physics ..help!?

Question

question
A student uses an audio oscillator of adjustable frequency to measure the depth of a water well. The student hears two resonances at 51.5Hz and 60.0Hz. How deep is the well ?

Can you give me an explanation and an answer .. the answer should be 21.5m

I don't know where to start and where to finish with this question.. the txt book is not helping .. just half of a page .. so am basically stuck.. being at it for 3 hrs...
Thank you...

Answer 1

The answer is correct.

I'll give you the basic approach.

1) cycles/second * meters/cycles = meters/second
i.e.) Freq * Wavelength = Velocity

Which means if you know the velocity of sound (~344 m/s) and the frequency you can solve for wavelength, which we will call "W"

Lets call D the depth of the well. Your book should tell you the resonances points for a pipe closed at one end will be:

D = [1+2n]*W/4

(Note W is usually expressed by the greek letter lambda)

n is the number of harmonics in the pipe. So if you knew n you could calculate D. But you do know that the next resonance point will occur at n+1. Which happends to be the next highter frequency 60hz. Which gives us:

A) D = [1+2n]*W1/4
B) D = [1+2(n+1)]*W2/4

where W1 is the wavelegnth for 51.5hz & W2 is the wavelegnth for for 60hz.

Since both equations equal D set them equal to each other and solve for 'n'. Substitue 'n' into eq.A and solve for D

Have fun.

Answer 2

You are working with the resonance of sound waves in a closed pipe (which see).