Question

The first overtone frequency in an open organ pipe is equal to the fundamental frequency of closed organ pipe. If length of the closed organ pipe is 40 cm, the length of the open organ pipe is

• A. 10 cm
• B. 20 cm
• C. 160 cm
• D. 80 cm

Answer 1

The correct option is C 160 cm

For open organ pipe $f_{n^{th}overtone}=(n+1)\frac{v}{2l}$

for closed organ pipe $f_0=\frac{v}{4l}$

For Fundamental mode of closed organ pipe $\frac{\lambda }{4}=l_1\Rightarrow \lambda =4l_1$

$f_0=\frac{v}{\lambda }=\frac{v}{4l_1}$

For first overtone in open organ pipe
$\lambda =l_2$

$f=\frac{v}{l_2}$

$f=f_0\Rightarrow \frac{\mathrm{v/}}{4l_1}=\frac{\mathrm{v/}}{l_2}$

$l_2=4l_1=4\times 40$
$l_2=160cm$