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
Is there an error in this question or solution?
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