Solved Examples

How to Use This Problem Bank

Nine sections of theory, and now the part that actually earns marks. What follows is 48 worked problems covering the whole chapter — classifying waves, reading an equation, snapshot and history graphs, phase differences, particle motion against wave motion, wave speed on strings and in gases, superposition, reflection, standing waves, pipes, beats and the Doppler effect.

They run easy first, hard last, in four parts that follow the order the chapter was taught in. Work them with a pen: cover the solution, try it, then compare — and read the takeaway at the end of each one, because that is where the marks usually leak away.

Notation for This Chapter

Symbol	Meaning	Unit
$y$	displacement of a particle of the medium	m
$a$	amplitude	m
$\lambda$	wavelength	m
$k$	angular wave number, $\frac{2\pi}{\lambda}$	rad/m
$\nu$	frequency	Hz
$\omega$	angular frequency, $2\pi\nu$	rad/s
$T$	time period, $\frac{1}{\nu}$	s
$T$	tension in a string (always named "tension")	N
$v$	wave speed	m/s
$\frac{\partial y}{\partial t}$	particle velocity	m/s
$\mu$	linear mass density of a string	kg/m
$\rho$	density	kg/m³
$\gamma$	ratio of specific heats	none
$\nu^{\,\prime}$	frequency an observer receives	Hz
$v_s$, $v_o$	signed velocities of source and observer	m/s

The Three Symbols That Get Confused

Key Point — read these before the first problem:

1. $T$ does two jobs. In $v = \sqrt{\frac{T}{\mu}}$ it is the tension, measured in newtons. Everywhere else it is the time period, measured in seconds. Every solution below names which one it means and writes the unit beside it.
2. $k$ is the angular wave number, $\frac{2\pi}{\lambda}$, in radians per metre. It is never a spring constant in this chapter.
3. $v$ is the wave speed; the particle speed is $\frac{\partial y}{\partial t}$. They are different quantities with different values. The greatest particle speed anywhere is $\omega a$, and it has nothing to do with $v = \nu\lambda$.

Key Point — every formula this section uses, in one place: $$y = a\sin(kx - \omega t + \phi) \ \text{ travels towards } +x, \qquad y = a\sin(kx + \omega t + \phi) \ \text{ travels towards } -x$$ $$k = \frac{2\pi}{\lambda}, \qquad \omega = 2\pi\nu = \frac{2\pi}{T}, \qquad v = \nu\lambda = \frac{\omega}{k}$$ $$\Delta\phi = \frac{2\pi}{\lambda}\Delta x = \frac{2\pi}{T}\Delta t, \qquad \left(\frac{\partial y}{\partial t}\right)_{\max} = \omega a, \qquad \left(\frac{\partial^2 y}{\partial t^2}\right)_{\max} = \omega^2 a$$ $$v = \sqrt{\frac{T_{\text{tension}}}{\mu}} \ \ \text{(string)}, \qquad v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T_{\text{kelvin}}}{M_0}} \ \ \text{(gas)}, \qquad v \propto \sqrt{T_{\text{kelvin}}}, \quad v \propto \frac{1}{\sqrt{M_0}}$$ $$a_{\text{res}} = 2a\cos\frac{\phi}{2}, \qquad y_{\text{standing}} = 2a\sin kx\cos\omega t, \qquad \nu_{\text{beat}} = \lvert \nu_1 - \nu_2 \rvert$$ $$\nu_n = \frac{nv}{2L} \ \ \text{(string, open pipe)}, \qquad \nu_n = \frac{nv}{4L}, \ n \text{ odd} \ \ \text{(closed pipe)}, \qquad \nu^{\,\prime} = \nu\left(\frac{v - v_o}{v - v_s}\right)$$

Two Conventions Used in Every Solution

Direction of travel. A minus sign between $kx$ and $\omega t$ means the wave travels towards $+x$; a plus sign means towards $-x$. A reflected wave must therefore have the opposite sign from the incident one.

The Doppler axis. Draw the line from the source to the observer and take that direction as positive. Then $v_s$ and $v_o$ are signed velocity components along that one line, and

$$\nu^{\,\prime} = \nu\left(\frac{v - v_o}{v - v_s}\right)$$

covers all four cases with nothing memorised. An observer moving towards the source is moving in the negative direction, so $v_o$ is negative and the numerator grows. Every Doppler solution below states the positive direction, then says in words whether the answer should come out higher or lower, and only then substitutes.

Harmonics and Overtones

System	Harmonics present	$n$th harmonic is	First overtone
String fixed at both ends	$n = 1, 2, 3, 4, \ldots$	the $(n-1)$th overtone	2nd harmonic
Pipe open at both ends	$n = 1, 2, 3, 4, \ldots$	the $(n-1)$th overtone	2nd harmonic
Pipe closed at one end	$n = 1, 3, 5, 7, \ldots$ only	the $\frac{n-1}{2}$th overtone	3rd harmonic

The Constants Used Throughout

Quantity	Value
Speed of sound in air	340 m/s, unless a problem states otherwise
$g$	9.8 m/s²
$\gamma$ for air	1.4
Gas constant $R$	8.314 J/(mol K)
Molar mass of air $M_0$	29.0 g/mol
$\pi$	3.1416

Where a problem gives a different speed of sound, a different $g$ or a different temperature, that value is used and no other. Temperatures go into every gas formula in kelvin.

[Board Important] Every solution below writes the formula on its own line before any number goes into it, converts centimetres to metres on a line of its own, and names whether it is quoting $\omega$ or $\nu$, tension or period, wave speed or particle speed. Do all three in the exam. A correct formula with an arithmetic slip still earns most of the marks; a $2\pi$ dropped between $\omega$ and $\nu$ loses them all.

Solved Examples

Part 1: Reading the Wave, and Writing It Down

Twelve warm-ups. No tensions and no pipes yet — just the vocabulary, the five constants that live inside $y = a\sin(kx - \omega t + \phi)$, and the business of turning a description or a pair of graphs into an equation and back again.

Constants for this part: $\pi = 3.1416$; speed of sound in air 340 m/s where sound appears.

Example 1: Six disturbances, sorted

For each, say whether the wave is transverse, longitudinal or a mixture, and whether it is mechanical, electromagnetic or a matter wave.

(a) A compression pulse sent along a slinky by pushing its end along its own length. (b) The vibration of a stretched tabla membrane. (c) A radio broadcast crossing 30 km of open country. (d) A seismic S wave crossing solid rock. (e) An X-ray beam in a hospital scanner. (f) The wave associated with an electron in a diffraction experiment.

Solution:

Two independent questions are being asked, and they must be answered separately. Transverse or longitudinal asks how the oscillation is oriented relative to the direction the wave travels. Mechanical, electromagnetic or matter asks what kind of thing is waving.

1. (a) Longitudinal, mechanical. The coils move back and forth along the slinky, which is the same direction the compression travels. A medium is required, so it is mechanical.
2. (b) Transverse, mechanical. Each element of the membrane moves perpendicular to the sheet while the disturbance spreads across it. A stretched membrane resists shear, which is what lets a transverse wave exist on it.
3. (c) Transverse, electromagnetic. The electric and magnetic fields oscillate at right angles to the direction of travel. No medium is needed — this is the one family in the list that crosses a vacuum.
4. (d) Transverse, mechanical. The "S" is for secondary, and also, conveniently, for shear. S waves need a shear modulus, which is why they cross rock and stop dead at the liquid outer core.
5. (e) Transverse, electromagnetic. An X-ray is light with a very short wavelength. Same family as (c), different frequency.
6. (f) A matter wave. It is neither mechanical nor electromagnetic; it is the wave nature of the electron itself, and the transverse/longitudinal labels do not apply to it in the way they do to the others.

Final Answer: (a) longitudinal, mechanical; (b) transverse, mechanical; (c) transverse, electromagnetic; (d) transverse, mechanical; (e) transverse, electromagnetic; (f) matter wave.

Takeaway: Answer the two questions separately. "Transverse or longitudinal" is about geometry; "mechanical, electromagnetic or matter" is about what is doing the waving. A single label can never carry both pieces of information, and examiners award the two halves separately.

Example 2: Every constant off one equation

A transverse wave on a long string is described by

$$y(x,t) = 0.04\sin(2.0\,x - 60\,t)$$

with all quantities in SI units. Find (a) the amplitude, (b) the angular wave number and the wavelength, (c) the angular frequency, the frequency and the period, (d) the wave speed and its direction, and (e) the greatest speed any particle of the string reaches.

Solution:

Symbols first: compare with $y = a\sin(kx - \omega t + \phi)$, where $k$ is the angular wave number in rad/m and $\omega$ the angular frequency in rad/s. Here $T$ will mean the time period in seconds; no tension appears in this problem at all.

1. Read off the three constants by matching brackets. $$a = 0.04 \ \text{m}, \qquad k = 2.0 \ \text{rad/m}, \qquad \omega = 60 \ \text{rad/s}$$

2. (b) Wavelength from $k$. $$\lambda = \frac{2\pi}{k} = \frac{6.2832}{2.0} = 3.14 \ \text{m}$$

3. (c) Frequency and period from $\omega$. $$\nu = \frac{\omega}{2\pi} = \frac{60}{6.2832} = 9.55 \ \text{Hz}, \qquad T = \frac{1}{\nu} = 0.105 \ \text{s}$$

4. (d) Wave speed, two ways. $$v = \frac{\omega}{k} = \frac{60}{2.0} = 30 \ \text{m/s}, \qquad v = \nu\lambda = 9.55 \times 3.14 = 30 \ \text{m/s}$$ The sign between $kx$ and $\omega t$ is a minus, so the wave travels towards $+x$.

5. (e) Maximum particle speed — a different quantity entirely. $$\left(\frac{\partial y}{\partial t}\right)_{\max} = \omega a = 60 \times 0.04 = 2.4 \ \text{m/s}$$

Final Answer: $a = 0.04$ m; $k = 2.0$ rad/m and $\lambda = 3.14$ m; $\omega = 60$ rad/s, $\nu = 9.55$ Hz, $T = 0.105$ s; $v = 30$ m/s towards $+x$; maximum particle speed 2.4 m/s.

Takeaway: The wave crosses the string at 30 m/s while no particle of the string ever exceeds 2.4 m/s. The first number is $\frac{\omega}{k}$, the second is $\omega a$. They are not related, they are not equal, and swapping them is the single commonest error in this chapter.

Example 3: Writing the equation from a description

A transverse harmonic wave of amplitude 5.0 mm and frequency 250 Hz travels at 100 m/s in the $-x$ direction along a string. At $t = 0$ the particle at $x = 0$ is passing through $y = 0$ and moving in the $+y$ direction. Write the equation of the wave.

Solution:

The standard form is $y = a\sin(kx \pm \omega t + \phi)$; a plus sign between $kx$ and $\omega t$ means travel towards $-x$.

1. Convert to SI first. $$a = 5.0 \ \text{mm} = 5.0 \times 10^{-3} \ \text{m}$$

2. Wavelength from the speed and the frequency. $$\lambda = \frac{v}{\nu} = \frac{100}{250} = 0.40 \ \text{m}$$

3. The two bracket constants. $$k = \frac{2\pi}{\lambda} = \frac{6.2832}{0.40} = 15.7 \ \text{rad/m}, \qquad \omega = 2\pi\nu = 6.2832 \times 250 = 1571 \ \text{rad/s}$$

4. Choose the sign for the direction. Travel towards $-x$ needs the plus form: $$y = a\sin(kx + \omega t + \phi)$$

5. Use the initial condition to fix $\phi$. At $x = 0$ the equation reads $y = a\sin(\omega t + \phi)$. $$y(0,0) = 0 \ \Rightarrow \ \sin\phi = 0 \ \Rightarrow \ \phi = 0 \ \text{or} \ \pi$$ Now the direction of motion decides between them. The particle velocity is $$\frac{\partial y}{\partial t} = a\omega\cos(\omega t + \phi), \qquad \left.\frac{\partial y}{\partial t}\right|_{t=0} = a\omega\cos\phi$$ This must be positive, and $\cos 0 = +1$ while $\cos\pi = -1$. So $\phi = 0$.

Final Answer: $$y(x,t) = \left(5.0 \times 10^{-3}\right)\sin\left(15.7\,x + 1571\,t\right) \ \text{m}$$

Takeaway: Two facts fix the phase constant, not one. "At $y = 0$" narrows $\phi$ to two choices $\pi$ apart; only the direction of the particle's motion picks the right one. Whenever a question tells you which way the particle is moving, it is handing you the second condition on purpose.

Example 4: A wave written with the time term first

A transverse harmonic wave is described by

$$y(x,t) = 2.5\sin\left(50\,t + 0.025\,x + \frac{\pi}{6}\right)$$

where $x$ and $y$ are in centimetres and $t$ is in seconds. The positive $x$ direction runs from left to right. (a) Is this a travelling wave or a standing wave, and if it is travelling, which way and how fast? (b) Give the amplitude and the frequency. (c) What is the initial phase at the origin? (d) What is the least distance between two successive crests? (e) Sketch in words the history graphs of the particles at $x = 0$, $x = 20$ cm and $x = 40$ cm — in what respect do they differ?

Solution:

Reading the symbols: $k$ is the angular wave number and $\omega$ the angular frequency. The bracket has been written with the time term first, which changes nothing: $\sin(\omega t + kx + \phi)$ is exactly $\sin(kx + \omega t + \phi)$.

1. (a) Travelling, and which way. The displacement is a single sinusoid in the combination $(kx + \omega t)$, not a product of a function of $x$ and a function of $t$, so it is a travelling wave. The sign between $kx$ and $\omega t$ is a plus, so it travels towards $-x$ — from right to left. $$k = 0.025 \ \text{rad/cm}, \qquad \omega = 50 \ \text{rad/s}$$ $$v = \frac{\omega}{k} = \frac{50}{0.025} = 2000 \ \text{cm/s} = 20 \ \text{m/s}$$

2. (b) Amplitude and frequency. $$a = 2.5 \ \text{cm} = 0.025 \ \text{m}, \qquad \nu = \frac{\omega}{2\pi} = \frac{50}{6.2832} = 7.96 \ \text{Hz}$$

3. (c) Initial phase at the origin. Put $x = 0$ and $t = 0$ into the bracket: $$\text{phase} = \frac{\pi}{6} \ \text{rad} = 30°$$

4. (d) Distance between successive crests is one wavelength: $$\lambda = \frac{2\pi}{k} = \frac{6.2832}{0.025} = 251 \ \text{cm} = 2.51 \ \text{m}$$ Check: $\nu\lambda = 7.96 \times 2.51 = 20$ m/s, which is $v$ again.

5. (e) The three history graphs. Each is a plot of $y$ against $t$ for one fixed $x$, so each is a sine curve of the same amplitude 2.5 cm and the same frequency 7.96 Hz. Only the starting phase differs, because $x$ enters the bracket as the constant $kx$: $$\Delta\phi = k\,\Delta x = 0.025 \times 20 = 0.50 \ \text{rad} \ \text{(between } x = 0 \text{ and } x = 20 \text{ cm)}$$ $$\Delta\phi = 0.025 \times 40 = 1.0 \ \text{rad} \ \text{(between } x = 0 \text{ and } x = 40 \text{ cm)}$$

Final Answer: (a) travelling, right to left, at 20 m/s; (b) 2.5 cm and 7.96 Hz; (c) $\frac{\pi}{6}$ rad; (d) 2.51 m; (e) identical in amplitude and frequency, differing only in phase — by 0.50 rad and 1.0 rad respectively.

Takeaway: Do the unit conversion on $k$ before you touch $\lambda$. Here $k$ came out in rad/cm because $x$ was in centimetres, so $\frac{2\pi}{k}$ is a length in centimetres. And note part (e): in a travelling wave every particle has the same amplitude and the same frequency; only the phase varies from place to place.

Example 5: The same wave, seen as a snapshot and as a history

A wave on a string is $y(x,t) = 0.05\sin(4\,x - 24\,t)$ in SI units. (a) A photograph is taken at $t = 0$ and $y$ is plotted against $x$. What is the horizontal spacing between two crests on that plot? (b) A sensor watches the single particle at $x = 0$ and $y$ is plotted against $t$. What is the horizontal spacing between two crests on that plot? (c) Which of the two graphs gives you the wave speed, and how?

Solution:

A snapshot graph has distance on the horizontal axis; a history graph has time on it. They look almost identical and they carry different information, which is exactly why the axis label must be read first.

1. Read the constants. $$a = 0.05 \ \text{m}, \qquad k = 4.0 \ \text{rad/m}, \qquad \omega = 24 \ \text{rad/s}$$

2. (a) The snapshot spacing is the wavelength. $$\lambda = \frac{2\pi}{k} = \frac{6.2832}{4.0} = 1.57 \ \text{m}$$

3. (b) The history spacing is the time period. $$T = \frac{2\pi}{\omega} = \frac{6.2832}{24} = 0.262 \ \text{s}$$ Here $T$ is a period in seconds, not a tension. Equivalently $\nu = \frac{1}{T} = 3.82$ Hz.

4. (c) Neither graph alone gives the speed; the pair does. $$v = \frac{\lambda}{T} = \frac{1.57}{0.262} = 6.0 \ \text{m/s}$$ and the direct check $v = \frac{\omega}{k} = \frac{24}{4.0} = 6.0$ m/s agrees. A snapshot on its own tells you nothing about how fast the pattern is sliding — you would need a second snapshot a known time later.

Final Answer: (a) 1.57 m, the wavelength; (b) 0.262 s, the period; (c) the speed needs both, $v = \frac{\lambda}{T} = 6.0$ m/s.

Takeaway: Look at the horizontal axis label before anything else. Distance on the axis means the spacing you measure is $\lambda$; time on the axis means it is $T$. Getting this backwards turns a one-line problem into a wrong answer that looks completely reasonable.

Example 6: Phase difference from a path difference

A sound wave of frequency 680 Hz travels through air at 340 m/s. Find the phase difference between the oscillations of two air particles separated, along the direction of travel, by (a) 0.125 m, (b) 1.75 m, (c) $\frac{\lambda}{2}$, (d) $\frac{3\lambda}{4}$. (e) What is the smallest separation for which the phase difference is $\frac{2\pi}{3}$?

Solution:

Symbols and units: $\Delta\phi$ is a phase difference in radians, $\Delta x$ a path difference in metres, and the link is $\Delta\phi = \frac{2\pi}{\lambda}\Delta x$ — one whole wavelength of separation is one whole cycle of phase.

1. Wavelength first. $$\lambda = \frac{v}{\nu} = \frac{340}{680} = 0.50 \ \text{m}$$

2. (a) $\Delta x = 0.125$ m, which is $\frac{\lambda}{4}$: $$\Delta\phi = \frac{2\pi}{0.50} \times 0.125 = \frac{\pi}{2} \ \text{rad} = 90°$$

3. (b) $\Delta x = 1.75$ m. In wavelengths that is $\frac{1.75}{0.50} = 3.5$: $$\Delta\phi = 2\pi \times 3.5 = 7\pi \ \text{rad}$$ Phase repeats every $2\pi$, so subtract $6\pi$: the two particles are $\pi$ rad apart — exactly out of phase, one at a crest whenever the other is at a trough.

4. (c) $\Delta x = \frac{\lambda}{2}$: $$\Delta\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi \ \text{rad} = 180°$$

5. (d) $\Delta x = \frac{3\lambda}{4}$: $$\Delta\phi = \frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = \frac{3\pi}{2} \ \text{rad} = 270°$$

6. (e) Working backwards. $$\Delta x = \frac{\lambda}{2\pi}\Delta\phi = \frac{0.50}{6.2832} \times \frac{2\pi}{3} = \frac{0.50}{3} = 0.167 \ \text{m}$$

Final Answer: (a) $\frac{\pi}{2}$; (b) $7\pi$, equivalent to $\pi$; (c) $\pi$; (d) $\frac{3\pi}{2}$; (e) 0.167 m.

Takeaway: Divide the path difference by the wavelength first and read the answer as a fraction of a cycle. Parts (c) and (d) never needed a number at all — the answer is fixed by the fraction of a wavelength, whatever $\lambda$ happens to be.

Example 7: Phase difference from a time difference

A harmonic wave of frequency 200 Hz travels at 300 m/s. Find (a) the phase difference between the displacements of one particular particle at two instants 0.50 ms apart; (b) the phase difference between two particles 0.375 m apart at one instant; (c) the path difference that is equivalent to the 0.50 ms delay in part (a).

Solution:

The rule in play: phase can be shifted by moving in space ($\Delta\phi = \frac{2\pi}{\lambda}\Delta x$) or by waiting in time ($\Delta\phi = \frac{2\pi}{T}\Delta t$). Here $T$ is the time period in seconds.

1. The two basic quantities. $$T = \frac{1}{\nu} = \frac{1}{200} = 5.0 \times 10^{-3} \ \text{s} = 5.0 \ \text{ms}, \qquad \lambda = \frac{v}{\nu} = \frac{300}{200} = 1.5 \ \text{m}$$

2. (a) Time difference of 0.50 ms. That is one tenth of a period: $$\Delta\phi = \frac{2\pi}{T}\Delta t = \frac{6.2832}{5.0} \times 0.50 = 0.628 \ \text{rad} = 36°$$

3. (b) Path difference of 0.375 m. That is a quarter of a wavelength: $$\Delta\phi = \frac{2\pi}{\lambda}\Delta x = \frac{6.2832}{1.5} \times 0.375 = \frac{\pi}{2} \ \text{rad} = 90°$$

4. (c) Converting a delay into a distance. In a time $\Delta t$ the wave advances $v\Delta t$: $$\Delta x = v\,\Delta t = 300 \times 0.50 \times 10^{-3} = 0.15 \ \text{m}$$ Check it: $\frac{2\pi}{1.5} \times 0.15 = 0.628$ rad, the same answer as (a). It has to be — the wave taking 0.50 ms to cover 0.15 m is the same statement.

Final Answer: (a) 0.628 rad (36°); (b) $\frac{\pi}{2}$ rad (90°); (c) 0.15 m.

Takeaway: A delay of $\Delta t$ and a separation of $v\Delta t$ produce exactly the same phase difference. That equivalence is why $\frac{\lambda}{2}$ of path is quoted as a "$\pi$ phase change" and why $\frac{T}{4}$ of waiting is quoted as "90° behind". One relation, two doors into it.

Example 8: One source, two microphones

A loudspeaker on a stand emits a steady 500 Hz note. Two microphones stand on the same straight line out from it, at 6.12 m and 7.14 m. Take the speed of sound as 340 m/s. Find (a) the wavelength, (b) the phase difference between the signals the two microphones pick up, and (c) the time lag between them.

Solution:

Working rule: phase difference from path difference, $\Delta\phi = \frac{2\pi}{\lambda}\Delta x$, with $\Delta x$ the extra distance the sound must cover to reach the further microphone.

1. (a) Wavelength. $$\lambda = \frac{v}{\nu} = \frac{340}{500} = 0.68 \ \text{m}$$

2. Path difference. $$\Delta x = 7.14 - 6.12 = 1.02 \ \text{m}$$

3. Express it in wavelengths — always do this before reaching for a calculator. $$\frac{\Delta x}{\lambda} = \frac{1.02}{0.68} = 1.5$$ One and a half wavelengths.

4. (b) Phase difference. $$\Delta\phi = 2\pi \times 1.5 = 3\pi \ \text{rad}$$ which is $\pi$ rad once whole cycles are stripped out: the two microphones receive signals exactly out of phase. When one sees a compression arriving, the other sees a rarefaction.

5. (c) Time lag. $$\Delta t = \frac{\Delta x}{v} = \frac{1.02}{340} = 3.0 \times 10^{-3} \ \text{s} = 3.0 \ \text{ms}$$ Sense check: the period is $\frac{1}{500} = 2.0$ ms, and $3.0$ ms is one and a half periods — matching the one and a half wavelengths exactly.

Final Answer: (a) 0.68 m; (b) $3\pi$ rad, that is $\pi$ rad or 180°, exactly out of phase; (c) 3.0 ms.

Takeaway: A phase difference of $3\pi$ and one of $\pi$ describe the same physical state. Always reduce to the interval from $0$ to $2\pi$ before answering "in phase or out of phase", and always cross-check the time lag against the period — the two must give the same number of cycles.

Example 9: Two graphs into one equation

A snapshot of a wave on a string shows exactly 5 complete waves in a length of 3.0 m. A history graph of one particle shows exactly 30 complete oscillations in 0.60 s. The amplitude is 6.0 mm, the wave travels towards $+x$, and at $t = 0$ the particle at $x = 0$ is at its greatest negative displacement. Find the equation of the wave, its speed, and the maximum particle speed.

Solution:

What each graph gives you: the snapshot supplies $\lambda$, the history supplies $\nu$; the standard form for travel towards $+x$ is $y = a\sin(kx - \omega t + \phi)$.

1. From the snapshot. $$\lambda = \frac{3.0}{5} = 0.60 \ \text{m}, \qquad k = \frac{2\pi}{\lambda} = \frac{6.2832}{0.60} = 10.5 \ \text{rad/m}$$

2. From the history. $$\nu = \frac{30}{0.60} = 50 \ \text{Hz}, \qquad \omega = 2\pi\nu = 6.2832 \times 50 = 314 \ \text{rad/s}$$

3. Wave speed, both ways. $$v = \nu\lambda = 50 \times 0.60 = 30 \ \text{m/s}, \qquad v = \frac{\omega}{k} = \frac{314.16}{10.472} = 30 \ \text{m/s}$$

4. Phase constant from the initial condition. At $x = 0$, $t = 0$ the displacement must be $-a$: $$a\sin\phi = -a \ \Rightarrow \ \sin\phi = -1 \ \Rightarrow \ \phi = -\frac{\pi}{2}$$

5. The equation (with $a = 6.0$ mm $= 6.0\times10^{-3}$ m): $$y(x,t) = \left(6.0\times10^{-3}\right)\sin\left(10.5\,x - 314\,t - \frac{\pi}{2}\right) \ \text{m}$$ which is the same as $y = -\left(6.0\times10^{-3}\right)\cos(10.5\,x - 314\,t)$, since $\sin\left(\theta - \frac{\pi}{2}\right) = -\cos\theta$.

6. Maximum particle speed. $$\left(\frac{\partial y}{\partial t}\right)_{\max} = \omega a = 314.16 \times 6.0\times10^{-3} = 1.88 \ \text{m/s}$$

Final Answer: $y = \left(6.0\times10^{-3}\right)\sin\left(10.5\,x - 314\,t - \frac{\pi}{2}\right)$ m; $v = 30$ m/s; maximum particle speed 1.88 m/s.

Takeaway: "So many waves in so many metres" is $\lambda$; "so many oscillations in so many seconds" is $\nu$. Nothing else in the question can supply either. And when the particle starts at an extreme, $\phi$ is $\pm\frac{\pi}{2}$ — the sign follows from which extreme.

Example 10: Travelling, standing, or neither?

Classify each of the following, all in SI units, as a single travelling wave, a standing wave, or neither. Give the speed where there is one.

(a) $y = 4\sin(6x)\cos(20t)$ (b) $y = 5\sin(3x - 12t) + 5\cos(3x - 12t)$ (c) $y = 2\cos x\,\sin t + 2\cos 2x\,\sin 2t$ (d) $y = 3\sin(4x + 10t)$

Solution:

The test is structural, and it is quick. A travelling wave contains $x$ and $t$ only through the single combination $(kx \mp \omega t)$. A standing wave is a product of a function of $x$ alone and a function of $t$ alone, with one frequency.

1. (a) Standing wave. It is $[\text{function of } x] \times [\text{function of } t]$ with one frequency. The pattern does not move; nodes sit wherever $\sin 6x = 0$. The two travelling waves it is built from each have $$v = \frac{\omega}{k} = \frac{20}{6} = 3.33 \ \text{m/s}$$ and they run in opposite directions. The standing wave itself has no speed — that number belongs to its two ingredients.

2. (b) A single travelling wave. Both terms share the identical bracket $(3x - 12t)$, so combine them: $$5\sin\theta + 5\cos\theta = 5\sqrt{2}\,\sin\left(\theta + \frac{\pi}{4}\right), \qquad \theta = 3x - 12t$$ So $y = 7.07\sin\left(3x - 12t + \frac{\pi}{4}\right)$, a travelling wave of amplitude 7.07 towards $+x$ with $$v = \frac{12}{3} = 4.0 \ \text{m/s}$$

3. (c) Neither. Each term on its own is a standing wave, but they have different frequencies (1 rad/s and 2 rad/s) and different wave numbers. Their sum is a superposition of two different normal modes: periodic, but not a single travelling wave and not a single standing wave.

4. (d) A single travelling wave towards $-x$ (plus sign in the bracket), with $$v = \frac{\omega}{k} = \frac{10}{4} = 2.5 \ \text{m/s}$$

Final Answer: (a) standing, its component waves running at 3.33 m/s; (b) travelling towards $+x$ at 4.0 m/s, amplitude 7.07; (c) neither; (d) travelling towards $-x$ at 2.5 m/s.

Takeaway: Product means standing, single bracket means travelling. And check (b) before you classify: two terms are not automatically two waves — if they carry the same bracket they merge into one. Two terms with different brackets, as in (c), genuinely cannot be reduced.

Example 11: What the slope of each graph means

For the wave $y = 0.04\sin(2.0\,x - 60\,t)$ of Example 2, find at $x = 0$, $t = 0$: (a) the slope of the snapshot graph, $\frac{\partial y}{\partial x}$; (b) the particle velocity, $\frac{\partial y}{\partial t}$; (c) verify the relation between them.

Solution:

Two speeds, kept apart: $v = 30$ m/s is the wave speed here (from Example 2). $\frac{\partial y}{\partial t}$ is the particle velocity — a completely different quantity, and the one that is measured in the same units.

1. (a) Slope of the snapshot. Differentiate with respect to $x$, holding $t$ fixed: $$\frac{\partial y}{\partial x} = 0.04 \times 2.0\cos(2.0x - 60t) = 0.08\cos(2.0x - 60t)$$ $$\left.\frac{\partial y}{\partial x}\right|_{(0,0)} = 0.08\cos 0 = 0.08$$ This is a pure number — a slope, not a speed. It has no units.

2. (b) Particle velocity. Differentiate with respect to $t$, holding $x$ fixed: $$\frac{\partial y}{\partial t} = -0.04 \times 60\cos(2.0x - 60t) = -2.4\cos(2.0x - 60t)$$ $$\left.\frac{\partial y}{\partial t}\right|_{(0,0)} = -2.4 \ \text{m/s}$$ The particle at the origin is moving downwards at that instant, at 2.4 m/s.

3. (c) The relation. Compare the two derivatives: every $\frac{\partial y}{\partial x}$ is $\frac{\omega}{k} = v$ times the corresponding $\frac{\partial y}{\partial t}$, with a sign flip. $$\frac{\partial y}{\partial t} = -v\,\frac{\partial y}{\partial x}$$ $$-30 \times 0.08 = -2.4 \ \text{m/s} \ \checkmark$$ And note that $2.4$ m/s is exactly $\omega a$, so at this instant the particle at the origin happens to be moving at its maximum speed — which fits, because $y = 0$ there.

Final Answer: (a) slope $= 0.08$, dimensionless; (b) particle velocity $= -2.4$ m/s, downwards; (c) $\frac{\partial y}{\partial t} = -v\frac{\partial y}{\partial x}$, and $-30 \times 0.08 = -2.4$ m/s.

Takeaway: Particle velocity = $-$(wave speed) $\times$ (slope of the snapshot). So on a snapshot, a particle on a rising part of the curve (positive slope) is moving down for a wave going right. That single sentence answers a whole family of "which way is this particle moving?" questions without any calculation.

Example 12: Four short answers that carry full marks

Answer briefly. (a) A violin and a sitar play the same note, at the same loudness. Why can you still tell which is which? (b) Why does a sharp pulse gradually lose its shape as it travels through a dispersive medium? (c) A wave's frequency is fixed by one thing and its speed by another. Which is which, and what does that force to happen when a wave crosses into a new medium? (d) Why does a standing wave transport no energy along the string?

Solution:

Units first: $\nu$ is a frequency in hertz, $\lambda$ a wavelength in metres, and $v$ the wave speed of the medium — no particle speeds appear in any of these four answers.

1. (a) Timbre — the harmonic mixture. Both instruments sound the same fundamental, which is what fixes the pitch, so the note is the same. But the relative strengths of the overtones are set by how the instrument is built and how it is played, and they differ completely between a bowed violin string and a plucked sitar string. The ear reads that mixture as the quality, or timbre, of the note.

2. (b) Dispersion means the speed depends on the frequency. A pulse of any shape other than a pure sinusoid is, by Fourier's theorem, a sum of many harmonic components. In a dispersive medium those components travel at different speeds, so after a while they are no longer lined up as they were at the start, and the sum they add up to has a different shape. The pulse spreads out.

3. (c) Frequency belongs to the source; speed belongs to the medium. The source shakes the medium at $\nu$ oscillations per second, and every boundary the wave crosses passes on that same $\nu$ — layers on the two sides of a boundary cannot oscillate at different rates without tearing apart. The speed, though, is fixed by the elastic property and the density of whatever the wave is now in. Since $v = \nu\lambda$ must still hold and $\nu$ cannot change, the wavelength takes the whole of the change.

4. (d) A standing wave has permanent nodes. A node never moves, so no energy can be carried across it: transporting energy past a point requires that point to do work on its neighbour, and work needs displacement. Energy sloshes between kinetic and potential within each loop, twice per cycle, but the net flow across any node is zero.

Final Answer: (a) different overtone mixtures, that is different timbre; (b) different frequencies travel at different speeds and fall out of step; (c) the source fixes $\nu$, the medium fixes $v$, so $\lambda$ changes at a boundary; (d) nodes are permanently at rest, and no energy crosses a point that never moves.

Takeaway: These four one-liners appear in Board papers almost every year, and each is worth two marks for two sentences. Learn the reason, not the sentence: overtone mixture, speed depending on frequency, the source owning $\nu$, and a motionless node blocking energy flow.

Part 2: Particle Motion, Strings, and the Speed of Sound

Twelve problems on the two speeds that live in every wave, and on what sets the wave speed in a string and in a gas. Watch the letter $T$ throughout this part: in $v = \sqrt{\frac{T}{\mu}}$ it is a tension in newtons, and in $\nu = \frac{1}{T}$ it is a period in seconds. Each solution says which one it means.

Constants for this part: $g = 9.8$ m/s²; speed of sound in air 340 m/s unless stated; $R = 8.314$ J/(mol K); $\gamma_{\text{air}} = 1.4$; $M_{0,\text{air}} = 29.0$ g/mol.

Example 13: Maximum particle speed, and where on the wave it happens

A transverse wave on a string is $y = 0.02\sin(5\,x - 250\,t)$ in SI units. Find (a) the wave speed, (b) the greatest speed reached by any particle of the string, (c) the points on the waveform at which a particle is moving fastest and those at which it is momentarily at rest, and (d) the amplitude the wave would need for the maximum particle speed to equal the wave speed.

Solution:

Name the two speeds: $v$ is the wave speed — the rate at which the pattern slides along the string. The particle speed is $\left\lvert\frac{\partial y}{\partial t}\right\rvert$, and its greatest value is $\omega a$.

1. Read the constants. $$a = 0.02 \ \text{m}, \qquad k = 5.0 \ \text{rad/m}, \qquad \omega = 250 \ \text{rad/s}$$

2. (a) Wave speed. $$v = \frac{\omega}{k} = \frac{250}{5.0} = 50 \ \text{m/s}$$ (Cross-check: $\lambda = \frac{2\pi}{5} = 1.257$ m, $\nu = \frac{250}{2\pi} = 39.79$ Hz, and $\nu\lambda = 50$ m/s.)

3. (b) Maximum particle speed. $$\left(\frac{\partial y}{\partial t}\right)_{\max} = \omega a = 250 \times 0.02 = 5.0 \ \text{m/s}$$ That is one tenth of the wave speed, and the two numbers have nothing to do with each other.

4. (c) Where each happens. Differentiating, $$\frac{\partial y}{\partial t} = -\omega a\cos(kx - \omega t)$$ This is largest in magnitude when the cosine is $\pm 1$, which is precisely when $\sin(kx - \omega t) = 0$, that is when $y = 0$. So:

• fastest at the mean position, $y = 0$ — every point where the curve crosses the axis;
• momentarily at rest at the crests and troughs, $y = \pm a$, where the cosine is zero.

This is simple harmonic motion seen sideways: each particle is an oscillator, fastest in the middle and stationary at the ends of its swing.

5. (d) Making the two equal. $$\omega a = v \ \Rightarrow \ a = \frac{v}{\omega} = \frac{50}{250} = 0.20 \ \text{m}$$ A 20 cm amplitude on a wave of wavelength 1.26 m — a violently distorted string, which is why the two speeds are so rarely comparable in practice.

Final Answer: (a) 50 m/s; (b) 5.0 m/s; (c) fastest at $y = 0$, at rest at the crests and troughs; (d) $a = 0.20$ m.

Takeaway: A crest is where the displacement is largest and the particle speed is zero. Students routinely say a particle "moves fastest at the crest" because the crest looks like the exciting part of the picture. It is the flattest, dullest part of the motion — the particle has just stopped and is about to turn round.

Example 14: Particle velocity and acceleration at one point

A wave on a string is $y = 0.05\sin(4\,x - 200\,t)$ in SI units. Find, for the particle at $x = 0.20$ m at the instant $t = 0.010$ s, (a) its displacement, (b) its velocity, and (c) its acceleration. Also give (d) the maximum particle speed and maximum particle acceleration, and (e) the wave speed.

Solution:

Whose motion is asked about: all three of $y$, $\frac{\partial y}{\partial t}$ and $\frac{\partial^2 y}{\partial t^2}$ describe the particle; only the last part of the question is about the wave.

1. Constants and the phase at the stated point and time. $$a = 0.05 \ \text{m}, \qquad k = 4.0 \ \text{rad/m}, \qquad \omega = 200 \ \text{rad/s}$$ $$\theta = kx - \omega t = 4.0(0.20) - 200(0.010) = 0.80 - 2.00 = -1.20 \ \text{rad}$$ Angles inside $\sin$ and $\cos$ are in radians — do not let a calculator sit in degree mode.

2. (a) Displacement. $$y = 0.05\sin(-1.20) = 0.05 \times (-0.9320) = -0.0466 \ \text{m}$$

3. (b) Particle velocity. $$\frac{\partial y}{\partial t} = -\omega a\cos\theta = -200 \times 0.05 \times \cos(-1.20) = -10 \times 0.3624 = -3.62 \ \text{m/s}$$ Negative, so the particle is moving in the $-y$ direction.

4. (c) Particle acceleration — and here the shortcut is worth more than the differentiation: $$\frac{\partial^2 y}{\partial t^2} = -\omega^2 y = -(200)^2 \times (-0.0466) = +1.86 \times 10^{3} \ \text{m/s}^2$$ The acceleration always points back towards $y = 0$, which is the signature of simple harmonic motion.

5. (d) The two maxima. $$\left(\frac{\partial y}{\partial t}\right)_{\max} = \omega a = 200 \times 0.05 = 10 \ \text{m/s}$$ $$\left(\frac{\partial^2 y}{\partial t^2}\right)_{\max} = \omega^2 a = 40000 \times 0.05 = 2.0 \times 10^{3} \ \text{m/s}^2$$

6. (e) Wave speed. $$v = \frac{\omega}{k} = \frac{200}{4.0} = 50 \ \text{m/s}$$

Final Answer: (a) $-0.0466$ m; (b) $-3.62$ m/s; (c) $+1.86\times10^3$ m/s²; (d) 10 m/s and $2.0\times10^3$ m/s²; (e) 50 m/s.

Takeaway: $\frac{\partial^2 y}{\partial t^2} = -\omega^2 y$ saves you the second differentiation every time. Get $y$ first, multiply by $-\omega^2$, and you have the acceleration in one line — with its sign already correct, pointing back towards the mean position.

Example 15: A jerk along a heavy rope

A rope of mass 3.0 kg and length 15.0 m is stretched taut with a tension of 250 N. One end is given a sharp transverse jerk. How long does the disturbance take to reach the other end?

Solution:

Mind the $T$: in $v = \sqrt{\frac{T}{\mu}}$ the symbol $T$ is the tension, in newtons, and $\mu$ is the mass per unit length in kg/m. No period appears in this problem.

1. Linear mass density. $$\mu = \frac{m}{L} = \frac{3.0}{15.0} = 0.20 \ \text{kg/m}$$

2. Wave speed. $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{250 \ \text{N}}{0.20 \ \text{kg/m}}} = \sqrt{1250} = 35.4 \ \text{m/s}$$

3. Time to cross the rope. $$t = \frac{L}{v} = \frac{15.0}{35.36} = 0.424 \ \text{s}$$

Final Answer: 0.424 s.

Takeaway: The mass of the rope enters only through $\mu$, never on its own. A rope twice as long with twice the mass has the same $\mu$, the same wave speed — and takes twice as long, because the pulse has twice as far to go.

Example 16: A stone dropped from a tower

A stone is dropped from rest at the top of a 200 m tower and splashes into a pond at its base. Taking $g = 9.8$ m/s² and the speed of sound in air as 340 m/s, when is the splash heard at the top?

Solution:

Two separate journeys, in sequence — the stone falling down under gravity, then the sound travelling up at a constant 340 m/s. The total is the sum, never the average.

1. Time for the stone to fall. It starts from rest, so $$h = \frac{1}{2}gt_1^2 \ \Rightarrow \ t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 200}{9.8}} = \sqrt{40.82} = 6.39 \ \text{s}$$

2. Time for the sound to climb back. Sound moves at a steady speed, so $$t_2 = \frac{h}{v} = \frac{200}{340} = 0.588 \ \text{s}$$

3. Total. $$t = t_1 + t_2 = 6.39 + 0.59 = 6.98 \ \text{s}$$

Final Answer: The splash is heard 6.98 s after the stone is released.

Takeaway: The falling stone accelerates; the sound does not. Use $h = \frac{1}{2}gt^2$ for one leg and $h = vt$ for the other, and never mix them. Notice too that the sound leg is under a tenth of the total here — but at 8% it is far too big to neglect if the question quotes three figures.

Example 17: Tuning a wire to the speed of sound

A steel wire is 10.0 m long and has a mass of 2.50 kg. What tension must it be under so that transverse waves travel along it at 340 m/s, the speed of sound in air?

Solution:

Symbol check: $T$ here is the tension in newtons, the unknown of the problem.

1. Linear mass density. $$\mu = \frac{m}{L} = \frac{2.50}{10.0} = 0.250 \ \text{kg/m}$$

2. Rearrange the speed formula. $$v = \sqrt{\frac{T}{\mu}} \ \Longrightarrow \ T = \mu v^2$$

3. Substitute. $$T = 0.250 \times (340)^2 = 0.250 \times 115600 = 2.89 \times 10^{4} \ \text{N}$$

Final Answer: A tension of $2.89\times10^4$ N, that is about 28.9 kN.

Takeaway: Because $T = \mu v^2$, the tension goes up as the square of the speed you want. That is why matching a wire's wave speed to the speed of sound needs a tension of nearly three tonnes-weight on a wire — and why this arrangement is a calculation, not an experiment.

Example 18: A hanging block sets the tension

A light string of linear mass density 2.0 g/m runs horizontally across a table, passes over a light frictionless pulley at the edge, and carries a 5.0 kg block hanging from its free end. Take $g = 9.8$ m/s². Find (a) the tension in the string, (b) the speed of transverse waves on it, (c) the time a pulse takes to cross the 1.6 m horizontal stretch, and (d) the hanging mass that would double the wave speed.

Solution:

Which $T$ is which: $T$ is the tension in newtons. It is set by the hanging block, and an ideal pulley passes it unchanged around the corner, so the horizontal stretch is under the same tension as the vertical one.

1. Convert. $$\mu = 2.0 \ \text{g/m} = 2.0 \times 10^{-3} \ \text{kg/m}$$

2. (a) Tension. The block hangs in equilibrium, so the string tension balances its weight: $$T = Mg = 5.0 \times 9.8 = 49 \ \text{N}$$

3. (b) Wave speed. $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{49}{2.0\times10^{-3}}} = \sqrt{24500} = 156.5 \ \text{m/s}$$

4. (c) Crossing time. $$t = \frac{\ell}{v} = \frac{1.6}{156.52} = 1.02 \times 10^{-2} \ \text{s} = 10.2 \ \text{ms}$$

5. (d) Doubling the speed. Since $v \propto \sqrt{T}$ and $T \propto M$, doubling $v$ needs four times the tension and hence four times the mass: $$M^{\,\prime} = 4 \times 5.0 = 20 \ \text{kg}$$ Check: $T^{\,\prime} = 196$ N, $v^{\,\prime} = \sqrt{\frac{196}{0.002}} = 313.0$ m/s, which is $2 \times 156.5$. Correct.

Final Answer: (a) 49 N; (b) 156.5 m/s; (c) 10.2 ms; (d) 20 kg.

Takeaway: A hanging mass is just a tension in disguise: write $T = Mg$ on its own line and the problem becomes an ordinary $v = \sqrt{T/\mu}$ question. And remember the square root — to double a wave speed you must quadruple the load, not double it.

Example 19: A wire of given radius and density

A steel wire of radius 0.40 mm is stretched by a tension of 160 N. The density of steel is 7800 kg/m³. Find (a) the linear mass density of the wire, (b) the speed of transverse waves on it, and (c) the frequency of a wave whose wavelength on the wire is 0.25 m.

Solution:

Two densities, two symbols: $\rho$ is the volume density in kg/m³ and $\mu$ the linear density in kg/m; they are linked by the cross-sectional area, $\mu = \rho A$. $T$ is the tension in newtons.

1. Cross-sectional area. Convert the radius first: $$r = 0.40 \ \text{mm} = 4.0 \times 10^{-4} \ \text{m}$$ $$A = \pi r^2 = 3.1416 \times \left(4.0\times10^{-4}\right)^2 = 5.027 \times 10^{-7} \ \text{m}^2$$

2. (a) Linear mass density. A one-metre length has volume $A \times 1$, so $$\mu = \rho A = 7800 \times 5.027\times10^{-7} = 3.92 \times 10^{-3} \ \text{kg/m}$$

3. (b) Wave speed. $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{160}{3.921\times10^{-3}}} = \sqrt{4.081\times10^{4}} = 202 \ \text{m/s}$$

4. (c) Frequency. $$\nu = \frac{v}{\lambda} = \frac{202.0}{0.25} = 808 \ \text{Hz}$$

Final Answer: (a) $3.92\times10^{-3}$ kg/m; (b) 202 m/s; (c) 808 Hz.

Takeaway: $\mu = \rho\pi r^2$ is the bridge between a wire you can measure with a screw gauge and a wave speed. Square the radius after converting it to metres — doing the conversion last is where the factors of $10^6$ go astray.

Example 20: Sound in carbon dioxide, twice over

For carbon dioxide at 0°C take $\gamma = 1.30$ and molar mass $M_0 = 44.0$ g/mol, with $R = 8.314$ J/(mol K). (a) Compute the speed of sound in it from Newton's isothermal formula. (b) Compute it with the Laplace correction. (c) By what percentage do they differ? (d) Compare with air at the same temperature, for which $\gamma = 1.40$ and $M_0 = 29.0$ g/mol.

Solution:

What goes into the formula: $\gamma$ is the ratio of specific heats, $M_0$ the molar mass in kg/mol, and the temperature goes in as $T$ in kelvin — this is the one place in the chapter where $T$ means neither a tension nor a period, so it is written $T_{\text{kelvin}}$ below.

1. Convert the inputs. $$M_0 = 44.0 \ \text{g/mol} = 0.0440 \ \text{kg/mol}, \qquad T_{\text{kelvin}} = 0 + 273 = 273 \ \text{K}$$

2. (a) Newton's formula treats the compressions as isothermal, so the relevant modulus is $P$ itself. Using $\frac{P}{\rho} = \frac{R T_{\text{kelvin}}}{M_0}$ from the gas law, $$v_{\text{N}} = \sqrt{\frac{P}{\rho}} = \sqrt{\frac{R T_{\text{kelvin}}}{M_0}} = \sqrt{\frac{8.314 \times 273}{0.0440}} = \sqrt{51584} = 227 \ \text{m/s}$$

3. (b) With the Laplace correction the compressions are adiabatic, the modulus is $\gamma P$, and $$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T_{\text{kelvin}}}{M_0}} = \sqrt{\frac{1.30 \times 8.314 \times 273}{0.0440}} = \sqrt{67059} = 259 \ \text{m/s}$$

4. (c) The gap. The ratio is exactly $\sqrt{\gamma}$: $$\frac{v}{v_{\text{N}}} = \sqrt{1.30} = 1.140, \qquad \text{a rise of } 14.0\%$$ For air, where $\gamma = 1.40$, the same argument gives $\sqrt{1.40} = 1.183$ — which is why Newton's prediction of roughly 280 m/s for air fell about 15% short of the measured 331 m/s.

5. (d) Air at 0°C, for comparison. $$v_{\text{air}} = \sqrt{\frac{1.40 \times 8.314 \times 273}{0.0290}} = \sqrt{109574} = 331 \ \text{m/s}$$ Carbon dioxide is slower, and the reason is mostly its mass: at 44 against 29 g/mol its molecules are more sluggish, and $v \propto \frac{1}{\sqrt{M_0}}$.

Final Answer: (a) 227 m/s; (b) 259 m/s; (c) 14.0% higher; (d) 331 m/s in air, so sound is about 22% slower in carbon dioxide.

Takeaway: The whole of the Laplace correction is one factor of $\sqrt{\gamma}$. Compressions in a sound wave happen far too fast for heat to leak out of them, so the process is adiabatic, not isothermal — and the correction is bigger for a gas with a bigger $\gamma$.

Example 21: A hot afternoon and a cold night

The speed of sound in air is 340 m/s at 15°C. Find it (a) at 40°C and (b) at $-10°C$. (c) At what temperature would it be 20% greater than its 15°C value?

Solution:

Before substituting: $v \propto \sqrt{T_{\text{kelvin}}}$, so every temperature must be converted to kelvin before any ratio is taken. Celsius temperatures in a ratio give nonsense — and a negative one gives an impossible square root.

1. Convert all three temperatures. $$15°C = 288 \ \text{K}, \qquad 40°C = 313 \ \text{K}, \qquad -10°C = 263 \ \text{K}$$

2. (a) At 40°C. $$v = 340\sqrt{\frac{313}{288}} = 340 \times \sqrt{1.0868} = 340 \times 1.0425 = 354 \ \text{m/s}$$

3. (b) At $-10°C$. $$v = 340\sqrt{\frac{263}{288}} = 340 \times \sqrt{0.9132} = 340 \times 0.9556 = 325 \ \text{m/s}$$

4. (c) Twenty per cent faster means $v = 1.20 \times 340 = 408$ m/s. Square the ratio to undo the root: $$\frac{T_{\text{kelvin}}}{288} = (1.20)^2 = 1.44 \ \Rightarrow \ T_{\text{kelvin}} = 288 \times 1.44 = 414.7 \ \text{K} = 141.7°C$$

5. A cross-check worth doing. At 0°C the same rule gives $340\sqrt{\frac{273}{288}} = 331$ m/s, which is the standard measured value for dry air. The starting figure of 340 m/s at 15°C is consistent with it.

Final Answer: (a) 354 m/s; (b) 325 m/s; (c) 414.7 K, that is about 142°C.

Takeaway: A 20% rise in speed needs a 44% rise in absolute temperature, because of the square root. Sound speed is a lazy function of temperature — a 25-degree swing between (a) and the starting point moves it by only 4%, which is why the rule of thumb "about 0.6 m/s per degree Celsius" works so well over ordinary ranges.

Example 22: Hydrogen against oxygen

At the same temperature, compare the speed of sound in hydrogen ($\gamma = 1.41$, $M_0 = 2.0$ g/mol) with that in oxygen ($\gamma = 1.40$, $M_0 = 32.0$ g/mol). (a) Find the ratio of the speeds. (b) Evaluate both at 0°C, with $R = 8.314$ J/(mol K).

Solution:

The formula to use: $v = \sqrt{\frac{\gamma R T_{\text{kelvin}}}{M_0}}$ with $M_0$ in kg/mol and $T_{\text{kelvin}}$ in kelvin. At a common temperature the $R T_{\text{kelvin}}$ cancels from any ratio, leaving only $\gamma$ and $M_0$.

1. (a) Take the ratio and cancel. $$\frac{v_{\text{H}}}{v_{\text{O}}} = \sqrt{\frac{\gamma_{\text{H}}/M_{0,\text{H}}}{\gamma_{\text{O}}/M_{0,\text{O}}}} = \sqrt{\frac{1.41/2.0}{1.40/32.0}} = \sqrt{\frac{0.7050}{0.04375}} = \sqrt{16.11} = 4.01$$ So sound travels about four times faster in hydrogen, and the two $\gamma$ values, being almost equal, contribute almost nothing — the factor of 4 is essentially $\sqrt{\frac{32}{2}} = 4$, pure molar mass.

2. (b) Absolute values at $0°C = 273$ K. $$v_{\text{H}} = \sqrt{\frac{1.41 \times 8.314 \times 273}{0.0020}} = \sqrt{1.600\times10^{6}} = 1265 \ \text{m/s}$$ $$v_{\text{O}} = \sqrt{\frac{1.40 \times 8.314 \times 273}{0.0320}} = \sqrt{9.930\times10^{4}} = 315 \ \text{m/s}$$ Their ratio is $\frac{1265}{315} = 4.01$, as it must be.

Final Answer: (a) about 4.01 to 1; (b) 1265 m/s in hydrogen, 315 m/s in oxygen.

Takeaway: Lighter gas, faster sound — and the dependence is $\frac{1}{\sqrt{M_0}}$, not $\frac{1}{M_0}$. Sixteen times lighter gives four times faster. When two gases have similar $\gamma$, you can do the whole comparison from the molar masses in your head.

Example 23: Ultrasound meeting a water surface

An ultrasonic source in air emits at 1.20 MHz. The beam strikes a flat water surface: part reflects back into the air, part is transmitted into the water. Take the speed of sound as 340 m/s in air and 1486 m/s in water. Find (a) the wavelength of the reflected sound and (b) of the transmitted sound. (c) A hospital scanner works at 4.0 MHz in tissue where the speed of sound is 1.70 km/s — what wavelength does it use?

Solution:

The frequency is fixed by the source and is the same in both media and in the reflected beam; the speed is fixed by the medium. Since $v = \nu\lambda$, the wavelength absorbs the entire change.

1. (a) The reflected sound never leaves the air. Its speed is still 340 m/s and its frequency is still 1.20 MHz: $$\lambda_{\text{air}} = \frac{v_{\text{air}}}{\nu} = \frac{340}{1.20\times10^{6}} = 2.83 \times 10^{-4} \ \text{m} = 0.283 \ \text{mm}$$

2. (b) The transmitted sound is in water, where the speed is 1486 m/s, but the frequency is unchanged at 1.20 MHz: $$\lambda_{\text{water}} = \frac{1486}{1.20\times10^{6}} = 1.24 \times 10^{-3} \ \text{m} = 1.24 \ \text{mm}$$ About 4.4 times longer, which is exactly the ratio of the two speeds.

3. (c) The scanner. With $v = 1.70$ km/s $= 1700$ m/s: $$\lambda = \frac{1700}{4.0\times10^{6}} = 4.25 \times 10^{-4} \ \text{m} = 0.425 \ \text{mm}$$

Final Answer: (a) 0.283 mm; (b) 1.24 mm; (c) 0.425 mm.

Takeaway: At a boundary, frequency is conserved and wavelength is not. Ask "which medium is this part of the beam in?" and read the speed off that; the frequency was decided back at the source and never changes. Part (c) also shows why scanners run at megahertz — you cannot resolve detail finer than about a wavelength, and 0.4 mm is fine enough to see inside a body.

Example 24: Pressure, temperature and humidity

Explain, using $v = \sqrt{\frac{\gamma P}{\rho}}$, why the speed of sound in air (a) does not depend on the pressure, (b) increases with temperature, and (c) increases with humidity. For (c), estimate the change when 3.0% of the air molecules (by mole) are replaced by water vapour, taking the molar mass of dry air as 29.0 g/mol and of water as 18.0 g/mol, and holding $\gamma$ fixed.

Solution:

What each symbol carries: $P$ is the pressure in pascals, $\rho$ the density in kg/m³, $\gamma$ the ratio of specific heats. The gas law lets $\frac{P}{\rho}$ be rewritten as $\frac{R T_{\text{kelvin}}}{M_0}$, and that rewriting is what makes all three parts obvious.

1. (a) Pressure. Compress a fixed mass of gas at constant temperature and the pressure rises — but so does the density, in exactly the same proportion, because the same mass now occupies a smaller volume. The ratio $\frac{P}{\rho}$ is untouched, so $v$ is untouched. Formally, $$\frac{P}{\rho} = \frac{R T_{\text{kelvin}}}{M_0}$$ in which $P$ does not appear at all. Sound at the top of a mountain travels no slower for the thinner air, provided the temperature is the same.

2. (b) Temperature. The same rewriting gives $$v = \sqrt{\frac{\gamma R T_{\text{kelvin}}}{M_0}} \ \Rightarrow \ v \propto \sqrt{T_{\text{kelvin}}}$$ Heat the gas and its molecules move faster; a disturbance is passed along by molecular collisions, so it is handed on more quickly.

3. (c) Humidity. Water vapour has $M_0 = 18.0$ g/mol, well below dry air's 29.0 g/mol. At a given pressure and temperature, a fixed number of molecules occupies a fixed volume whatever they are, so swapping heavy nitrogen and oxygen molecules for lighter water molecules makes the air less dense — and $v \propto \frac{1}{\sqrt{M_0}}$.

4. The estimate. The average molar mass of the mixture is the mole-weighted average: $$M_0^{\,\prime} = 0.97 \times 29.0 + 0.03 \times 18.0 = 28.13 + 0.54 = 28.67 \ \text{g/mol}$$ $$\frac{v^{\,\prime}}{v} = \sqrt{\frac{29.0}{28.67}} = \sqrt{1.0115} = 1.0057$$ a rise of 0.57%, or about 2 m/s on a speed of 340 m/s.

Final Answer: (a) $P$ cancels out of $\frac{P}{\rho}$ at fixed temperature; (b) $v \propto \sqrt{T_{\text{kelvin}}}$; (c) moist air is less dense, and 3% water vapour raises the speed by about 0.57%, roughly 2 m/s.

Takeaway: Rewrite $\sqrt{\frac{\gamma P}{\rho}}$ as $\sqrt{\frac{\gamma R T_{\text{kelvin}}}{M_0}}$ and all three answers fall out at once. The second form contains no pressure, an explicit temperature, and an explicit molar mass — which is precisely the list of things the question asked about.