Maharashtra State BoardHSC Science (General) 11th
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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. - Physics

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.

  1. If she makes 60 claps in 1 minute, what value should be the speed of sound in air?
  2. 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:

  1. velocity of sound (v)
  2. distance (s)

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

Calculation:

  1. 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
  2. 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
  1. When the girl makes 60 claps in 1 minute, the value of speed of is 340 m/s.
  2. 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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APPEARS IN

Balbharati Physics 11th Standard Maharashtra State Board
Chapter 8 Sound
Exercises | Q 3. (iv) | Page 158
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