Answer to Question #163191 in Optics for Takam Praise

Question

A student uses an audio oscillator of adjustable frequency to measure the depth of water well. The student reports hearing two successive resonances at 51.87Hz and 59.85Hz 

1. How deep is the well?
2. How many antinodes are in the standing wave at 51.87Hz?

Answer 1

(a) The two resonance conditions are

$L=(2n_1+1)\dfrac{\lambda_1}{4} \text {and } L=(2n_2+1)\dfrac{\lambda_2}{4}$

Since the first frequency is smaller than the second frequency, the first wavelength is larger than

the second wavelength. This means

$n_1<n_2 \text { or } n_2=n_1+1$

Together,

 $(2n_1+1)\dfrac{v}{f_1}=(2n_1+3)\dfrac{v}{f_2}$

 $\Rightarrow (2n_1+1)f_2=(2n_1+3)f_1$

   

$\Rightarrow (2n_1+1)=(2n_1+3)\dfrac{f_1}{f_2}$

  $\Rightarrow (2n_1+1)=2n_1\dfrac{f_1}{f_2}+3\dfrac{f_1}{f_2}$

  $\Rightarrow 2n_1(1-\dfrac{f_1}{f_2})=3\dfrac{f_1}{f_2}-1$

  $\Rightarrow n_1(0.2666)=1.6\Rightarrow n_1=\dfrac{1.6}{0.266}=6.01$

 $n_1=6 \text { and } n_2=7$

The depth of the well is

 $L=(2n_1+1)\dfrac{\lambda_1}{4}$

   $=\dfrac{13}{4}\times \dfrac{343}{51.87}=21.491m$

(b) The number of anti-nodes for $n = 6 \text { is }7.$