An echo is sound that reflects from a surface and returns to the listener. The measured time covers the journey there and back, so the distance to the surface is half of speed × time.
This skill appears in electromagnetic spectrum and sound in questions about cliffs, valleys, sonar and ultrasound, usually with speed and time given.
How do I handle a return journey?
Draw the path first. A line from the source to the wall, then the same line back. Label the one-way distance d.
Use the general equation for a constant speed: distance = speed × time. The distance in that equation is the total path, which is 2d. So 2d = speed × time, and d = speed × time ÷ 2.
Worked example
A student shouts towards a cliff and hears the echo 1.4 s later. (Invented example data.) The speed of sound is 330 m/s. How far away is the cliff?
Step 1, draw and label: the sound goes to the cliff and back. Call the one-way distance d.
Step 2, total path: distance = speed × time = 330 × 1.4 = 462 m.
Step 3, halve it: d = 462 ÷ 2 = 231 m.
Step 4, check: 2 × 231 = 462 m, and 462 ÷ 330 = 1.4 s, which matches the time given. The cliff is 231 m away.
The answer is also plausible: a cliff a couple of hundred metres away gives an echo after about a second, which fits everyday experience.
The mistake to watch for
A common slip is to use the total path as the distance to the wall. A student calculates 330 × 1.4 = 462 and writes “the cliff is 462 m away”.
Mistaken answer: The cliff is 462 m away.
This is the distance the sound travelled, there and back.
The correction is one line in the working: “total path = 462 m, so distance to cliff = 462 ÷ 2 = 231 m”. Before you write the final answer, ask “was this a return journey?”
The same habit works in reverse. If you know the distance to the wall and need the time, double the distance first.
Check yourself
1. An echo returns after 0.50 s. The speed of sound is 330 m/s. How far is the reflecting wall?
Show answer
Total path = 330 × 0.50 = 165 m. Distance = 165 ÷ 2 = 82.5 m. Check: 2 × 82.5 ÷ 330 = 0.50 s.
2. A ship’s sonar sends a pulse down. The echo from the sea bed returns after 0.80 s. The speed of sound in water is 1500 m/s. Find the depth.
Show answer
Total path = 1500 × 0.80 = 1200 m. Depth = 1200 ÷ 2 = 600 m. Check: 2 × 600 ÷ 1500 = 0.80 s.
3. A student stands 85 m from a wall. The speed of sound is 340 m/s. How long until she hears the echo?
Show answer
Total path = 2 × 85 = 170 m. Time = 170 ÷ 340 = 0.50 s.
Where this leads next
The last lesson in the module is interpreting pitch and loudness separately. For the speeds used here, see comparing sound travel in different media. Quoting your answer with the right number of significant figures is easier with the bounds and rounding explainer.
Students who lose marks on return journeys usually understand the physics but rush the diagram. A teacher watching the working live can build the drawing habit, as in online one-to-one Physics tuition.