned}\]

Again, the shape is sinusoidal with the same amplitude and frequency, but a further phase shift of \(0.072\) rad relative to \(x = 0\).

All three y–t graphs are sine curves of identical amplitude and identical frequency. The graphs are simply shifted along the time axis — each is a horizontal translation of the others. The only difference is the phase of oscillation, which increases with increasing \(x\) (since the wave travels in the \(-x\) direction, points at larger \(x\) are ahead in phase).

The given travelling harmonic wave is \( y(x,t) = 2.0 \cos 2\pi(10t - 0.0080x + 0.35) \). Expanding, the phase of the wave is the full argument of the cosine. The spatial part \(-0.0080x\) (with \(x\) in cm) determines how phase varies with position. The phase difference between two points separated by \(\Delta x\) is

\[\Delta \phi = 2\pi \times 0.0080\, \Delta x \quad (\Delta x \text{ in cm})\]

The wavelength is found by noting that one full cycle of phase (\(2\pi\)) corresponds to one wavelength:

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

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

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

When the separation equals half a wavelength, substituting \(\Delta x = \lambda/2 = 62.5\ \text{cm}\):

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

This confirms the general result: points separated by \(\lambda/2\) are always exactly in opposite phase (\(\Delta\phi = \pi\)) — one is at a crest when the other is at a trough.

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

The displacement \( y(x,t) = 0.06\sin\!\left(\frac{2\pi}{3}x\right)\cos(120\pi t) \) is expressed as a product of a function of \(x\) alone and a function of \(t\) alone. The spatial and temporal parts are fully separated, which is the defining characteristic of a stationary wave. The string has fixed nodes whose positions never change with time, confirming this interpretation. The boundary conditions (clamped at both ends) require nodes at \(x = 0\) and \(x = 1.5\ \text{m}\), which is consistent with \(\sin(2\pi/3 \times 1.5) = \sin(\pi) = 0\). ✓

Using the trigonometric product-to-sum identity,

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

the given equation is rewritten as

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

This shows that the stationary wave is formed by two travelling waves, each of amplitude \(0.03\ \text{m}\): one moving in the positive \(x\)-direction and one in the negative \(x\)-direction.

From the wave equation, the wave number is \(k = \frac{2\pi}{3}\ \text{rad m}^{-1}\) and angular frequency is \(\omega = 120\pi\ \text{rad s}^{-1}\).

Wavelength of each travelling wave:

\[\lambda = \frac{2\pi}{k} = \frac{2\pi}{2\pi/3} = 3\ \text{m}\]

Frequency of each travelling wave:

\[f = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60\ \text{Hz}\]

Speed of each travelling wave:

\[v = f\lambda = 60 \times 3 = 180\ \text{m s}^{-1}\]

The linear mass density of the string is

\[\mu = \frac{m}{L} = \frac{3.0\times10^{-2}}{1.5} = 2.0\times10^{-2}\ \text{kg m}^{-1}\]

Using \(v = \sqrt{T/\mu}\), the tension is

\[\begin{aligned}T &= \mu v^2 = (2.0\times10^{-2})(180)^2 = 648\ \text{N}\end{aligned}\]

In the stationary wave \( y(x,t) = 0.06\sin\!\left(\frac{2\pi}{3}x\right)\cos(120\pi t) \), the time-dependent factor \(\cos(120\pi t)\) is common to all positions. Therefore, every point on the string oscillates with the same angular frequency \(\omega = 120\pi\ \text{rad s}^{-1}\), i.e., the same frequency of 60 Hz. Yes — all points share the same frequency.

The phase of oscillation depends on the sign of \(\sin(2\pi x/3)\). Within any single loop between two consecutive nodes, this function maintains the same sign, so all points in that loop oscillate in phase with each other. However, across a node the function changes sign, so particles on either side of a node are in opposite phase. Therefore, the phase is not the same for all points — it is uniform within a loop but reverses at every node.

The amplitude of oscillation at position \(x\) is given by the magnitude of the spatial factor:

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

This varies continuously from zero at the nodes to a maximum of 0.06 m at the antinodes. The amplitude is not the same for all points — it depends strongly on position.

Substituting \(x = 0.375\ \text{m}\) into the amplitude expression:

\[\begin{aligned}A &= 0.06\sin\!\left(\frac{2\pi}{3}\times 0.375\right) \\&= 0.06\sin\!\left(\frac{\pi}{4}\right) \\&= 0.06\times\frac{1}{\sqrt{2}} \approx 4.24\times10^{-2}\ \text{m}\end{aligned}\]

An elastic wave must be finite, single-valued, and well-defined over the entire propagation domain. We now examine each function.

The displacement is written as a product of \(\cos(3x)\) (purely spatial) and \(\sin(10t)\) (purely temporal). The variables are completely separated, which is the hallmark of a stationary wave. Fixed nodes exist wherever \(\cos(3x) = 0\), confirming the stationary character.

Although the argument is the travelling-wave combination \((x - vt)\), the square-root function is undefined (not real) whenever \(x - vt < 0\). A physical wave must be well-defined over the entire medium — this condition fails for all points where \(x < vt\). Therefore, this function represents no valid physical wave.

Both terms share the identical argument \((5x - 0.5t)\). Using the harmonic addition formula,

\[\begin{aligned}y &= 3\sin(5x-0.5t) + 4\cos(5x-0.5t) \\&= \sqrt{3^2+4^2}\,\sin(5x - 0.5t + \phi) \\&= 5\sin(5x - 0.5t + \phi)\end{aligned}\]

where \(\tan\phi = 4/3\). The combined expression depends only on \((x - vt)\) and is a single sinusoidal function — this is a travelling wave of amplitude 5 m, wave number 5 rad m⁻¹, travelling in the positive \(x\)-direction with speed \(v = 0.5/5 = 0.1\ \text{m s}^{-1}\).

Each term individually represents a stationary wave (\(\cos x\sin t\) and \(\cos 2x\sin 2t\)), but they have different angular frequencies (1 rad s⁻¹ and 2 rad s⁻¹) and different wave numbers. Their superposition does not produce a single stationary wave with fixed, time-invariant nodes, nor does it reduce to any function of \((x \pm vt)\). It represents none of the pure wave types — it is a superposition of two independent standing waves forming a complex, non-periodic displacement pattern.

The wire is stretched between two rigid supports and vibrates in its fundamental mode. We first determine the length of the wire from the given mass and linear mass density.

\[\begin{aligned}L &= \frac{m}{\mu} = \frac{3.5\times10^{-2}}{4.0\times10^{-2}} = 0.875\ \text{m}\end{aligned}\]

In the fundamental mode, the length of the string equals half the wavelength of the stationary wave:

\[\lambda = 2L = 2\times 0.875 = 1.75\ \text{m}\]

The speed of the transverse wave is

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

Using the relation \(v = \sqrt{T/\mu}\), the tension is

\[\begin{aligned}T &= \mu v^2 = (4.0\times10^{-2})(78.75)^2 \approx 248\ \text{N} \approx 2.48\times10^{2}\ \text{N}\end{aligned}\]

The tube is open at one end and closed at the other by a movable piston. The two resonant lengths are \(L_1 = 25.5\ \text{cm}\) and \(L_2 = 79.3\ \text{cm}\). For a pipe closed at one end, the difference between two successive resonant lengths equals half the wavelength.

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

Therefore, 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 at the experimental temperature is obtained using \(v = f\lambda\):

\[\begin{aligned}v &= 340\times 1.076 \approx 3.66\times10^{2}\ \text{m s}^{-1}\end{aligned}\]

The steel rod has a total length of \(L = 100\ \text{cm} = 1.0\ \text{m}\) and is clamped at its middle. When clamped at the centre, longitudinal vibrations produce a node at the clamp and antinodes at both free ends. In the fundamental mode, the length of the rod equals half the wavelength:

\[\lambda = 2L = 2\times 1.0 = 2.0\ \text{m}\]

The fundamental frequency is \(f = 2.53\ \text{kHz} = 2.53\times10^{3}\ \text{Hz}\). The speed of sound in steel is

\[\begin{aligned}v &= f\lambda = (2.53\times10^{3})\times 2.0 = 5.06\times10^{3}\ \text{m s}^{-1}\end{aligned}\]

The pipe has length \(L = 20\ \text{cm} = 0.20\ \text{m}\).

The fundamental frequency (first harmonic) of a closed pipe is

\[\begin{aligned}f_1 &= \frac{v}{4L} = \frac{340}{4\times0.20} = \frac{340}{0.80} = 425\ \text{Hz}\end{aligned}\]

The source frequency is 430 Hz, which is very close to 425 Hz. Let us check if 430 Hz matches a higher harmonic more exactly. The resonant frequencies of the closed pipe are \(425\ \text{Hz},\ 1275\ \text{Hz},\ 2125\ \text{Hz},\ldots\) None exactly matches 430 Hz, but 425 Hz is close enough to consider the fundamental as resonantly excited — the small deviation (5 Hz) arises due to the neglected end correction. For the purposes of this problem, 430 Hz excites the fundamental mode (first harmonic) of the closed pipe.

The fundamental frequency of an open pipe of the same length is

\[\begin{aligned}f_1 &= \frac{v}{2L} = \frac{340}{2\times0.20} = \frac{340}{0.40} = 850\ \text{Hz}\end{aligned}\]

The resonant frequencies of the open pipe are \(850\ \text{Hz},\ 1700\ \text{Hz},\ 2550\ \text{Hz},\ldots\) The source frequency 430 Hz does not match any of these. Therefore, the 430 Hz source will not produce resonance when the pipe is open at both ends.

Let the original frequency of string A be \(f_A = 324\ \text{Hz}\) and that of string B be \(f_B\). The initial beat frequency is 6 Hz, so

\[|f_A - f_B| = |324 - f_B| = 6\]

This gives two possible values:

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

Now, the tension in string A is slightly reduced. Since \(f \propto \sqrt{T}\), reducing the tension lowers the frequency of string A — its new frequency is less than 324 Hz. The observed beat frequency decreases from 6 Hz to 3 Hz. This means the frequency of A is approaching that of B — i.e., A and B are getting closer in frequency. Since A's frequency is falling, B's frequency must be below the original 324 Hz.

Therefore, the only consistent value is

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

Verification: after reducing tension, if \(f_A\) falls to, say, 321 Hz, then \(|321 - 318| = 3\ \text{Hz}\). ✓ This matches the observed new beat frequency.