Measuring the speed of sound means measuring one distance and one time

1. Sound travels at a finite speed, so if you can measure how far it goes and how long it takes to get there, the speed follows.
2. Every way of finding the speed of sound in air reduces to those two measurements, a distance in metres and a time in seconds.
   1. $$v=\frac{d}{t}$$
   2. $v$ is the speed in $m/s$, $d$ is the distance the sound travels in $m$, and $t$ is the time taken in $s$.
3. Measure the distance with a metre ruler or a tape measure, along the straight path the sound actually follows from the source to the detector.
4. The timing is the awkward part, because sound gets through roughly $340m$ of air every second.
   1. Across a $3m$ bench the sound arrives after about $0.009s$, far too quick for anybody to catch with a hand-held stopwatch.
5. Either stretch the distance out to tens or hundreds of metres so the time grows into something a person can time, or hand the timing to an electronic timer that starts and stops itself.
   1. A timer that reads in milliseconds gives values such as $8.8ms$, which has to be written as $0.0088s$ before it goes anywhere near the equation.
6. Take every reading more than once and use the mean, because one timing on its own carries random error you cannot see.

A hard flat surface sends the sound straight back to you

1. When a sound wave meets a wall, some of its energy is reflected back into the air instead of passing into the wall.
2. The reflected wave carries on at the same speed as the original, so it reaches your ear a short time after the sound you made.
3. Shout across a valley and the rock face on the far side hands your voice back to you a second or so later.
   1. Clap towards the flat end wall of a sports hall, or towards a large wall across a playground, and the clap comes back as one sharp repeat.
4. You notice the reflection as a separate sound only when the surface is far enough away, because two sounds that arrive within about $0.1s$ of each other are heard as one.
   1. At $340m/s$ the wave covers $34m$ in that $0.1s$, so the wall needs to be more than roughly $17m$ away before you hear the delay.
   2. Nearer than that, the reflection merges into the original sound and only makes the space sound hard and ringing.
5. A large, hard, flat surface such as a cliff, a brick wall or a tiled corridor gives the clearest return, because it sends nearly all the sound energy back in one direction.
6. A soft surface such as a curtain, a carpet or a bank of long grass absorbs the sound energy into its fibres, so the reflection that returns is weak or missing altogether.
7. A broken-up surface such as a hedge, a line of trees or a pile of rubble scatters the reflection in all directions, so what gets back to you is too spread out in time and too quiet to pick out.

The sound in an echo covers twice the distance to the surface

1. Stand a measured distance $d$ from a wall and make one sharp sound, and the wave has two journeys to finish before you hear it again.
   1. It travels the distance $d$ out to the wall, then the same distance $d$ back to where you are standing.
2. The distance that belongs in the equation is that whole path, $2d$, and not the gap between you and the wall.
   1. $$v=\frac{2d}{t}$$
   2. Here $t$ is the time from making the sound to hearing the reflection, so it is the time for the whole out-and-back trip.
3. One way to make the timing manageable by hand is to clap in a steady rhythm so that each clap lands exactly on the reflection of the clap before it.
   1. Time $20$ claps, divide the total by $20$, and you have a value of $t$ in which your reaction time has been shared out and largely cancelled.
4. The same there and back reasoning turns up wherever a pulse is sent out and its reflection is timed.

Turning the equation round gives a distance from a known speed

1. When the speed of sound is already known and the distance is what you are after, rearrange the same relationship.
   1. $$d=\frac{vt}{2}$$
   2. Multiplying the speed by the echo time gives the whole path the sound covered, and halving it gives the one-way distance to the reflecting surface.
2. A walker shouts towards a cliff and hears the reply $1.2s$ later, on a day when the speed of sound in air is $340m/s$.
   1. $$d=\frac{340\times 1.2}{2}$$
   2. $$d=204m$$
   3. The cliff stands $204m$ away, and the sound itself covered $408m$.
3. You can halve the time first instead, putting the one-way time of $0.6s$ into $d=vt$, which gives the same $204m$.
   1. Pick one of those two routes and stay in it, because halving the time and doubling the distance in the same calculation corrects for the return journey twice.