Question #163191


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?
1
Expert's answer
2021-02-23T10:06:42-0500

(a) The two resonance conditions are


L=(2n1+1)λ14and L=(2n2+1)λ24L=(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


n1<n2 or n2=n1+1n_1<n_2 \text { or } n_2=n_1+1


Together,


 (2n1+1)vf1=(2n1+3)vf2(2n_1+1)\dfrac{v}{f_1}=(2n_1+3)\dfrac{v}{f_2}


 (2n1+1)f2=(2n1+3)f1\Rightarrow (2n_1+1)f_2=(2n_1+3)f_1


   

(2n1+1)=(2n1+3)f1f2\Rightarrow (2n_1+1)=(2n_1+3)\dfrac{f_1}{f_2}


  (2n1+1)=2n1f1f2+3f1f2\Rightarrow (2n_1+1)=2n_1\dfrac{f_1}{f_2}+3\dfrac{f_1}{f_2}


  2n1(1f1f2)=3f1f21\Rightarrow 2n_1(1-\dfrac{f_1}{f_2})=3\dfrac{f_1}{f_2}-1


  n1(0.2666)=1.6n1=1.60.266=6.01\Rightarrow n_1(0.2666)=1.6\Rightarrow n_1=\dfrac{1.6}{0.266}=6.01


 n1=6 and n2=7n_1=6 \text { and } n_2=7


The depth of the well is


 L=(2n1+1)λ14L=(2n_1+1)\dfrac{\lambda_1}{4}


   =134×34351.87=21.491m=\dfrac{13}{4}\times \dfrac{343}{51.87}=21.491m


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



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