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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A student user an audio oscillator of adjustable frequency to measure the depth of a water well. He reports hearing two successive resonances at 52.0 Hz and 60.0 Hz. How deep is the well?

To determine
The depth of a water well.

Explanation

Given Info: The speed of the sound in air is 343 m/s and the frequency of the hearing report in first resonance is 52.0 Hz

Formula to calculate the depth of the well under the resonance condition

L=(2n1)λ14 (1)

  • L is the depth of the well.
  • n is the integer.
  • λ1 is the wavelength of the first resonance condition.

The relation between the wavelength and frequency is,

λ1=vf1 (2)

  • v is the speed of the sound in air.
  • f1 is the frequency of the hearing report in first resonance.

Use equation (2) to rewrite equation (1) we get,

L=(2n1)v4f1

Substitute 343 m/s for v and 52.0 Hz for f1 .

L=(2n1)343m/s4(52.0Hz)=1.65m(2n1) (3)

Thus, student measure an audio oscillator under the depth of the water well for first successive resonance is 1.65m(2n1) .

Formula to calculate the depth of the well under the resonance condition

L=(2n+1)v4f2 (4)

  • f2 is the frequency of the hearing report in second resonance.

Substitute 343 m/s for v and 60.0 Hz for f2 .

L=(2n+1)343m/s4(60.0Hz)=1.43m(2n+1) (5)

Under the resonance condition, the student measure an audio oscillator under the depth of the water well for first successive resonance is equal to the second resonance (L=L)

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