The Speed of Sound
Simple Explanation
The speed of sound depends on the medium it travels through — about 340 m/s in air at room temperature, faster in liquids, and faster still in solids. The time delay of an echo can be used to measure distance: since the sound travels to a surface AND back before the echo is heard, d = vt/2.
Why Do We Need It?
Measuring distance using echoes is the basis of sonar (used by ships and submarines) and how bats and dolphins navigate using echolocation.
Formula
Distance from an Echo
d = vt / 2
A sound wave sent toward a distant surface travels there AND back before the echo is heard, so the total time measured corresponds to twice the distance — this formula divides that out.
- d
- — distance to the reflecting surface, in metres (m)
- v
- — speed of sound in the medium, in metres per second (m/s)
- t
- — total time between the sound being sent and the echo being heard, in seconds (s)
When to use it: Whenever the distance to a reflecting surface (a cliff, the sea floor, an obstacle) needs to be found from the round-trip time of an echo.
Worked Example
Find the distance to a cliff using an echo
A person shouts toward a cliff and hears the echo 3 s later. Using the speed of sound in air as 340 m/s, find the distance to the cliff.
Why Does This Work?
The measured time t is for the ENTIRE round trip — sound travelling to the cliff, then back again — so the actual one-way distance is only covered in half that time, which is exactly why the formula divides vt by 2.
Real-Life Example
Sonar on ships
A ship uses sonar to measure the depth of the sea floor beneath it.
The ship sends a sound pulse downward and times how long it takes for the echo to return from the sea floor, then uses exactly this echo-distance formula (with the speed of sound in water) to calculate the depth.
Practice
A sonar pulse sent from a ship returns as an echo after 0.8 s. Using the speed of sound in water as 1500 m/s, find the depth of the sea floor.
MediumCommon mistake
Forgetting to divide by 2 — using the full time t directly (instead of t/2, or equivalently dividing the final answer by 2) gives a distance TWICE too large, since the sound travelled the distance there and back.
Quick Review
- d = vt/2 for an echo (round-trip distance).
- Speed of sound in air ≈ 340 m/s; faster in liquids and solids.
- This is the basis of sonar and echolocation.