Why You Hear a Train Through the Rail Before You Hear It Through the Air

Tighten a sitar string and its note rises; knock on an iron gate and the sound races through the metal far faster than through the air. A wave's speed is set by the medium it travels in — its stiffness and its inertia.

When two waves arrive at the same place, they simply add, which can make sound louder or cancel it out.

This part covers wave speed on a string, the speed of sound, the principle of superposition, and the resultant of two waves.

**A wave on a string with tension $T$ and mass per unit length $\mu$ travels at $v=\sqrt{\frac{T}{\mu }}$ — faster for a tighter or lighter string.

Derivation outline.** Move along with a pulse at speed $v$, so the string slides through the crest. A tiny piece of length $\Delta l$ and mass $\mu \Delta l$ at the top follows an arc of radius $R$. The two tension forces pull it towards the centre with net force $T\frac{\Delta l}{R}$, which must supply the centripetal force:

$$T\frac{\Delta l}{R}=\frac{(\mu \Delta l)v^2}{R}\Rightarrow v=\sqrt{\frac{T}{\mu }}$$

Dimensional analysis gives the same form.

Worked example 1. A string has $\mu =0.01$ kg/m under $100$ N tension:

$$v=\sqrt{\frac{100}{0.01}}=100\text{ m/s}$$

**Worked example 2 — finding $\mu$.** A $2$ m string of mass $5$ g has $\mu =2.5\times 10^{-3}$ kg/m. Under $90$ N:

$$v=\sqrt{\frac{90}{2.5\times 10^{-3}}}=\sqrt{36000}\approx 190\text{ m/s}$$

Worked example 3. Quadrupling the tension doubles the speed.

An everyday example. Turning the tuning pegs of a veena raises the string tension, speeding up its waves and raising the pitch.

The substance. Wave speed on a string does not depend on amplitude or frequency — only on $T$ and $\mu$.

**The speed of a longitudinal wave is $\sqrt{\frac{\text{elastic modulus}}{\text{density}}}$: $v=\sqrt{\frac{Y}{\rho }}$ in a solid rod and $v=\sqrt{\frac{B}{\rho }}$ in a fluid, where for a gas the adiabatic bulk modulus $\gamma P$ gives $v=\sqrt{\frac{\gamma P}{\rho }}$.

Worked example 1 — air.** With $P=1.013\times 10^5$ Pa and $\rho =1.293$ kg/m${}^3$ at $0$ °C:

$$\text{isothermal (Newton’s formula): }\sqrt{\frac{P}{\rho }}\approx 280\text{ m/s}\text{adiabatic (Laplace correction): }\sqrt{\frac{1.4P}{\rho }}\approx 331\text{ m/s}$$

The adiabatic value matches measurement, because compressions happen too fast for heat to flow.

Worked example 2 — temperature. Since $v\propto \sqrt{T}$, at $30$ °C:

$$v=331\sqrt{\frac{303}{273}}\approx 349\text{ m/s}$$

Worked example 3 — steel and water.

$$v_{steel}=\sqrt{\frac{2\times 10^{11}}{7800}}\approx 5.1\times 10^3\text{ m/s}{v}_{water}=\sqrt{\frac{2.2\times 10^9}{1000}}\approx 1.5\times 10^3\text{ m/s}$$

An everyday example. Tap one end of a long iron gate while a friend listens with an ear pressed to the far end — they hear the tap twice, first through the metal, then through the air.

The substance. Changing air pressure at constant temperature does not change the speed of sound, because $\frac{P}{\rho }$ stays the same.

**When two or more waves overlap, the resultant displacement at any point and instant is the algebraic sum of the displacements each wave would produce on its own: $y=y_1+y_2+\dots$.

After overlapping, each wave carries on unchanged, as if the other had never been there.

Worked example 1 — two pulses.** A pulse of height $+3$ mm meets one of $-2$ mm on a string. Where they overlap exactly, the string is displaced by

$$y=3+(-2)=+1\text{ mm}$$

Moments later, the $+3$ mm and $-2$ mm pulses move apart with their shapes restored.

Worked example 2 — two waves in step. $y_1=0.02\sin (3x-6t)$ and $y_2=0.01\sin (3x-6t)$ add to

$$y=0.03\sin (3x-6t)\text{ m}$$

Worked example 3 — two waves out of step. $y_1=0.02\sin (3x-6t)$ and $y_2=0.02\sin (3x-6t+\pi )$ cancel everywhere, since $\sin (\theta +\pi )=-\sin \theta$.

An everyday example. In a noisy classroom, the sound waves from many voices overlap at your ear, yet you can still pick out a friend calling your name.

The substance. Waves pass through each other instead of bouncing off, which particles cannot do.

**Two waves of amplitudes $a_1$ and $a_2$ and the same frequency, with phase difference $\varphi$, combine into a wave of amplitude $A=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \varphi }$; for equal amplitudes $a$ this becomes $A=2a\cos \frac{\varphi }{2}$.

Derivation for equal amplitudes.** Using $\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$:

$$a\sin (kx-\omega t)+a\sin (kx-\omega t+\varphi )=2a\cos \frac{\varphi }{2}\sin \left(kx-\omega t+\frac{\varphi }{2}\right)$$

**Worked example 1 — equal amplitudes $a=3$ mm.**

- $\varphi =0$: $A=6$ mm — constructive interference
- $\varphi =\frac{\pi }{2}$: $A=6\cos 45°\approx 4.24$ mm
- $\varphi =\frac{2\pi }{3}$: $A=6\cos 60°=3$ mm
- $\varphi =\pi$: $A=0$ — destructive interference

Worked example 2 — unequal amplitudes $4$ mm and $3$ mm:

$$\varphi =\frac{\pi }{2}:A=\sqrt{16+9}=5\text{ mm}\varphi =\pi :A=4-3=1\text{ mm}$$

An everyday example. Noise-cancelling headphones play a sound wave that is half a cycle out of step with the outside noise, so the two nearly cancel.

The substance. Intensity goes as amplitude squared — doubling the amplitude by constructive interference makes the sound four times as intense.