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

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(B) For a closed organ pipe,the frequencies of harmonics are given by $f_n = (2n-1)f_0$,where $n=1, 2, 3, ...$ corresponds to the fundamental,$1^{st}$

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$a$

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(B) For a closed organ pipe,the frequencies of harmonics are given by $f_n = (2n-1)f_0$,where $n=1, 2, 3, ...$ corresponds to the fundamental,$1^{st}$ overtone,$2^{nd}$ overtone,and so on.For the $3^{rd}$ overtone,$n=4$,so the frequency is $f_4 = (2(4)-1)f_0 = 7f_0$.The length of the pipe $l$ in terms of wavelength $\lambda$ for the $n^{th}$ harmonic is $l = \frac{(2n-1)\lambda}{4}$.For the $3^{rd}$ overtone $(n=4)$,$l = \frac{7\lambda}{4}$,which implies $\lambda = \frac{4l}{7}$.The displacement amplitude $y(x)$ in a closed organ pipe at a distance $x$ from the closed end (node) is given by $y(x) = a \sin(kx)$,where $k = \frac{2\pi}{\lambda}$.Substituting $k = \frac{2\pi}{(4l/7)} = \frac{7\pi}{2l}$,we get $y(x) = a \sin\left(\frac{7\pi x}{2l}\right)$.At $x = l/7$,the amplitude is $y(l/7) = a \sin\left(\frac{7\pi (l/7)}{2l}\right) = a \sin\left(\frac{\pi}{2}\right) = a(1) = a$.

$A$ closed organ pipe has length $l$. The air in it is vibrating in the $3^{rd}$ overtone with a maximum displacement amplitude $a$. The displacement amplitude at a distance $l/7$ from the closed end of the pipe is:

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