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A open pipe of length $l$ is vibrating in $3 \mathrm{rd}$ overtone with maximum amplitude $A$. The amplitude at a distance of $\frac{l}{16}$ from any open end is
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Verified Answer
The correct answer is:
$\frac{A}{\sqrt{2}}$
For open organ pipe, wavelength of $n^{\text {th }}$ overtone
$$
\begin{aligned}
& \lambda_n=\frac{2 l}{n+1} \quad(n=3) \\
& \lambda=\frac{2 l}{4}=\frac{l}{2}
\end{aligned}
$$
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, $R=A \cos \phi$
where, $\phi=$ phase angle.
$$
\begin{gathered}
\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}=\frac{\pi}{4}=45^{\circ} \\
R=A \cos 45^{\circ} \\
R=\frac{A}{\sqrt{2}}
\end{gathered}
$$
$$
\begin{aligned}
& \lambda_n=\frac{2 l}{n+1} \quad(n=3) \\
& \lambda=\frac{2 l}{4}=\frac{l}{2}
\end{aligned}
$$
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, $R=A \cos \phi$
where, $\phi=$ phase angle.
$$
\begin{gathered}
\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}=\frac{\pi}{4}=45^{\circ} \\
R=A \cos 45^{\circ} \\
R=\frac{A}{\sqrt{2}}
\end{gathered}
$$
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