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A student uses an audio oscillator of adjustable frequency to measure the depth of a water...

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

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Answer #1

Given,

f1 = 52 Hz and f2 = 60 Hz

Let L be the depth

We know that

L = nv/4f

for both the frequencies the sounf travelled the same lenth

L1 = L2

n1 v/4 f1 = n2 v/4 f2

n1 f2 = n2 f2

n1f2 = (n1 + 2)f2

n1 = 2f1/(f2 - f1)

n1 = 2 x 52/(60 - 52) = 13

n2 = 15

L = 13 x 343/4 x 52 = 21.44 m

Hence, L = 21.44 m

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