An open pipe of length l is vibrating in the | vedclass

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(C) For an open organ pipe,the wavelength of the $n^{th}$ overtone is given by $\lambda_n = \frac{2l}{n+1}$.For the $3^{rd}$ overtone,$n=3$,so $\lambd

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$\frac{A}{\sqrt{2}}$

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(C) For an open organ pipe,the wavelength of the $n^{th}$ overtone is given by $\lambda_n = \frac{2l}{n+1}$.For the $3^{rd}$ overtone,$n=3$,so $\lambda = \frac{2l}{3+1} = \frac{2l}{4} = \frac{l}{2}$.Since the pipe is open at both ends,antinodes (maximum amplitude $A$) are formed at the open ends.The displacement amplitude $R$ at a distance $x$ from an antinode is given by $R = A \cos(kx)$,where $k = \frac{2\pi}{\lambda}$.Given $x = \frac{l}{16}$,we calculate the phase angle $\phi = kx = \frac{2\pi}{\lambda} \cdot \frac{l}{16}$.Substituting $\lambda = \frac{l}{2}$,we get $\phi = \frac{2\pi}{(l/2)} \cdot \frac{l}{16} = \frac{4\pi}{l} \cdot \frac{l}{16} = \frac{\pi}{4}$.Therefore,the amplitude $R = A \cos(\frac{\pi}{4}) = A \cdot \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}}$.

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

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