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

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) 200s- 1 (B) 100s- 1 (C) 250 s- 1 (D) 150 s- 1

Tardigrade Admin · Accepted Answer

Frequency of second overtone (fifth harmonic) of close pipe

= \frac{5 v /4}{l}

Frequency of first overtone (second harmonic) of open pipe

= \frac{2 v}{2 l}

Accordingly,

\frac{5 v}{4 l} - \frac{2 v}{2 l} = 100

Or

\frac{v}{4 l} = 100

Or

v = 400 l

Fundamental frequency of open pipe

= \frac{v}{2 l} = \frac{400 l}{2 l} = 200 s^{- 1}

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

$200 s^{- 1}$

$100 s^{- 1}$

$250 s^{- 1}$

$150 s^{- 1}$

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

200s−1

100s−1

250s−1

150s−1

Frequency of second overtone (fifth harmonic) of close pipe =l5v/4​ Frequency of first overtone (second harmonic) of open pipe =2l2v​ Accordingly, 4l5v​−2l2v​=100 Or 4lv​=100 Or v=400l Fundamental frequency of open pipe =2lv​=2l400l​=200s−1

$200 s^{- 1}$

$100 s^{- 1}$

$250 s^{- 1}$

$150 s^{- 1}$