A closed organ pipe has length L. The air in | vedclass

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(B) For a closed organ pipe,the frequency of the $n^{th}$ overtone is given by $f_n = (2n+1) \frac{v}{4L}$.For the third overtone,$n=3$,so the frequen

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$a$

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(B) For a closed organ pipe,the frequency of the $n^{th}$ overtone is given by $f_n = (2n+1) \frac{v}{4L}$.For the third overtone,$n=3$,so the frequency is $f_3 = (2(3)+1) \frac{v}{4L} = \frac{7v}{4L}$.The wavelength $\lambda$ is given by $\lambda = \frac{v}{f_3} = \frac{4L}{7}$.The displacement amplitude $A(x)$ at a distance $x$ from the closed end (which acts as a node for displacement) is given by $A(x) = A_{max} \sin(Kx)$,where $K = \frac{2\pi}{\lambda}$.Here,$A_{max} = a$ and $K = \frac{2\pi}{4L/7} = \frac{7\pi}{2L}$.Substituting $x = \frac{L}{7}$ into the equation:$A(\frac{L}{7}) = a \sin\left(\frac{7\pi}{2L} \cdot \frac{L}{7}\right) = a \sin\left(\frac{\pi}{2}\right) = a(1) = a$.

$A$ closed organ pipe has length $L$. The air in it is vibrating in the third overtone with a maximum amplitude $a$. The amplitude at a distance $\frac{L}{7}$ from the closed end of the pipe is:

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