Question

A standing wave in second overtone is maintained in an open organ pipe of length $l$. The distance between consecutive displacement node and pressure node is $l/x$. Find $x$.

Answer 1

Frequency of second overtone $(i.en=3)$ in an open organ pipe,

${\nu }_3=\frac{3v}{2l}$

Now as $v={\nu }_3\lambda$

${\nu }_3=\frac{3{\nu }_3\lambda }{2l}⟹\lambda =\frac{2l}{3}$

The distance between the consecutive displacement node and pressure node is equal to $\frac{\lambda }{4}$ where $\lambda$ is the wavelength of the wave.

Thus $\frac{l}{x}=\frac{\lambda }{4}$

$\frac{l}{x}=\frac{\frac{2l}{3}}{4}⟹x=6$