Question

Fundamental frequency of an open organ pipe is in resonance with first overtone of a closed pipe. The ratio of their lengths is

• A. 4:1
• B. 3:2
• C. 2:3
• D. 1:4

Answer 1

Answer: (C)

2:3

First overtone of a closed pipe $\nu =3{\nu }_c=\frac{3\nu }{4L_1}$

Fundamental frequency of open pipe is ${\nu }_0=\frac{\nu }{2L_2}$

So, by problem $3{\nu }_c={\nu }_0$

$\Rightarrow \frac{3\nu }{4L_1}=\frac{\nu }{2L_2}$

$\Rightarrow \frac{L_2}{L_1}=\frac{4}{2\times 3}=\frac{2}{3}$