Question
A man stands m from a tall cliff and claps his hands. He hears the echo s later. Then he walks m closer and claps again — when does he hear this echo? Take the speed of sound in air as m/s. The trap is that students confuse “distance to cliff” with “distance travelled by sound.”
Solution — Step by Step
Sound travels to the cliff and back — that’s m. Time s. Matches the given echo time, so the speed value is consistent.
The new distance to the cliff m. Sound now travels m round-trip.
The echo is heard s after the second clap.
Why This Works
Echo time depends on the round-trip path of sound — to the obstacle and back. The man’s distance halves only the one-way trip. We always multiply by 2.
A second concept hidden in this problem: for an echo to be heard distinctly, the round-trip time must be at least s (the persistence of hearing). Both clap echoes here are well above that threshold, so both are audible.
Alternative Method
Direct ratio: if distance to cliff scales by , the echo time scales by the same factor. So new s. This shortcut works when speed of sound is unchanged.
Common Mistake
The classic CBSE trap: students compute s and report it. They forgot to double for the return trip. Always sketch the path — outgoing arrow + return arrow — before plugging into .