Question

A open pipe of length $l$ is vibrating in $3\text{rd}$ overtone with maximum amplitude $A$. The amplitude at a distance of $\frac{l}{16}$ from any open end is

• A. A
• B. 0
• C. $\frac{A}{\sqrt{2}}$
• D. $\frac{\sqrt{3}A}{2}$

Answer 1

Answer: (C)

For open organ pipe, wavelength of $n^{\text{th }}$ overtone
${\lambda }_n=\frac{2l}{n+1}(n=3)$
$\lambda =\frac{2l}{4}=\frac{l}{2}$
As pipe is open, so antinode (maximum amplitude) will form at open ends.
At $l=0$, amplitude $=A$
At $\frac{l}{16}$ or $\frac{\lambda }{8}$ distance, amplitude, $R=A\cos \varphi$
where, $\varphi =$ phase angle.
$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{8}=\frac{\pi }{4}=45^{\circ }$
$R=A\cos 45^{\circ }$
$R=\frac{A}{\sqrt{2}}$