Mechanical waves

A mechanical wave carries energy through matter without carrying the matter along with it. Sound is the mechanical wave you meet every day, and this short unit answers one question about it: what fixes its speed? The answer turns out to depend on only two properties of the medium — how stiff it is and how heavy it is — plus, for a gas, one subtlety about heat that took physics a century to get right.

7.1 Speed of wave motion; velocity of sound in solid and liquid

Every wave obeys the kinematic relation

but this only relates speed, frequency and wavelength — it does not say what the speed is. The speed is decided by the medium, not by the source. If you sound a horn louder or at a higher pitch, the sound still arrives at the same time.

A mechanical wave travels because a disturbed part of the medium pulls its neighbour back (elasticity) and because that neighbour resists being moved (inertia). Newton showed that these two combine as

where \rho is the density of the medium and E is the modulus of elasticity appropriate to the kind of deformation the wave produces. Choosing the right modulus is the whole skill.

Medium and wave	Deformation	Modulus used	Speed
Long thin solid rod, longitudinal	stretch/squeeze along rod	Young's modulus Y	v=\sqrt{Y/\rho}
Liquid, longitudinal	change of volume	Bulk modulus B	v=\sqrt{B/\rho}
Gas, longitudinal	change of volume	Bulk modulus of gas	v=\sqrt{B/\rho}
Stretched string, transverse	change of shape	tension T, mass per length \mu	v=\sqrt{T/\mu}

A liquid and a gas have no rigidity, so they cannot be sheared: sound in a fluid is always longitudinal. A solid resists both squeezing and shearing, so a solid carries both longitudinal and transverse waves — which is exactly why an earthquake such as Gorkha 2015 produces two body waves, the fast longitudinal P-wave and the slower transverse S-wave, and why seismologists at the National Seismological Centre can locate an epicentre from the gap between their arrivals.

Notice from the table that steel is about eight times denser than water, yet sound goes more than three times faster in steel. Density alone never decides the answer: Young's modulus of steel is roughly a hundred times the bulk modulus of water, and stiffness wins.

7.2 Velocity of sound in gas

For a gas the deformation is a change of volume, so the modulus needed is the bulk modulus

The minus sign is there because an increase in pressure produces a decrease in volume, and it makes B positive.

Newton (1686) now had to decide how the gas is compressed. He argued that the compressions and rarefactions are slow enough, and the gas thin enough, for the heat generated in a compression to leak away at once, so the temperature of the gas stays constant. That is an isothermal change.

Test it at normal temperature and pressure, where P = 1.013\times10^{5}\ \text{Pa} and \rho = 1.293\ \text{kg m}^{-3} for air:

The measured value at 0 °C is about 332 m s⁻¹. Newton's answer is too small by 52 m s⁻¹ — an error of about 16 %. No experiment of that era was that bad, so the assumption, not the measurement, had to be wrong.

7.3 Laplace's correction

In 1816 Laplace located the faulty assumption. Sound of even the lowest audible frequency, 20 Hz, compresses a given layer of air 20 times a second; at 1000 Hz the compression lasts about half a millisecond. Air is also a very poor conductor of heat. In that time practically no heat leaves a compression or enters a rarefaction. The changes are therefore not isothermal but adiabatic.

For air \gamma = 1.4, so the corrected value at NTP is

which agrees with experiment. The graph below shows why \gamma appears: an adiabatic curve through a given state is steeper than the isothermal curve through the same state, by exactly the factor \gamma, so the gas is harder to squeeze and the wave travels faster.

	Newton	Laplace
Assumed process	isothermal	adiabatic
Reason given	heat escapes instantly	air is a bad conductor, changes are fast
Bulk modulus	B = P	B = \gamma P
Speed formula	v=\sqrt{P/\rho}	v=\sqrt{\gamma P/\rho}
Value for air at 0 °C	280 m s⁻¹	331 m s⁻¹
Agreement with experiment	16 % too low	agrees

7.4 Effect of temperature, pressure, humidity on velocity of sound

To see which quantities really matter, eliminate \rho using the ideal gas equation. For n moles of gas of molar mass M in volume V, \rho = nM/V and PV = nRT, so P/\rho = RT/M and

This single result settles every case, because it contains no pressure at all.

Pressure — no effect

At a fixed temperature, doubling the pressure of a gas doubles its density too (Boyle's law), so the ratio P/\rho is unchanged and the speed is unchanged. This is why sound does not travel measurably faster at sea level in Biratnagar than high in Kathmandu valley for the same temperature — any difference you actually hear comes from the temperature difference, not the pressure.

Temperature — the one that matters

From v=\sqrt{\gamma RT/M}, at constant composition

So if v_0 is the speed at 0 °C (273 K) and v_t the speed at t °C,

For small t, using the binomial approximation (1+x)^{1/2}\approx 1+x/2,

With v_0 = 332\ \text{m s}^{-1} this gives v_0/546 = 0.61, so

The velocity of sound in air increases by about 0.61 m s⁻¹ for each 1 °C rise in temperature. The graph shows how good the straight-line rule is over ordinary Nepali temperatures.

Humidity — a small increase

Water vapour has molar mass 18 g mol⁻¹ while dry air averages about 28.8 g mol⁻¹. When water vapour is added to air it replaces some of the heavier nitrogen and oxygen molecules, so at the same pressure and temperature moist air is less dense than dry air. Since v\propto 1/\sqrt{\rho}, the speed rises. The effect is small — a few metres per second between bone-dry and saturated air — but it is real, and it is one reason distant sounds seem to carry unusually well in the humid air just before a monsoon shower.

Factor changed	Effect on v	Reason
Pressure (at constant T)	none	P/\rho fixed by Boyle's law
Temperature	increases, v\propto\sqrt{T}	v=\sqrt{\gamma RT/M}
Humidity	increases slightly	moist air is less dense than dry air
Wind	adds vectorially	the medium itself is moving
Amplitude (loudness)	none	speed is a property of the medium
Frequency (pitch)	none	air is non-dispersive for sound

Wind does not change the speed relative to the air; it moves the air itself. If a wind of speed w blows at an angle \theta to the direction in which the sound travels, the speed relative to the ground is

Sound therefore carries further downwind than upwind.

Chapter summary

• Every wave obeys v=f\lambda, but the speed itself is set by the medium: v=\sqrt{E/\rho}, elasticity over inertia.
• Solid rod: v=\sqrt{Y/\rho} (\approx 5060\ \text{m s}^{-1} in steel). Liquid: v=\sqrt{B/\rho} (\approx 1480\ \text{m s}^{-1} in water). Fluids carry only longitudinal sound; solids carry longitudinal and transverse waves.
• Newton's formula: assuming isothermal compressions, B_{iso}=P and v=\sqrt{P/\rho}, giving 280 m s⁻¹ for air — 16 % too low.
• Laplace's correction: the compressions are adiabatic, so B_{adi}=\gamma P and v=\sqrt{\gamma P/\rho} = 331\ \text{m s}^{-1}, which matches experiment.
• Eliminating density gives v=\sqrt{\gamma RT/M}: the speed depends on temperature and on the gas, never on the pressure.
• v_t=v_0\sqrt{1+t/273}\approx v_0+0.61t for air: +0.61 m s⁻¹ per °C.
• Humidity raises v slightly (moist air is lighter); wind adds vectorially, v_{eff}=v+w\cos\theta; amplitude and frequency have no effect.

Practice questions

Group A — Multiple choice (1 mark each)

1. The velocity of sound in a gas is independent of [1]atemperaturebpressurecdensitydnature of the gas
2. Laplace's correction to Newton's formula multiplies the velocity by [1]a\gammab\sqrt{\gamma}c1/\gammad\gamma^{2}
3. Sound travels fastest in [1]avacuumbaircwaterdsteel
4. If the absolute temperature of a gas is made four times, the velocity of sound in it becomes [1]ahalfbtwicecfour timesdunchanged
5. The velocity of sound in moist air is [1]agreater than in dry airbless than in dry aircequal to that in dry airdzero

Answers to Group A

1. (b) — v=\sqrt{\gamma RT/M} contains no P; at constant T, P/\rho is fixed.

2. (b) — the modulus changes from P to \gamma P, so v changes by \sqrt{\gamma}.

3. (d) — Young's modulus of steel is enormous compared with the moduli of fluids.

4. (b) — v\propto\sqrt{T}, and \sqrt{4}=2.

5. (a) — water vapour (M = 18) is lighter than air (M = 28.8), so \rho falls and v rises.

Group B — Short answer (5 marks each)

1. Derive Newton's formula for the velocity of sound in a gas and state the assumption on which it rests. Why does it fail? [5]
2. State and explain Laplace's correction. Show that it raises the calculated velocity in air at NTP from about 280 m s⁻¹ to about 331 m s⁻¹. (Take P=1.013\times10^{5}\ \text{Pa}, \rho=1.293\ \text{kg m}^{-3}, \gamma=1.4.) [5]
3. Show that the velocity of sound in air increases by about 0.61 m s⁻¹ for a rise of 1 °C, given that it is 332 m s⁻¹ at 0 °C. [5]
4. The velocity of sound in a certain liquid is 1500 m s⁻¹ and its density is 1200 kg m⁻³. Find the bulk modulus of the liquid. [5]
5. Explain why sound travels faster in solids than in liquids, and faster in liquids than in gases, even though solids are the densest of the three. [5]

Answers to Group B

1. Isothermal assumption, PV= const, P\,dV+V\,dP=0, hence B=P and v=\sqrt{P/\rho}. It gives 280 m s⁻¹ against a measured 332 m s⁻¹ — a 16 % error — because sound compressions are far too rapid, and air far too poor a conductor, for the heat to escape.

2. Adiabatic: PV^{\gamma}= const gives B=\gamma P, so v=\sqrt{\gamma P/\rho}. Numerically \sqrt{1.013\times10^{5}/1.293}=280\ \text{m s}^{-1}, and 280\times\sqrt{1.4}=280\times1.183=331\ \text{m s}^{-1}.

3. v_t=v_0(1+t/273)^{1/2}\approx v_0(1+t/546), so v_t-v_0 = v_0 t/546 = (332/546)t = 0.61t. For t=1\ ^{\circ}\text{C} the increase is 0.61 m s⁻¹.

4. v=\sqrt{B/\rho}\Rightarrow B=\rho v^{2}=1200\times(1500)^{2} = 1200\times2.25\times10^{6} = 2.7\times10^{9}\ \text{Pa}.

5. Speed is \sqrt{E/\rho}. Going gas → liquid → solid the density rises by a factor of roughly 10^{3} and then 8, but the elastic modulus rises by a far larger factor (about 10^{4} from air to water and about 10^{2} again from water to steel). Stiffness therefore wins over inertia.

Group C — Long answer (8 marks each)

1. (a) Starting from v=\sqrt{E/\rho}, derive Newton's formula for the speed of sound in a gas and explain why it disagrees with experiment. [4](b) State Laplace's correction, derive the corrected formula, and hence show that v=\sqrt{\gamma RT/M}. Use this to explain the effect of pressure, temperature and humidity on the speed of sound. [4]
2. (a) A gun is fired in a valley at Pokhara and the echo from a cliff is heard after 2.5 s on a day when the temperature is 20 °C. Taking the speed of sound at 0 °C as 332 m s⁻¹, find the distance of the cliff. [4](b) The same gun is fired when the temperature has fallen to 5 °C. By how much does the echo time change? [4]

Answer to Group C question 2

(a) v_{20} = 332 + 0.61\times20 = 332+12.2 = 344.2\ \text{m s}^{-1}. Path length =2d = 344.2\times2.5 = 860.5\ \text{m}, so d = 430\ \text{m}.

(b) v_{5} = 332 + 0.61\times5 = 335.05\ \text{m s}^{-1}. The cliff has not moved, so 2d = 860.5\ \text{m} still and

t = \frac{860.5}{335.05} = 2.57\ \text{s}

The echo now takes about 0.07 s longer — a small but measurable delay, and a reminder that any echo or resonance-tube measurement of v must record the air temperature.