Q. 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) $200s^{- 1}$
B) $100s^{- 1}$
C) $300\ s^{- 1}$
D) $250\ s^{- 1}$
Explanation

(a) Frequency of second overtone (fifth harmonic) of close pipe
$= \frac{5v}{fl}$
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
$= \frac{v}{2l} = \frac{400l}{2l} = 200s^{- 1}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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