Question

A closed organ pipe has length 'l'. The air in it is vibrating in $3^{rd}$ overtone with maximum amplitude 'a'. Find the amplitude at a distance of $l/7$ from closed end of the pipe.

Answer 1

Now from concepts of standing waves, me know,

$y=Acos\frac{2\pi x}{\lambda }\sin \frac{2\pi vt}{\lambda }$

Amplitude $\to$ at $x=L/7$ is then,

$a=A\cos \frac{2\pi }{\lambda }\left(\frac{L}{7}\right)=A\cos \frac{2\pi L}{7\lambda }$

Now, again we have, $\lambda =\frac{v}{f}$

$\therefore a=A\cos \frac{2\pi L}{7}\left(\frac{f}{v}\right)$ ....(1)

for closed organ pipe, (open at one end closed at the other)

the permissible modes of vibration is giuenas,

$f=(2n-1)\frac{v}{4L}$, and for $3^{rd}$ overtone, $n=4$

$f=\frac{7v}{4L}$, and comparing with equation (1),

we get $\to a=A\cos \frac{2\pi L}{7v}\left(\frac{7v}{4L}\right)$

$\Rightarrow a=A\cos \frac{\pi }{2}=0$

$\Rightarrow a=0$