Question

The first overtone frequency of a closed organ pipe $P_1$, is equal to the fundamental frequency of an open organ pipe $P_2.$ If the length of the pipe $P_1$ is 30 cm, what will be the length of $P_2?$

• A. 2 cm
• B. 15 cm
• C. 20 cm
• D. 30 cm

Answer 1

The correct option is C 20 cm

For an organ pipe open at both ends,
$f_n=\frac{nv}{2L}$ .....(1) where, $f$ stands for for pipe open at both ends
For an organ pipe closed at one end,
$f_n^{\prime }=\frac{(2n-1)v}{4L^{\prime }}$ .....(2) where $f^{\prime }$ stands for for pipe closed at one end

From eqn(1):
Fundamental frequency: $f_1=\frac{v}{2L}$
From eqn(2):
First overtone: $f_2^{\prime }=\frac{3v}{4L^{\prime }}$

Given: $f_2^{\prime }=f_1$
$\Rightarrow L=\frac{2}{3}{L}^{\prime }=\frac{2}{3}\times 30=20cm$