Waves and Sound

Sound is the wave you can hear, and almost every question on it is a ratio or a sign. The Waves and Optics area carries 8 of the 50 Physics questions (official MEC weightage), and we estimate about 3 of them come from this chapter. They cluster on five things: reading y = A\sin(\omega t - kx), the speed of sound in a gas, the harmonic series of a pipe, the decibel, and the Doppler formula — with the sign in the Doppler formula the single most common place to lose the mark.

11.1 Progressive waves and the wave equation

A wave is a disturbance that carries energy and momentum from one place to another without carrying matter. A mechanical wave needs a material medium (sound, waves on a string, ripples); an electromagnetic wave does not (light, radio). In a transverse wave the particles vibrate perpendicular to the direction of travel; in a longitudinal wave they vibrate along it.

	Transverse	Longitudinal
Particle motion	perpendicular to the wave direction	parallel to the wave direction
Form	crests and troughs	compressions and rarefactions
Can be polarized	yes	no
Can travel in a solid	yes	yes
Can travel in a liquid or gas	no (no shear)	yes
Examples	light, wave on a string, ripple	sound in air, sound in steel

Sound in air is longitudinal: the air layers vibrate back and forth along the direction the sound travels, producing alternate regions of high pressure (compression) and low pressure (rarefaction).

The wave quantities

Quantity	Symbol	Definition	SI unit
Amplitude	A	maximum displacement of a particle from its mean position	m
Wavelength	\lambda	distance between two nearest particles in the same phase	m
Frequency	f	number of complete waves passing a point per second	Hz
Period	T	time for one complete wave; T = 1/f	s
Wave speed	v	distance travelled by a given phase per second	m s⁻¹
Angular frequency	\omega	\omega = 2\pi f = 2\pi/T	rad s⁻¹
Angular wave number	k	k = 2\pi/\lambda	rad m⁻¹

One complete wave passes a point in one period and occupies one wavelength, so

The progressive wave equation

For a wave of amplitude A travelling in the positive x direction,

Every symbol: y is the displacement of the particle at position x at time t; A is the amplitude; \omega = 2\pi f is the angular frequency; k = 2\pi/\lambda is the angular wave number; and (\omega t - kx) is the phase of that particle, measured in radians.

Why the minus sign means "towards +x": follow a point of constant phase. If \omega t - kx = constant, then differentiating gives \omega - k\,dx/dt = 0, so dx/dt = \omega/k = v > 0 — the phase pattern moves to the right. With a plus sign, y = A\sin(\omega t + kx), the same argument gives dx/dt = -v: the wave travels towards -x. y = A\sin(kx - \omega t) is the same wave as the first, since \sin(-\theta) = -\sin\theta only flips the sign of y.

Phase difference. Two points a distance \Delta x apart differ in phase by \Delta\phi = \frac{2\pi}{\lambda}\Delta x; one point at two times \Delta t apart differs by \frac{2\pi}{T}\Delta t. A path difference of \lambda is a phase difference of 2\pi; a path difference of \lambda/2 is \pi.

Particle velocity is not wave velocity

This is the single most examined distinction in the chapter. Differentiate y = A\sin(\omega t - kx):

and the slope of the wave profile at that instant is

Dividing, \dfrac{\partial y/\partial t}{\partial y/\partial x} = -\dfrac{\omega}{k} = -v, so

The minus sign is the physics: where the profile slopes up to the right, the particle is moving down, because the wave is arriving from the left and the crest has not reached it yet.

	Wave speed v	Particle speed v_p
What moves	the phase pattern (energy)	the medium's particles
Value	\omega/k, fixed by the medium	A\omega\cos(\omega t - kx), changes with time
Maximum	v	A\omega
Direction	along the wave	along the vibration (perpendicular for a transverse wave)
Depends on	medium only	amplitude and frequency as well

The ratio of the two maxima is \dfrac{A\omega}{v} = Ak = \dfrac{2\pi A}{\lambda} — usually a small number. In a sound wave in air the particles move a few millimetres per second while the wave moves at 340 m s⁻¹.

Superposition and reflection

The principle of superposition states that when two or more waves arrive at a point, the resultant displacement is the algebraic sum of the separate displacements. Waves pass through one another unchanged.

When a wave meets a boundary:

• at a rigid (fixed) end it reflects with a phase change of \pi — a crest comes back as a trough, and the displacement at the boundary is always zero;
• at a free (open) end it reflects with no phase change — a crest comes back as a crest, and the displacement at the boundary is maximum.

This difference is exactly why a string fixed at both ends and a pipe open at both ends have different harmonic series, and why a pipe closed at one end has none of the even harmonics.

11.2 Speed of sound in solids, liquids and gases

Sound is an elastic wave, so its speed is fixed by two properties of the medium: how stiff it is (the elastic modulus E) and how heavy it is (the density \rho):

Stiffness pushes the disturbed layer back quickly; inertia slows it down. That one formula explains every comparison in the table below.

Medium	Modulus to use	Formula	Typical speed (m s⁻¹)
Solid (long thin rod)	Young's modulus Y	v = \sqrt{Y/\rho}	steel 5000, aluminium 5100, glass 5000
Liquid	Bulk modulus B	v = \sqrt{B/\rho}	water 1480 (20 °C), sea water 1530
Gas	Bulk modulus, adiabatic	v = \sqrt{\gamma P/\rho}	air 331 (0 °C), 343 (20 °C)

Solids are fastest because their elastic moduli are enormous compared with their densities; gases are slowest because a gas is easy to squeeze.

Newton's value and Laplace's correction

The factors that change the speed of sound

Factor	Effect on v in a gas	Reason
Temperature	increases; v \propto \sqrt{T} (kelvin)	v = \sqrt{\gamma RT/M}
Pressure	none, at constant temperature	\rho \propto P, so P/\rho is unchanged
Molar mass	decreases as M rises; v \propto 1/\sqrt{M}	lighter gas, faster sound
Humidity	increases slightly in air	water vapour (M = 18) is lighter than air (M \approx 29)
Wind	adds vectorially to v	the medium itself is moving
Frequency	none (in air, over the audible range)	air is non-dispersive for sound
Amplitude	none, for ordinary sounds	small-amplitude elasticity is linear

Near 0 °C, v = 331\sqrt{1 + t/273} \approx 331 + 0.61t, so the speed rises by about 0.61 m s⁻¹ for each °C.

11.3 Stationary waves: harmonics and overtones in strings and pipes

How a stationary wave is made

Send a wave along a string and let it reflect at the fixed end. The incident and reflected waves have the same amplitude and frequency and travel in opposite directions, so they superpose:

The result is a stationary (standing) wave. It is no longer a travelling wave: the x and t parts have separated, so every particle vibrates with the same frequency \cos(\omega t) but with an amplitude 2A\sin(kx) that depends on where it is.

Feature	Where	What happens there
Node	\sin(kx) = 0, i.e. x = 0, \lambda/2, \lambda,\dots	permanently at rest
Antinode	\sin(kx) = \pm1, i.e. x = \lambda/4, 3\lambda/4,\dots	vibrates with amplitude 2A

Distance between two neighbouring nodes = \lambda/2; between a node and the next antinode = \lambda/4. All particles between two consecutive nodes vibrate in phase; particles in adjacent loops vibrate in opposite phase. Energy shuttles back and forth between the loops but there is no net transfer of energy along the wave — that is what "stationary" means.

Strings fixed at both ends

Both ends are nodes, so the string holds a whole number of half-wavelengths:

All harmonics are present. The fundamental f_1 = v/2L is the first harmonic; 2f_1 is the second harmonic or first overtone; 3f_1 the third harmonic or second overtone. For a string of mass per unit length \mu under tension T, v = \sqrt{T/\mu}, so

The three laws of the vibrating string follow at once: f \propto 1/L (length), f \propto \sqrt{T} (tension), f \propto 1/\sqrt{\mu} (mass per unit length).

Pipes

A pipe has a node of displacement at a closed end (the air cannot move against the solid cap) and an antinode at an open end (the air is free to swing).

	Open pipe (both ends open)	Closed pipe (one end closed)
Boundary conditions	antinode at each end	node at the closed end, antinode at the open end
Length in the nth mode	L = n\lambda/2	L = (2n-1)\lambda/4
Wavelength	\lambda_n = 2L/n	\lambda_n = 4L/(2n-1)
Frequency	f_n = nv/2L	f_n = (2n-1)v/4L
Harmonics present	all: f_1, 2f_1, 3f_1,\dots	odd only: f_1, 3f_1, 5f_1,\dots
Fundamental	v/2L	v/4L
First overtone	2v/2L (2nd harmonic)	3v/4L (3rd harmonic)
Second overtone	3v/2L (3rd harmonic)	5v/4L (5th harmonic)
Nodes / antinodes in the nth mode	n+1 nodes, n antinodes	n nodes, n antinodes
Effective length with end correction	L + 2e	L + e

For a 0.5 m pipe with v = 340 m s⁻¹ the numbers are:

Mode	Open pipe	Closed pipe
Fundamental	340 Hz	170 Hz
1st overtone	680 Hz	510 Hz
2nd overtone	1020 Hz	850 Hz

Notice the closed pipe has half the fundamental of the open pipe of the same length, and every second harmonic is missing.

End correction

The antinode at an open end does not sit exactly at the cut edge — the air just outside the pipe moves too, so the effective length is slightly longer. For a pipe of diameter d, each open end adds

so a closed pipe has effective length L + e and an open pipe L + 2e. End correction lowers every frequency slightly.

The resonance tube measures it. A tuning fork of frequency f is held over a tube whose water level can be raised. Resonance (loudest sound) occurs when the air column is a quarter-wavelength, then three-quarters:

Subtracting, l_2 - l_1 = \lambda/2, so

The subtraction kills e, which is why the tube gives a good value of v even though the end correction is unknown.

Beats

Two waves of slightly different frequencies travelling in the same direction superpose. At a point the two displacements are A\cos(2\pi f_1 t) and A\cos(2\pi f_2 t); their sum is

a wave of the average frequency whose amplitude is slowly modulated. The amplitude is maximum twice per modulation cycle, so the beat frequency is

Beats are heard only when f_b is small — below about 10 Hz the ear hears separate pulses; above about 15 Hz it hears a rough, dissonant note instead.

11.4 Pressure amplitude, intensity and loudness, pitch and quality

The pressure wave

A sound wave can be described either by the displacement of the air particles, y = A\sin(\omega t - kx), or by the excess pressure it creates. The two are a quarter-cycle out of phase: where the particles are crowded together (displacement zero, changing fastest) the pressure is greatest.

p_m is the pressure amplitude, in pascal. Because p_m \propto \omega A = 2\pi fA, a high-frequency wave of the same displacement carries far more pressure variation — and far more energy.

A consequence worth memorising: at a closed end the air cannot move, so it is a displacement node and a pressure antinode; at an open end the air swings freely, a displacement antinode and a pressure node.

Intensity

Intensity is the energy crossing unit area per second, perpendicular to the direction of travel: I = P/A, unit W m⁻².

So for the same medium and frequency, I \propto A^2; for the same amplitude, I \propto f^2.

Inverse-square law. A point source radiating power P equally in all directions spreads it over a sphere of area 4\pi r^2:

Double the distance and the intensity falls to a quarter; the amplitude falls to a half, because A \propto \sqrt{I}.

Loudness in decibels

The ear responds to intensity over about thirteen powers of ten, so a logarithmic scale is used. The loudness level in decibels is

I_0 is the threshold of hearing at 1 kHz. Some landmarks:

Source	L (dB)	I (W m⁻²)
threshold of hearing	0	10^{-12}
rustling leaves	10	10^{-11}
whisper	20	10^{-10}
normal conversation	60	10^{-6}
busy street in Kathmandu	80	10^{-4}
rock concert	110	10^{-1}
jet engine at 30 m	130	10

Two rules that answer most dB questions:

• +10 dB means ten times the intensity; +20 dB means a hundred times, and also ten times the pressure amplitude and ten times the displacement amplitude.
• Two identical sources sounding together give +3 dB, because the intensities add (2I), not the amplitudes.

Pitch, loudness and quality

Sensation	Physical quantity it follows	How it changes
Pitch	frequency	higher frequency → higher pitch
Loudness	intensity (and frequency, through the ear's response)	more intensity → louder
Quality (timbre)	the number and relative strength of the overtones	same pitch, different waveform → different instrument

A violin and a flute playing the same 440 Hz note have the same pitch and can have the same loudness, but they sound different because their overtone mixtures differ. That mixture is the quality of the note.

11.5 Doppler effect

The Doppler effect is the apparent change in the frequency of a wave when the source, the observer, or both, move relative to the medium.

Source	Observer	Apparent frequency
at rest	at rest	f
at rest	moving towards, v_o	f\dfrac{v + v_o}{v} (higher)
at rest	moving away, v_o	f\dfrac{v - v_o}{v} (lower)
moving towards, v_s	at rest	f\dfrac{v}{v - v_s} (higher)
moving away, v_s	at rest	f\dfrac{v}{v + v_s} (lower)
moving towards, v_s	moving towards, v_o	f\dfrac{v + v_o}{v - v_s}
moving away, v_s	moving away, v_o	f\dfrac{v - v_o}{v + v_s}

Where the formula comes from

Moving source. The source emits f waves per second and at the same time moves towards the observer at v_s. During one period T = 1/f it advances v_sT, so the waves it has emitted are squeezed into the shorter distance

These waves still travel at v through the air, so the observer receives

Moving observer. The source is at rest, so the wavelength in the air is the normal \lambda = v/f. The observer meets the waves at the relative speed v + v_o, so

Both. Put the two together: the wavelength is set by the source's motion and the rate of meeting is set by the observer's motion, giving the combined formula above.

Echoes and reflections

When the sound reflects from a wall, treat the problem in two steps: first the wall is an observer receiving f_1; then the wall is a stationary source re-emitting f_1. If the source itself moves towards the wall at v_s, the wave arriving at the wall has f_1 = f\dfrac{v}{v - v_s}, and the source (still moving) hears the echo as

so the beat frequency between the emitted note and its echo is f'' - f = f\dfrac{2v_s}{v - v_s}.

Where it is used

• Ambulance and train sirens: the pitch drops abruptly as the vehicle passes.
• Police speed radar and weather radar: the shift in the reflected signal gives the speed of the target (for light, the same idea with c).
• Doppler echocardiography: ultrasound reflected from moving blood measures flow speed and direction.
• Astronomical redshift: light from receding galaxies is shifted to longer wavelengths — evidence that the universe is expanding.

Formula and fact sheet

Topic	Formula or fact
Wave equation	y = A\sin(\omega t - kx) travels +x; \omega = 2\pi f, k = 2\pi/\lambda
Wave speed	v = f\lambda = \omega/k
Phase difference	\Delta\phi = 2\pi\Delta x/\lambda = 2\pi\Delta t/T
Particle velocity	v_p = A\omega\cos(\omega t - kx) = -v\times\text{slope}; maximum A\omega
Speed in a medium	v = \sqrt{E/\rho}: solid \sqrt{Y/\rho}, liquid \sqrt{B/\rho}, gas \sqrt{\gamma P/\rho}
Gas	v = \sqrt{\gamma RT/M}; v \propto \sqrt{T}, v \propto 1/\sqrt{M}; independent of P at constant T
Air	331 m s⁻¹ at 0 °C, 343 at 20 °C, +0.61 m s⁻¹ per °C
String	v = \sqrt{T/\mu}; f_n = \frac{n}{2L}\sqrt{T/\mu}, all n
Open pipe	f_n = nv/2L; fundamental v/2L; all harmonics
Closed pipe	f_n = (2n-1)v/4L; fundamental v/4L; odd harmonics only
End correction	e = 0.3d per open end; closed L+e, open L+2e
Resonance tube	v = 2f(l_2 - l_1); e = (l_2 - 3l_1)/2
Beats	f_b = f_1 - f_2 (the difference, taken positive); T_b = 1/f_b; heard for f_b < 10 Hz
Pressure amplitude	p_m = \rho v\omega A = BAk; \Delta p leads displacement by 90^{\circ}
Intensity	I = \frac{1}{2}\rho v\omega^2A^2 = 2\pi^2\rho vf^2A^2 = p_m^2/2\rho v
Inverse square	I \propto 1/r^2, A \propto 1/r
Decibel	L = 10\log_{10}(I/I_0), I_0 = 10^{-12} W m⁻²; +10 dB = 10\times I; +3 dB = 2\times I
Doppler	f' = f\frac{v \pm v_o}{v \mp v_s}; upper sign for approach
Echo	f'' = f\frac{v + v_s}{v - v_s}; beat = f\frac{2v_s}{v - v_s}
Node / antinode	node–node \lambda/2; node–antinode \lambda/4

Chapter summary

• A progressive wave is y = A\sin(\omega t - kx), with v = \omega/k = f\lambda; the minus sign means it travels towards +x.
• Particle velocity is v_p = -v\times\text{slope} of the profile, with maximum A\omega — a completely different number from the wave speed.
• Sound speed is set by stiffness and density: \sqrt{Y/\rho} in solids, \sqrt{B/\rho} in liquids, \sqrt{\gamma P/\rho} = \sqrt{\gamma RT/M} in gases. It rises with temperature, falls with molar mass, and does not change with pressure at constant temperature.
• A string fixed at both ends and an open pipe give every harmonic, f_n = nv/2L; a closed pipe gives odd harmonics only, f_n = (2n-1)v/4L, and has half the fundamental of the open pipe of the same length.
• End correction e = 0.3d per open end; the resonance tube removes it by using l_2 - l_1 = \lambda/2.
• Beats: f_b = |f_1 - f_2|, audible only for small differences; waxing a fork lowers its frequency, filing raises it.
• Intensity is 2\pi^2\rho vf^2A^2, falls as 1/r^2, and converts to loudness by L = 10\log(I/I_0); +10 dB is ten times the intensity, +3 dB is two sources.
• Doppler: f' = f(v \pm v_o)/(v \mp v_s) with the upper sign for approach; a source moving away lowers the pitch, and motion across the line of sight gives no shift.

Practice questions

Set A — Recall (28 questions)

1. In y = A\sin(\omega t - kx), the quantity \omega/k is theawave speedbparticle speedcfrequencydwavelength
2. The SI unit of the intensity of a wave isaWbW m⁻¹cW m⁻²dJ m⁻²
3. Sound cannot travel throughasteelbvacuumcwaterdhydrogen
4. A pipe closed at one end can produceaall harmonicsbonly even harmonicsconly the fundamentaldonly odd harmonics
5. The beat frequency of two waves of frequencies f_1 and f_2 isa|f_1 - f_2|bf_1 + f_2c(f_1 + f_2)/2d\sqrt{f_1f_2}
6. The loudness level in decibels is defined with the reference intensitya1 W m⁻²b10⁻⁶ W m⁻²c10⁻¹² W m⁻²d10⁻¹⁶ W m⁻²
7. The pitch of a musical note is determined by itsaintensitybfrequencycwaveformdspeed
8. The number of nodes in the nth harmonic of a string fixed at both ends isanbn - 1c2ndn + 1
9. At constant temperature, the speed of sound in a gas isaindependent of pressurebproportional to pressurecproportional to \sqrt{P}dinversely proportional to pressure
10. Laplace corrected Newton's formula by treating the propagation of sound asaisothermalbisobariccadiabaticdisochoric
11. For a gas at absolute temperature T, the speed of sound isa\sqrt{RT/M}b\sqrt{\gamma RT/M}c\gamma\sqrt{RT/M}d\sqrt{\gamma M/RT}
12. An observer moving towards a stationary source with speed v_o hears a frequencyaf(v - v_o)/vbfv/(v + v_o)cfv/(v - v_o)df(v + v_o)/v
13. The distance between two consecutive nodes in a stationary wave isa\lambda/2b\lambdac\lambda/4d2\lambda
14. Two points on a wave separated by \lambda/4 differ in phase bya\pib2\pic\pi/2d\pi/4
15. The relation between wave speed, frequency and wavelength isav = f/\lambdabv = f\lambdacv = \lambda/fdv = f^2\lambda
16. In a stationary waveaenergy is carried at speed vball particles have equal amplitudecthe wave travels in one directiondthere is no net transfer of energy
17. The end correction of a pipe of diameter d, open at one end, is abouta0.3db0.6dc0.15ddd
18. The intensity of a wave is proportional toathe amplitudebthe square root of the amplitudecthe square of the amplitudedthe wavelength
19. The unit of loudness level is theawattbdecibelchertzdpascal
20. A sound wave of frequency 170 Hz travels at 340 m s⁻¹. Its wavelength isa0.5 mb1 mc4 md2 m
21. Which of the following is a longitudinal wave?asound in airblight in vacuumca wave on a stringda ripple on water
22. In a pipe open at both ends, the frequencies produced areaodd multiples of the fundamental onlybeven multiples onlycall integral multiples of the fundamentaldunrelated to the fundamental
23. The frequency of a stretched string is proportional toathe tensionbthe square root of the tensioncthe lengthdthe square root of the length
24. When a source of sound moves away from a stationary observer, the frequency heard isahigherbthe samecdoubleddlower
25. The speed of sound is greatest inaairbwatercsteeldvacuum
26. Beats are clearly heard when the frequencies of the two sources differ by less than abouta10 Hzb100 Hzc1000 Hzd10000 Hz
27. The third harmonic of a closed pipe containsa1 node and 1 antinodeb2 nodes and 2 antinodesc3 nodes and 3 antinodesd2 nodes and 3 antinodes
28. The speed of a wave is determined byaits frequencybits amplitudecits wavelengthdthe medium

Answers and explanations — Set A

1. (a) — v = \omega/k; multiplying \omega by t and k by x leaves a phase, and \omega/k has the unit rad s⁻¹ ÷ rad m⁻¹ = m s⁻¹.

2. (c) — intensity is power per unit area, W m⁻². The watt alone is power; J m⁻² is energy per area, not per area per second.

3. (b) — sound is a mechanical wave and needs particles to compress; in vacuum there is nothing to vibrate, so no sound (light still passes).

4. (d) — the closed end forces a node and the open end an antinode, which fits an odd number of quarter-wavelengths: L = (2n-1)\lambda/4. Even harmonics would need a node at both ends.

5. (a) — the envelope repeats |f_1 - f_2| times a second. The sum f_1 + f_2 is the frequency you would hear inside the beats, not the beat rate.

6. (c) — I_0 = 10^{-12} W m⁻² is the threshold of hearing at 1 kHz, the zero of the dB scale.

7. (b) — pitch is the ear's judgement of frequency. Intensity gives loudness and waveform gives quality.

8. (d) — n half-wavelengths need n+1 nodes: two ends plus n-1 interior nodes.

9. (a) — doubling P doubles \rho as well, so P/\rho and hence v = \sqrt{\gamma P/\rho} are unchanged. Only temperature (or molar mass) changes the speed.

10. (c) — compressions follow too quickly for heat to flow, so PV^{\gamma} is constant and B = \gamma P; that raises 280 m s⁻¹ to 331 m s⁻¹.

11. (b) — v = \sqrt{\gamma P/\rho} and P = \rho RT/M give v = \sqrt{\gamma RT/M}. Option (a) omits \gamma, which is about 1.4 for air.

12. (d) — the observer meets the waves at relative speed v + v_o while the wavelength stays v/f, so f' = f(v + v_o)/v. Option (c) is the moving-source formula.

13. (a) — the displacement repeats every half-wavelength, so nodes are \lambda/2 apart; node to antinode is \lambda/4.

14. (c) — a quarter of a wavelength is a quarter of 2\pi, that is \pi/2.

15. (b) — one wave passes per period and occupies one wavelength, so v = f\lambda.

16. (d) — the two travelling waves carry equal energy in opposite directions; energy oscillates between kinetic and elastic forms but does not travel.

17. (a) — the standard value is e = 0.3d (equivalently 0.6r) for each open end.

18. (c) — I = 2\pi^2\rho vf^2A^2, so doubling the amplitude quadruples the intensity.

19. (b) — the decibel. The watt is power, the hertz frequency and the pascal pressure.

20. (d) — \lambda = v/f = 340/170 = 2 m.

21. (a) — sound in air is a pressure (compression) wave; light is transverse, and so are waves on a string and water ripples.

22. (c) — an open pipe holds any number of half-wavelengths, so every harmonic nv/2L appears.

23. (b) — f = \frac{1}{2L}\sqrt{T/\mu}, so f \propto \sqrt{T}; quadrupling the tension only doubles the frequency.

24. (d) — a receding source stretches the wavelength, f' = fv/(v + v_s) < f.

25. (c) — steel has by far the largest elastic modulus for its density; gases are slowest because they are easy to compress.

26. (a) — above about 10–15 Hz the separate pulses merge into a rough note and the beats are no longer heard as beats.

27. (b) — the third harmonic is 3\lambda/4 long: nodes at the closed end and at 2L/3, antinodes at L/3 and at the open end — two of each.

28. (d) — v is fixed by the elasticity and inertia of the medium; f and \lambda can change, but their product stays at the value the medium allows.

Set B — Understanding (17 questions)

1. A wave y = A\sin(\omega t - kx) travels towards +x. At a point where the profile slopes upward, the particle there is movingaupwardbtowards +xcdownwarddwith the wave
2. Sound passes from air into water. Which quantity is unchanged?afrequencybspeedcwavelengthdamplitude
3. Two pipes of equal length, one open at both ends and one closed at one end, have fundamental frequencies in the ratio (open : closed)a1 : 2b1 : 1c4 : 1d2 : 1
4. The first overtone of a closed pipe is theasecond harmonicbthird harmoniccfourth harmonicdfifth harmonic
5. The tension in a sonometer wire is made four times larger. The frequency of the wire becomesa4 timesbhalfc2 timesd\sqrt{2} times
6. An unknown fork gives 4 beats per second with a 256 Hz fork. On loading the unknown fork with wax the beat rate falls to 2 per second. Its frequency isa260 Hzb252 Hzc254 Hzd264 Hz
7. A source moves away from a stationary observer at half the speed of sound. The frequency heard isaf/2b3f/2cfd2f/3
8. An observer moves away from a stationary source at half the speed of sound. The frequency heard isa2f/3bf/2c3f/2df
9. The intensity of sound 2 m from a point source is I. At 8 m from the same source it isaI/16bI/4cI/8dI/64
10. An increase of 10 dB in the loudness level means the intensity has increased by a factor ofa2b5c10d100
11. An increase of 20 dB means the pressure amplitude has increased by a factor ofa2b20c100d10
12. Two identical sources sounding together give a level 3 dB above one alone. Their combined intensity isa3 timesb2 timesc1.5 timesd10 times
13. The speed of sound in a gas doubles when the absolute temperature isaincreased four timesbdoubledchalveddincreased eight times
14. On a humid day the speed of sound in airadecreasesbstays the samecincreasesdbecomes zero
15. At a point where the displacement of a sound wave is zero, the pressure variation isazerobmaximumchalf its maximumdalways negative
16. In a stationary wave, particles in two adjacent loopsavibrate in phasebare at restchave different frequenciesdvibrate in opposite phase
17. The second resonance length in a resonance tube is aboutathree times the firstbtwice the firstcfour times the firstdhalf the first

Answers and explanations — Set B

1. (c) — v_p = -v\times\text{slope}; a positive slope with a wave moving towards +x gives a negative (downward) particle velocity. The crest is still to the left, so the particle is on its way down.

2. (a) — the frequency is fixed by the source and cannot change at a boundary; the higher speed in water forces a longer wavelength (\lambda = v/f).

3. (d) — open v/2L, closed v/4L, so the ratio is 2 : 1. The closed pipe sounds an octave lower.

4. (b) — closed pipes skip the second harmonic, so the first overtone is the third harmonic, 3v/4L.

5. (c) — f \propto \sqrt{T}, and \sqrt{4} = 2. Quadrupling the tension doubles the frequency, not quadruples it.

6. (a) — wax lowers the unknown frequency. If it was 260 Hz it falls to about 258 and the beats with 256 fall from 4 to 2, as stated. Had it been 252, waxing would have taken it further from 256 and the beats would have risen.

7. (d) — f' = f\dfrac{v}{v + v/2} = \dfrac{2}{3}f. The denominator grows, so the frequency falls.

8. (b) — the observer meets fewer waves: f' = f\dfrac{v - v/2}{v} = \dfrac{f}{2}.

9. (a) — I \propto 1/r^2; the distance is four times larger, so the intensity is 1/16.

10. (c) — 10\log_{10}x = 10 gives x = 10. Option (d) would be 20 dB.

11. (d) — 20 dB is a factor 10^2 in intensity, and amplitude goes as the square root of intensity, so a factor 10.

12. (b) — two incoherent identical sources add their intensities, 2I, which is 10\log 2 = 3 dB.

13. (a) — v \propto \sqrt{T}, so doubling v needs T four times larger. Doubling T only multiplies v by \sqrt{2}.

14. (c) — water vapour has molar mass 18 against about 29 for dry air, so humid air is less dense and sound travels slightly faster.

15. (b) — displacement and pressure are a quarter-cycle apart: a displacement node (particles crowded, zero displacement but maximum compression change) is a pressure antinode.

16. (d) — the loops on either side of a node move in opposite directions at every instant; particles inside one loop are in phase.

17. (a) — l_2 \approx 3l_1, because the second resonance is at three quarter-wavelengths against one at the first.

Set C — Application (12 questions)

1. A wave is y = 0.02\sin(100t - 2x) in SI units. Its speed isa2 m s⁻¹b50 m s⁻¹c100 m s⁻¹d200 m s⁻¹
2. For the same wave, the maximum speed of a particle of the medium isa50 m s⁻¹b0.04 m s⁻¹c100 m s⁻¹d2 m s⁻¹
3. The speed of sound in air at 0 °C is 331 m s⁻¹. At 27 °C it is abouta347 m s⁻¹b331 m s⁻¹c356 m s⁻¹d363 m s⁻¹
4. The ratio of the speed of sound in hydrogen to that in oxygen at the same temperature isa2b2.8c4d16
5. A pipe closed at one end is 0.5 m long. Taking v = 340 m s⁻¹, its fundamental frequency isa85 Hzb170 Hzc340 Hzd510 Hz
6. A pipe open at both ends is 0.5 m long. Taking v = 340 m s⁻¹, its fundamental frequency isa85 Hzb170 Hzc680 Hzd340 Hz
7. A train sounding a 500 Hz horn approaches a stationary listener at 34 m s⁻¹ (v = 340 m s⁻¹). The frequency heard isa556 Hzb550 Hzc455 Hzd444 Hz
8. The same train, now receding at 34 m s⁻¹, is heard ata556 Hzb545 Hzc455 Hzd500 Hz
9. A source of 680 Hz moves at 20 m s⁻¹ directly towards a wall. A listener standing at the source hears the echo with a beat frequency of (v = 340 m s⁻¹)a20 Hzb40 Hzc42 Hzd85 Hz
10. The intensity of a sound is 10^{-8} W m⁻². Its loudness level isa8 dBb40 dBc80 dBd120 dB
11. Two tuning forks of 512 Hz and 516 Hz are sounded together. The beat frequency isa2 Hzb8 Hzc4 Hzd514 Hz
12. A 500 Hz fork resonates with air columns of 17 cm and 51 cm in a resonance tube. The speed of sound isa340 m s⁻¹b330 m s⁻¹c350 m s⁻¹d170 m s⁻¹

Answers and explanations — Set C

1. (b) — v = \omega/k = 100/2 = 50 m s⁻¹. The amplitude 0.02 plays no part in the wave speed.

2. (d) — v_{p,\max} = A\omega = 0.02\times100 = 2 m s⁻¹. Option (a) is the wave speed, which is 25 times larger.

3. (a) — v = 331\sqrt{300/273} = 331(1.048) = 347 m s⁻¹. Adding 27 to the Celsius reading is not the same as increasing T by 10 %.

4. (c) — both diatomic, so \gamma cancels: \sqrt{M_{O_2}/M_{H_2}} = \sqrt{32/2} = 4.

5. (b) — closed pipe: f_1 = v/4L = 340/(4\times0.5) = 170 Hz.

6. (d) — open pipe: f_1 = v/2L = 340/(2\times0.5) = 340 Hz, twice the closed-pipe value.

7. (a) — f' = 500\times\dfrac{340}{340 - 34} = 500\times\dfrac{340}{306} = 556 Hz. Option (b) is the observer-moving answer, not the source-moving one.

8. (c) — f' = 500\times\dfrac{340}{340 + 34} = 500\times\dfrac{340}{374} = 455 Hz.

9. (d) — the wall receives 680\times340/320 = 722.5 Hz and re-emits it; the moving source hears 722.5\times360/340 = 765 Hz. The beat with the emitted 680 Hz is 85 Hz.

10. (b) — L = 10\log(10^{-8}/10^{-12}) = 10\log 10^4 = 40 dB.

11. (c) — f_b = |512 - 516| = 4 Hz.

12. (a) — v = 2f(l_2 - l_1) = 2(500)(0.34) = 340 m s⁻¹. The end correction cancels in the subtraction, which is why the tube works.