Sound: Real-World Scenarios (4)

easy 2 min read
Tags Sound

Question

A boy stands 200200 m from a tall wall and claps his hands. He hears the echo 1.21.2 s later. Calculate the speed of sound in air on that day.

Solution — Step by Step

The clap sound travels from the boy to the wall, reflects, and returns. So the total distance covered by the sound is 2×200=4002 \times 200 = 400 m, not 200200 m.

Speed = distance / time:

v=4001.2=333.3 m/sv = \frac{400}{1.2} = 333.3 \text{ m/s}

To the precision of the data, v333v \approx 333 m/s. This is close to the standard textbook value of 343343 m/s at 20°20°C; the slightly lower number suggests a cooler day.

Final Answer: v333v \approx 333 m/s.

Why This Works

Sound travels in straight lines until it hits a surface, reflects, and returns. The “echo time” is the round-trip time, so the distance the sound covers is twice the boy-to-wall distance. This is the same logic SONAR and ultrasound use.

The speed of sound depends on temperature: it increases by roughly 0.60.6 m/s per °°C rise. So the same experiment on a hot summer day in Chennai would give a slightly higher value than in a cool Delhi winter.

Alternative Method

Some textbooks set this up using “time for sound to reach wall = t/2t/2”. The boy hears the echo when the sound completes the round trip, so each leg takes 0.60.6 s. Then v=200/0.6=333v = 200/0.6 = 333 m/s. Same answer, slightly different framing.

The classic error: v=200/1.2=166.7v = 200 / 1.2 = 166.7 m/s. This forgets that the echo time covers a round trip. Half the speed of real sound — physically wrong, and an instant red flag if you sanity-check against the standard value of 343\sim 343 m/s.

For CBSE Class 9, examiners love asking the minimum distance for an echo to be heard distinctly: 17.2\geq 17.2 m, since the human ear distinguishes sounds separated by at least 0.10.1 s and sound covers 34.4\sim 34.4 m in that time. Memorize this number.

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