A student measures the depth of a water well with an adjustable frequency audio oscillator. 2 successive resonant frequencies are heard at 40Hz and 50Hz. What is the depth of the well?

17.15 m

Explanation:

Assuming the well is empty (full of air instead of water), the speed of sound is v = 343 m/s.

The water well acts as a pipe closed at one end.  Therefore, the depth of the water well must be an odd multiple of a quarter of the resonant wavelength.

L = (2n − 1) λ/4, n = 1, 2, 3, etc.

v = λf, so λ = v/f.  Substituting:

L = (2n − 1) v/(4f)

Solving for frequency:

f = (2n − 1) v/(4L)

The difference between two successive resonant frequencies is therefore:

Δf = (2(n+1) − 1) v/(4L) − (2n − 1) v/(4L)

Δf = (2n + 1) v/(4L) − (2n − 1) v/(4L)

Δf = 2 v/(4L)

Δf = v/(2L)

Plugging in values:

50 Hz − 40 Hz = 343 m/s / (2L)

2L = 34.3 m

L = 17.15 m

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