14.2 The Speed of a Travelling Wave
The speed of a wave is set by the medium — an elastic property against an inertial one: $v=\sqrt{T/\mu}$ on a stretched string, $v=\sqrt{B/\rho}$ for sound, and $v=\sqrt{\gamma p/\rho}$ in a gas once Laplace's correction is made to Newton's formula.
Turn the tuning peg of a guitar or a sitar and the note rises: a tighter string carries its waves faster, and a faster wave on a string of fixed length means a higher frequency. The thick bass strings, on the other hand, are heavy, and their waves are slow — which is why they give the low notes. Sound in air is far faster, about $343\ \text{m/s}$ on a mild day, so that thunder takes roughly $3$ seconds to come from lightning a kilometre away. In water it is more than four times faster again, and in steel faster still. What decides these speeds?
This section shows that for every mechanical wave the speed is fixed by two properties of the medium — an elastic property that pulls a displaced element back, and an inertial property that resists its being accelerated — and that it does not depend on the frequency or the amplitude. You will find the speed of a transverse wave on a string, $\sqrt{T/\mu}$, then the speed of sound in fluids and solids, and finally follow one of the great stories of physics: Newton's calculation of the speed of sound, which came out $15\%$ too low, and Laplace's explanation of why, more than a century later.
What decides the speed of a wave?
From Result 14.2, $v=\nu\lambda$. That does not mean a higher frequency gives a faster wave: shake the end of a rope faster and the waves come out shorter, with the speed unchanged. In a given medium (at a given temperature) the speed is fixed; the frequency is chosen by the source; and the wavelength adjusts to $\lambda=\dfrac v\nu$. A wave passing from one medium to another keeps its frequency (the source still sets it) but changes speed and wavelength.
For a stretched string the obvious candidates are the tension $T$ (the restoring property) and the mass per unit length (the inertia). The mass of the string as a whole cannot matter — a longer string of the same kind carries waves at the same speed — so it is the mass of each metre that counts.
The linear mass density of a uniform string or wire of mass $m$ and length $L$ is its mass per unit length,
$$\mu=\frac mL\qquad(\text{unit: kg/m}).$$For a wire of density $\rho$ and cross-sectional area $A$, $\mu=\rho A$.
A dimensional guess. Suppose $v=C\,T^a\mu^b$ with $C$ a pure number. In dimensions (Section 1.3), $[\text{L}\,\text{T}^{-1}]=[\text{M}\text{L}\text{T}^{-2}]^a[\text{M}\text{L}^{-1}]^b$. Matching powers: M: $0=a+b$; T: $-1=-2a$; L: $1=a-b$. So $a=\tfrac12$, $b=-\tfrac12$ and $v=C\sqrt{T/\mu}$. Dimensions cannot give $C$; the mechanics below shows that $C=1$.
Transverse waves on a stretched string
Send a single pulse along a string under tension $T$ and watch it from a frame that moves with the pulse at speed $v$. In this frame the pulse shape stands still and the string streams through it from right to left at speed $v$ (both frames are inertial, so Newton's laws hold in either). Look at a short element of length $\Delta l$ at the top of the pulse. If the amplitude is small, the element is very nearly an arc of a circle of some radius $R$; it subtends an angle $2\theta$ at the centre, so $\Delta l=2R\theta$.
The tension pulls along the string at each end of the element, at an angle $\theta$ below the horizontal. The horizontal components cancel; the vertical components add to a net force towards the centre of $2T\sin\theta\approx2T\theta=\dfrac{T\,\Delta l}{R}$ (small angle, $\sin\theta\approx\theta$). The element has mass $\mu\,\Delta l$ and moves round the arc at speed $v$, so it needs a centripetal force $\dfrac{\mu\,\Delta l\,v^2}{R}$ (Section 3.4). Equating the two, $\dfrac{T\Delta l}{R}=\dfrac{\mu\Delta l\,v^2}{R}$, and both $\Delta l$ and $R$ cancel: the pulse shape does not matter.
On a string or wire of linear mass density $\mu$ held at tension $T$, transverse waves of small amplitude travel at
$$v=\sqrt{\frac T\mu},$$whatever their shape, amplitude or frequency. For a wire of density $\rho$ under tensile stress $\sigma=\dfrac TA$, this is $v=\sqrt{\sigma/\rho}$.
The two assumptions are worth remembering: the string is perfectly flexible (it resists stretching but not bending), and the displacements are small, so that the tension is the same everywhere and the angles are small. Thick, stiff piano wires break the first assumption slightly, which is one reason pianos are tuned by ear rather than by formula.
The speed of sound in fluids and solids
A longitudinal wave squeezes and stretches the medium, so the restoring property is now its resistance to compression — the bulk modulus $B=-\dfrac{\Delta p}{\Delta V/V}$ (Section 8.2) — and the inertial property is the density $\rho$. Dimensions again fix the form: $[B]=[\text{M}\text{L}^{-1}\text{T}^{-2}]$ and $[\rho]=[\text{M}\text{L}^{-3}]$, so $\sqrt{B/\rho}$ is the only speed that can be made from them.
A piston argument. A long tube of cross-section $A$ holds a fluid of density $\rho$ at rest. At $t=0$ a piston at one end starts pushing in at a small steady speed $u$. A compression front runs ahead at the wave speed $v$: behind it the fluid moves with the piston at $u$, ahead of it the fluid is still at rest. After a time $\Delta t$ the front has gone $v\Delta t$ and the piston $u\Delta t$, so a column of original length $v\Delta t$ has been squeezed by $u\Delta t$. Its volume strain is $\dfrac{\Delta V}{V}=-\dfrac uv$, and its pressure is raised by $\Delta p=B\dfrac uv$. In the time $\Delta t$ the excess force $\Delta p\,A$ has set a mass $\rho A v\Delta t$ moving at speed $u$. By the impulse–momentum theorem (Section 4.1):
$$B\frac uv\,A\,\Delta t=(\rho A v\Delta t)\,u\quad\Longrightarrow\quad v^2=\frac B\rho .$$In a fluid of bulk modulus $B$ and density $\rho$, longitudinal waves travel at
$$v=\sqrt{\frac B\rho}.$$Along a thin solid rod, which can bulge sideways as it is squeezed, the bulk modulus is replaced by Young's modulus $Y$:
$$v=\sqrt{\frac Y\rho}.$$Typical values: about $343\ \text{m/s}$ in air at $20\ ^\circ\text{C}$, about $1.5\times10^3\ \text{m/s}$ in water and about $5\times10^3\ \text{m/s}$ along a steel rod. Steel is some $6000$ times denser than air, but it is more than a million times harder to compress, and the modulus wins. Solids usually carry sound fastest and gases slowest — which is why you can hear a distant train in the rails before you hear it through the air.
Sound in a gas: Newton's formula and Laplace's correction
To use $v=\sqrt{B/\rho}$ for a gas we need its bulk modulus, and that depends on how the gas is compressed. In 1687 Newton assumed that the temperature of the gas stays constant as a sound wave passes. For an ideal gas at constant temperature $pV=\text{constant}$ (Section 12.1). Change $V$ by a small $\Delta V$: $(p+\Delta p)(V+\Delta V)=pV$, so, dropping the tiny product $\Delta p\,\Delta V$, $V\Delta p=-p\,\Delta V$ and the isothermal bulk modulus is $B=-\dfrac{\Delta p}{\Delta V/V}=p$.
Newton's formula is therefore $v=\sqrt{p/\rho}$. For air at $0\ ^\circ\text{C}$ and $1\ \text{atm}$ ($\rho=1.293\ \text{kg/m}^3$) it gives $280\ \text{m/s}$, but the measured speed is $331\ \text{m/s}$ — about $15\%$ more. The error puzzled physicists for more than a century.
In 1816 Laplace pointed out what Newton had missed. A compression warms the gas a little and a rarefaction cools it; for the temperature to stay constant, heat would have to flow from each compression to the neighbouring rarefaction, half a wavelength away, within half a period. Gases conduct heat very poorly, and the compressions and rarefactions alternate far too quickly for this to happen. The changes are adiabatic (Section 11.2): $pV^\gamma=\text{constant}$, with $\gamma=\dfrac{C_p}{C_v}$. For a small change, $(V+\Delta V)^\gamma\approx V^\gamma\left(1+\gamma\dfrac{\Delta V}{V}\right)$ (the binomial approximation of Section 7.2), and the same steps as before give the adiabatic bulk modulus $B=\gamma p$ — larger than Newton's by the factor $\gamma$.
Sound waves compress and expand a gas adiabatically, so in an ideal gas of pressure $p$, density $\rho$ and ratio of specific heats $\gamma$
$$v=\sqrt{\frac{\gamma p}{\rho}} .$$With $\dfrac p\rho=\dfrac{RT}{M}$ for an ideal gas of molar mass $M$ at absolute temperature $T$, this is
$$v=\sqrt{\frac{\gamma RT}{M}} .$$Newton's isothermal formula $v=\sqrt{p/\rho}$ is too small by the factor $\sqrt\gamma$.
What changes the speed of sound in a gas?
- Pressure — no effect at constant temperature: doubling $p$ doubles $\rho$, and $\dfrac p\rho=\dfrac{RT}M$ is unchanged. Sound is no faster at sea level than on a mountain at the same temperature.
- Temperature — $v\propto\sqrt T$ (in kelvin). For air near room temperature, $v$ rises by about $0.6\ \text{m/s}$ per degree.
- Molar mass — at the same $T$, lighter gases carry sound faster: in helium your voice resonances shift up and you sound like a cartoon character.
- Humidity — water molecules ($M=18\ \text{g/mol}$) are lighter than the nitrogen and oxygen they replace, so moist air is less dense and sound travels slightly faster in it.
- Wind carries the whole medium, adding its velocity to that of the sound relative to the ground.
1. "Higher frequency, faster wave." The medium fixes $v$; a higher frequency gives a shorter wavelength.
2. Using the total mass of the string. $\mu$ is mass per unit length: a $4.0\ \text{g}$ string $0.80\ \text{m}$ long has $\mu=5.0\ \text{g/m}$ — and convert to $\text{kg/m}$.
3. $v\propto T$. The speed goes as the square root of the tension: four times the tension for twice the speed.
4. Sound faster at higher pressure. At constant temperature, $p$ and $\rho$ rise together and $v$ is unchanged.
5. Celsius in $\sqrt T$. $v\propto\sqrt T$ with $T$ in kelvin. Going from $10\ ^\circ\text{C}$ to $40\ ^\circ\text{C}$ does not multiply $v$ by $2$, only by $\sqrt{313/283}=1.05$.
6. Laplace "corrected the density". He corrected the bulk modulus: adiabatic $\gamma p$ in place of isothermal $p$.
In JEE Main, NEET and JEE Advanced (as part of the speed of sound, rms speed and specific heats of gases); not required by the CBSE syllabus.
Sound and the molecules. The rms speed of the molecules of an ideal gas is $v_\text{rms}=\sqrt{\dfrac{3RT}M}$ (Section 12.2). Dividing,
$$\frac{v_\text{sound}}{v_\text{rms}}=\sqrt{\frac\gamma3}.$$Since $\gamma\le\tfrac53$, sound is always a little slower than the molecules themselves ($0.68\,v_\text{rms}$ in air) — sensible, because a pressure disturbance is passed on by molecules colliding.
A mixture of gases. For a mixture, use the average molar mass and the mixture's own $\gamma$. For $n_1$ moles with $C_{v1}$ and $n_2$ moles with $C_{v2}$: $C_v=\dfrac{n_1C_{v1}+n_2C_{v2}}{n_1+n_2}$, $\gamma=1+\dfrac{R}{C_v}$ and $M=\dfrac{n_1M_1+n_2M_2}{n_1+n_2}$. For one mole each of helium ($C_v=\tfrac32R$) and hydrogen ($\tfrac52R$): $C_v=2R$, $\gamma=1.5$, $M=3.0\ \text{g/mol}$, and at $300\ \text{K}$ $v=\sqrt{\dfrac{1.5\times8.31\times300}{0.0030}}=1.1\times10^3\ \text{m/s}$.
Exercises
Take $g=9.8\ \text{m/s}^2$, $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$ and $0\ ^\circ\text{C}=273\ \text{K}$. Give answers to 2–3 significant figures.
- 1 A string of mass $5.0\ \text{g}$ and length $1.0\ \text{m}$ is under a tension of $80\ \text{N}$. Find the speed of transverse waves on it.
- 2 Waves travel at $200\ \text{m/s}$ along a wire of linear mass density $2.0\ \text{g/m}$. Find the tension.
- 3 By what factor does the wave speed change if the tension is made four times as large? By what percentage must the tension be raised to increase the speed by $10\%$?
- 4 Two wires of the same material carry the same tension; one has twice the diameter of the other. Compare the wave speeds.
- 5 A rope of length $12\ \text{m}$ and mass $3.0\ \text{kg}$ is pulled taut with a tension of $400\ \text{N}$. How long does a pulse take to travel along it?
- 6 On a wire under a tension of $60\ \text{N}$, a $300\ \text{Hz}$ wave has a wavelength of $0.50\ \text{m}$. Find the linear mass density of the wire.
- 7 Assume $v=C\,T^a\mu^b$ and use dimensions to find $a$ and $b$.
- 8 A steel wire (density $7.8\times10^3\ \text{kg/m}^3$) carries a tensile stress of $2.0\times10^8\ \text{Pa}$. Find the speed of transverse waves along it.
- 1 Find the speed of sound in water ($B=2.2\times10^9\ \text{Pa}$, $\rho=1.0\times10^3\ \text{kg/m}^3$).
- 2 Find the speed of longitudinal waves along an aluminium rod ($Y=7.0\times10^{10}\ \text{Pa}$, $\rho=2.7\times10^3\ \text{kg/m}^3$).
- 3 A $1.0\ \text{kHz}$ sound passes from air ($340\ \text{m/s}$) into water ($1480\ \text{m/s}$). Which of $\nu$, $v$, $\lambda$ change? Find the wavelength in each medium.
- 4 A ship's sonar pulse returns from the sea floor $0.80\ \text{s}$ after it is sent. Sound travels at $1530\ \text{m/s}$ in sea water. How deep is the sea?
- 5 A hammer strikes one end of a $1.0\ \text{km}$ steel rail. A listener at the other end hears two sounds, through the rail ($5.0\times10^3\ \text{m/s}$) and through the air ($340\ \text{m/s}$). Find the time between them.
- 6 Sound travels at $1200\ \text{m/s}$ in a liquid of density $800\ \text{kg/m}^3$. Find its bulk modulus.
- 7 Steel is about $6000$ times denser than air, yet sound is about $15$ times faster in steel. Explain.
- 1 Nitrogen at $0\ ^\circ\text{C}$ and $1.013\times10^5\ \text{Pa}$ has density $1.25\ \text{kg/m}^3$ and $\gamma=1.40$. Find the speed of sound from Newton's formula and from Laplace's.
- 2 Why did Newton assume the compressions in a sound wave are isothermal, and why is that assumption wrong?
- 3 The speed of sound in air is $331\ \text{m/s}$ at $0\ ^\circ\text{C}$. Find it at $40\ ^\circ\text{C}$.
- 4 At what temperature in °C is the speed of sound in air $1.5$ times its value at $0\ ^\circ\text{C}$?
- 5 The pressure of the air in a sealed room is doubled at constant temperature. What happens to the speed of sound?
- 6 Find the speed of sound in helium ($\gamma=1.67$, $M=4.0\ \text{g/mol}$) at $300\ \text{K}$.
- 7 Find the ratio of the speeds of sound in hydrogen ($M=2.0\ \text{g/mol}$) and oxygen ($32\ \text{g/mol}$) at the same temperature (both diatomic).
- 8 Is sound faster in dry air or in moist air at the same temperature and pressure? Explain.
- 1 (Modelling) Thunder is heard $4.5\ \text{s}$ after a lightning flash on a day when sound travels at $343\ \text{m/s}$. How far away was the flash? (Light takes a negligible time.)
- 2 (Modelling) A sitar player raises the tension of a string from $60\ \text{N}$ to $72\ \text{N}$. By what percentage does the wave speed on it rise?
- 3 (Modelling) Ultrasound travels at $1540\ \text{m/s}$ in soft tissue of density $1060\ \text{kg/m}^3$. Find the bulk modulus of the tissue.
- 4 A heavy uniform rope $2.0\ \text{m}$ long hangs from a ceiling. Show that the speed of transverse waves at a height $y$ above its lower end is $\sqrt{gy}$, and find the speed at the midpoint of the rope.
- 5 Two wires of the same material and length, of radii $r$ and $2r$, carry the same tension. Find the ratio of the times a pulse takes to travel along them.
- 6 A steel wire ($Y=2.0\times10^{11}\ \text{Pa}$, $\alpha=1.2\times10^{-5}\ \text{K}^{-1}$, $\rho=7.8\times10^3\ \text{kg/m}^3$) is clamped between rigid supports with a stress of $1.0\times10^8\ \text{Pa}$. It is then cooled by $20\ \text{K}$ (thermal stress, Section 10.1). Find the wave speed before and after.
E1. Board · 3 marks Write Newton's formula for the speed of sound in a gas and show that it gives about $280\ \text{m/s}$ for air at $0\ ^\circ\text{C}$ and $1.013\times10^5\ \text{Pa}$ (density $1.293\ \text{kg/m}^3$). What correction did Laplace make, and why?
E2. Board · 2 marks A wire $1.5\ \text{m}$ long of mass $6.0\ \text{g}$ is under a tension of $90\ \text{N}$. Find the speed of transverse waves on it, and the tension needed to make this speed $1.5$ times as large.
E3. NEET The speed of sound in a gas is $v$ at $27\ ^\circ\text{C}$. At $127\ ^\circ\text{C}$ (same pressure) it is (a) $\dfrac{2v}{\sqrt3}$ (b) $\dfrac{\sqrt3\,v}{2}$ (c) $\dfrac{4v}3$ (d) $v\sqrt{\dfrac{127}{27}}$
E4. JEE Main (Numerical value) Transverse waves travel at $200\ \text{m/s}$ on a stretched wire. The wire is replaced by one of the same material with half the radius, and the tension is halved. Find the new wave speed in m/s, to the nearest integer.
E5. JEE Advanced (One or more correct) A uniform heavy rope of length $L$ hangs freely from a ceiling. A small transverse pulse is started at its lower end. Which statements are correct? (A) At height $y$ above the lower end the pulse travels at $\sqrt{gy}$. (B) The pulse takes a time $2\sqrt{L/g}$ to reach the ceiling. (C) If a steady wave of fixed frequency is sent up the rope, its wavelength increases as it climbs. (D) The pulse travels up the rope at constant speed.
E6. NEET The ratio of the speed of sound in helium ($\gamma=\tfrac53$, $M=4\ \text{g/mol}$) to that in hydrogen ($\gamma=\tfrac75$, $M=2\ \text{g/mol}$) at the same temperature is (a) $\dfrac{5}{\sqrt{42}}$ (b) $\dfrac{\sqrt{42}}{5}$ (c) $\dfrac1{\sqrt2}$ (d) $\sqrt2$
Research question. Can two smartphones measure the speed of sound well enough to find $\gamma$ for air — and to see Laplace beat Newton?
Why it matters. The speed of sound is a precision thermometer: ocean scientists map the temperature of whole ocean basins from the travel times of low-frequency sound, and acoustic thermometers set the international temperature scale near room temperature. Meteorologists use "sodar" (sound radar) to profile winds and temperature above airports, and engineers correct every ultrasonic range-finder and parking sensor for air temperature.
What to do. Use the "acoustic stopwatch" in the free phyphox app on two phones.
- Place the phones a measured distance $d$ apart ($5$–$20\ \text{m}$) in a quiet corridor or playground. Each phone starts its timer on a loud clap and stops on the next.
- Clap next to phone A; then clap next to phone B. The difference of the two phones' readings is $\dfrac{2d}{v}$ (the sound crosses twice, and each phone's own delay cancels). Repeat $10$ times and average.
- Record the air temperature. Compare your $v$ with $331\sqrt{T/273}\ \text{m/s}$.
- Use $v=\sqrt{\gamma RT/M}$ with $M=29.0\ \text{g/mol}$ to find $\gamma$. Is it closer to $1$ (Newton) or $1.40$ (Laplace)?
- Repeat on a hot afternoon and a cool morning. Is the change in $v$ what $\sqrt T$ predicts?
Tools: Two smartphones, a long tape measure, a thermometer, a spreadsheet.
Hand in: Your table of readings; $v$ with its uncertainty; your value of $\gamma$; a short paragraph on whether your data rule out Newton's formula.
Going further. Measure in a humid bathroom after a shower and outside on a dry day. Can you detect the effect of humidity, or is it smaller than your uncertainty?
🧪 Self-check quiz
Q1. The speed of a transverse wave on a stretched string depends on
Q2. The tension in a string is made four times as large. The speed of waves on it
Q3. Laplace corrected Newton's formula for the speed of sound by assuming that the compressions and rarefactions are
Q4. At constant temperature, the pressure of the air is doubled. The speed of sound
Q5. The speed of sound in air at $0\ ^\circ\text{C}$ is $v$. It becomes $2v$ at about