A student uses an audio oscillator of adjustable frequency to measure the depth of a water well. Two successive resonant frequencies are heard at 187 Hz and 209 Hz. What is the depth of the well? The speed of sound in air is 344 m/s.

7.8m

Explanation:

Given

frequency f1 = 187Hz

frequency f2 = 209Hz

speed of sound in air 'v'=343m/s

lets assume a cylindrical well and consider 'L ' be the depth

L = nv/4f (where n=1,3,5, ...)

Since two successive resonant frequencies are heard, so n2=n1+2.

lets determine the value of n1 & n2 first

L1 = L2

n1 x v/4f1 = n2 x v/4f2 - --> (cancelling out the common)

n1 x f2 = n2 x f1

=>n1 x f2 = (n1+2) f1

=>n1 (f2-f1) = 2 f1

=>n1 = 2f1 / (f2-f1) = 2 (187) / (209-187) = 17

therefore, n2 = n1+2 = > 17+2 = > 19

next is to find 'L'

L = nv/4f

L = (17) x (344) / 4 x (187) = 15.88 m

L = 7.8m

Therefore, the depth of the well is 7.8m