Question

The frequency of the first overtone of an open organ pipe is $f$, then that of the closed organ pipe will be

• A. $3f/4$
• B. $f$
• C. $f/4$
• D. $f/2$

Answer 1

The correct option is A $3f/4$

Let $v$ be the velocity of sound.

Open organ pipe of length $L$:

Frequency of mth harmonic $f_m=\frac{mv}{4L}$

For 1st overtone i.e $m=2$ $f_2=\frac{2v}{2L}=\frac{v}{L}=f$ ..................(1)

Closed organ pipe of length $L$:

Frequency of different modes of vibration $f_n^{\prime }=\frac{(2n-1)v}{4L}$ where $n=1,2,\mathrm{3..............}$

First overtone i.e $n=2$, $f_2^{\prime }=\frac{3v}{4L}=\frac{3f}{4}$