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Physics Question on Waves

An open organ pipe is closed suddenly with the result that the second overtone of the closed pipe is found to be higher in frequency by 100100 than the first overtone of the original pipe. Then the fundamental frequency of the open pipe is

A

200 s−1200 \,{{s}^{-1}}

B

100 s−1100\,{{s}^{-1}}

C

300 s−1300\,{{s}^{-1}}

D

250 s−1250\,{{s}^{-1}}

Answer

200 s−1200 \,{{s}^{-1}}

Explanation

Solution

Frequency of second overtone (fifth harmonic) of closed pipe =5v4l=\frac{5v}{4l} .
Frequency of first overtone (second harmonic) of open pipe =2v2l=\frac{2v}{2l}
Accordingly, 5v4l−2v2l=100\frac{5v}{4l}-\frac{2v}{2l}=100
Or v4l=100\frac{v}{4l}=100
Or v=400 lv=400\,l
Fundamental frequency of open pipe =v2l=400l2l=200s−1=\frac{v}{2l}=\frac{400l}{2l}=200{{s}^{-1}}