As a take-home lab exercise, a couple of physics students are measuring the dept

Question

As a take-home lab exercise, a couple of physics students are measuring the depth of a water well using
sound resonance. They have a function generator, amplifier, and speaker through which they can play
tones between 75.0 Hz to 150.0 Hz. They notice distinct resonances at 91.0 Hz, 117 Hz, and 143 Hz, but
no other frequencies in the 75-150 Hz range cause resonance within the well. Note that a well is
effectively a tube open on one end (the top) and closed at the other end (the bottom, by water usually.)
If the speed of sound in air is 340.0 m/s, what is the depth of the well?

Explanation / Answer

the formula for the resonance frequencies in a resonating air column closed at one end is

f = ( n*v)/(4*L)

"n" here is an odd number (1, 3, 5...). This type of tube produces only odd harmonics and has its fundamental frequency an octave lower than that of an open cylinder (that is, half the frequency).

L is the depth of the well and v is the speed of sound in air given as 340 m/s

let f' : f": f = 91:117:143

=> n':n":n = 91 : 117: 143

all are odd prime numbers

hence n' = 91 n"=117 n=143

=> f' = n'*v/4*L

=> 91 = 91*340/4*L

=> L = 340/4 =85

hence the depth of the well is 85 m

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