Sum

**Solve the following problem.**

A girl stands 170 m away from a high wall and claps her hands at a steady rate so that each clap coincides with the echo of the one before.

- If she makes 60 claps in 1 minute, what value should be the speed of sound in air?
- Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.

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#### Solution

**Given: **distance (s) = 170 m

**To find:**

- velocity of sound (v)
- distance (s)

**Formula:** Speed = `"distance"/"time"`

**Calculation:**

- Girl produces 60 claps in 1 minute i.e., 1 clap in second From formula,

v = `(2("s"))/"t"` .....(for an echo)

∴ v = `(2 xx 170)/1` =**340 m/s** - Girl produces 45 claps per minute i.e., 1 clap in `4/3` s

From formula,

v = `(2("s"))/"t"` ....(for an echo)

∴ 2s = v × t

∴ s = `(340 xx 4)/(2 xx 3) = (340 xx 2)/3`=**226.67 m**

- When the girl makes 60 claps in 1 minute, the value of speed of is
**340 m/s**. - The girl is at a distance of
**226.67 m**from the wall when she produces 45 claps per minute.

Concept: Echo, Reverberation and Acoustics

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