Question

A closed organ pipe has a length $l$. The air in it is vibrating in third overtone with maximum displacement amplitude of $a$. The displacement amplitude at distance $l/7$ from the closed end of the pipe will be

• A. $0$
• B. $a$
• C. $a/2$
• D. $a/5$

Answer 1

The correct option is B $a$

We know that for closed organ pipe, resonate frequency,
$f=\frac{nv}{4l}$
For third overtone, $n=7$
So, $f=\frac{7v}{4l}$
$\Rightarrow \frac{v}{f}=\frac{4l}{7}$
$\Rightarrow \lambda =\frac{4l}{7}$
We know that the distance between two successive displacement nodes is $\frac{\lambda }{2}$.
So, $d=\frac{\lambda }{2}=\frac{2l}{7}$

From figure, we can see that, at distance $\frac{l}{7}$ from closed end, there is displacement antinode$\left(AN\right)$.
Thus, the displacement amplitude at distance $l/7$ from the closed end of the pipe will be $a$.

Why this question? Tips: In case of closed organ pipe, the particles of the medium cannot vibrate back and forth at the closed end. Hence, the displacement node is formed at the closed end of an organ pipe.