The speed of a travelling wave

What sets the speed?

Pluck a thick string and a thin string on a guitar, both tight. Waves on them travel at different speeds. Shout in cold air and in warm air. Sound travels a little faster in the warm air. The speed of a mechanical wave does not depend on how you make the wave. It depends on the medium.

Two properties of the medium matter. One is how strongly it pulls back when it is disturbed. A stronger pull back gives a faster wave. The other is its inertia. More inertia gives a slower wave.

The source sets the frequency. The medium sets the speed. The wavelength then follows from $\lambda = v/f$. When a wave passes from one medium into another, its frequency stays the same. Its speed and its wavelength change.

Waves on a stretched string

On a string, the pull back comes from the tension $T$, in newtons. Here $T$ is the tension, not the period. The inertia comes from the mass per unit length $\mu$, in kg/m. It is the mass of each metre of string.

Which formula can join $v$, $T$ and $\mu$? Look at the dimensions. Tension is $[\text{M}\,\text{L}\,\text{T}^{-2}]$ and $\mu$ is $[\text{M}\,\text{L}^{-1}]$. So $T/\mu$ has the dimensions $[\text{L}^2\,\text{T}^{-2}]$, the square of a speed. A full calculation shows that the number in front is exactly $1$:

A tighter string carries faster waves. A heavier string carries slower ones. A steel string under a tension of $90\ \text{N}$ with $\mu = 3.6\ \text{g/m}$ carries waves at $\sqrt{\dfrac{90\ \text{N}}{0.0036\ \text{kg/m}}} = 158\ \text{m/s}$.

How do we find $\mu$ for a wire? A wire of radius $r$ and density $\rho$ has a volume of $\pi r^2$ in each metre. So $\mu = \rho\pi r^2$.

Sound: a longitudinal wave

In a longitudinal wave the medium is squeezed and stretched. The pull back now comes from how hard it is to squeeze, its bulk modulus $B$. The inertia comes from its density $\rho$. The same kind of argument gives

For a long solid rod, the Young's modulus $Y$ takes the place of $B$: $v = \sqrt{Y/\rho}$. Steel is very stiff. So sound runs along a steel railway track about fifteen times faster than through the air. Put your ear to the rail, and you hear a far train long before the sound reaches you through the air.

Newton's formula and Laplace's correction

What is $B$ for a gas? Newton assumed that the temperature of the air stays the same as the sound passes. For such an isothermal change, $pV$ stays constant, and the bulk modulus is just the pressure $p$. So Newton's formula is

For air at $0\ ^\circ\text{C}$ and normal pressure, this gives about $280\ \text{m/s}$. But the measured speed is about $331\ \text{m/s}$. Newton was too low by about $15$ percent.

Laplace found the error. The compressions and rarefactions of a sound wave are very quick. Heat has no time to flow in or out. So the change is adiabatic, not isothermal. For an adiabatic change $pV^\gamma$ stays constant, and the bulk modulus is $\gamma p$. Here $\gamma$ is the ratio of the specific heats of the gas. So

For air $\gamma = 1.40$, and this gives $331\ \text{m/s}$, just as measured. We call this Laplace's correction.

Pressure, temperature and humidity

For an ideal gas, $p/\rho = RT/M$, where $M$ is the molar mass. So we can also write

This tells us three things.

• Pressure. At a fixed temperature, $p$ and $\rho$ rise together. Their ratio stays the same, so pressure alone does not change the speed of sound.
• Temperature. $v$ grows as the square root of the absolute temperature $T$, in kelvin.
• Humidity. Water vapour has a smaller molar mass than dry air. Damp air is less dense, so sound travels slightly faster in it.

Helium has $\gamma = 1.67$ and $M = 4\ \text{g/mol}$. So sound in helium is about three times as fast as in air.

Practise with new numbers

Try it in Python

The program finds the speed of sound in air at three temperatures, by Newton's formula and by Laplace's.

Common mistakes

• Using the total mass of a string for $\mu$. $\mu$ is the mass of each metre: the total mass divided by the length.
• Leaving $\mu$ in g/m. Change it to kg/m before you use $v = \sqrt{T/\mu}$.
• Using degrees Celsius in $\sqrt{T}$. The temperature must be in kelvin.
• Thinking a higher pressure speeds up sound. At a fixed temperature, $p/\rho$ stays the same.
• Thinking the frequency changes in a new medium. The frequency is set by the source. The speed and the wavelength change.

Remember these

• The medium sets the speed. The source sets the frequency.
• String: $v = \sqrt{\dfrac{T}{\mu}}$.
• Fluid: $v = \sqrt{\dfrac{B}{\rho}}$. Solid rod: $v = \sqrt{\dfrac{Y}{\rho}}$.
• Gas: Newton $\sqrt{p/\rho}$ is too low. Laplace: $v = \sqrt{\dfrac{\gamma p}{\rho}} = \sqrt{\dfrac{\gamma R T}{M}}$.
• $v \propto \sqrt{T}$, with $T$ in kelvin. Pressure alone has no effect.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The tension in a string is made four times as large. What happens to the speed of waves on it?

2. A rope $5.0\ \text{m}$ long has a mass of $2.0\ \text{kg}$ and a tension of $80\ \text{N}$. Find the speed of waves on it.

3. Why was Newton's value for the speed of sound in air too low?

4. The speed of sound in air is $330\ \text{m/s}$ at $0\ ^\circ\text{C}$. At what temperature is it twice as large?