Thank you for reporting, we will resolve it shortly
Q.
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
For open organ pipe, wavelength of $n^{\text {th }}$ overtone
$\lambda_{n}=\frac{2 l}{n+1} (n=3) $
$\lambda=\frac{2 l}{4}=\frac{l}{2}$
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, $R=A \cos \phi$
where, $\phi=$ phase angle.
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}=\frac{\pi}{4}=45^{\circ}$
$R=A \cos 45^{\circ} $
$R=\frac{A}{\sqrt{2}} $