Sound: Edge Cases and Subtle Traps (1)

easy 2 min read
Tags Sound

Question

A man stands 170170 m from a tall cliff and claps his hands. He hears the echo 1.01.0 s later. Then he walks 3434 m closer and claps again — when does he hear this echo? Take the speed of sound in air as 340340 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 2×170=3402 \times 170 = 340 m. Time =340/340=1.0= 340/340 = 1.0 s. Matches the given echo time, so the speed value is consistent.

The new distance to the cliff =17034=136= 170 - 34 = 136 m. Sound now travels 2×136=2722 \times 136 = 272 m round-trip.

t=272340=0.8 st = \frac{272}{340} = 0.8 \text{ s}

The echo is heard 0.80.8 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 0.1\sim 0.1 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 136/170=0.8136/170 = 0.8, the echo time scales by the same factor. So new t=0.8×1.0=0.8t = 0.8 \times 1.0 = 0.8 s. This shortcut works when speed of sound is unchanged.

Common Mistake

The classic CBSE trap: students compute t=136/340=0.4t = 136/340 = 0.4 s and report it. They forgot to double for the return trip. Always sketch the path — outgoing arrow + return arrow — before plugging into v=d/tv = d/t.

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