Question

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 $ 100 $ than the first overtone of the original pipe. Then the fundamental frequency of the open pipe is

• A. $ 200 \,{{s}^{-1}} $
• B. $ 100\,{{s}^{-1}} $
• C. $ 300\,{{s}^{-1}} $
• D. $ 250\,{{s}^{-1}} $
• E. $ 150\,{{s}^{-1}} $

Answer 1

Answer: (A)

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