Question

In a closed organ pipe of length $105\text{cm}$, standing waves are set up corresponding to the third overtone. At what minimum distance from the closed end a pressure node is formed?

• A. $18\text{cm}$
• B. $15\text{cm}$
• C. $25\text{cm}$
• D. $30\text{cm}$

Answer 1

The correct option is B $15\text{cm}$

Third overtone frequency for a closed organ pipe is $7^{th}$ harmonic frequency$(n=7)$.
$\Rightarrow f=\frac{7v}{4l}$
$\Rightarrow v=\frac{4lf}{7}$
From, $v=f\lambda$
$\Rightarrow \lambda =\frac{4l}{7}$
$\Rightarrow \lambda =\frac{4\times 105}{7}=60\text{cm}$
Pressure node will be formed, where there is displacement antinode.


Minimum distance from the closed end at which pressure node will be formed,
$d_{\text{min}}=\frac{\lambda }{4}$
$\Rightarrow d_{\text{min}}=\frac{60}{4}=15\text{cm}$