A pipe closed at one end has a length of 83 | vedclass

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(C) For a pipe closed at one end,the natural frequencies are given by $f_k = (2k - 1) \frac{V}{4L}$,where $k = 1, 2, 3, \dots$ is the mode number.Give

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$5$

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(C) For a pipe closed at one end,the natural frequencies are given by $f_k = (2k - 1) \frac{V}{4L}$,where $k = 1, 2, 3, \dots$ is the mode number.Given $V = 332 \ m/s$ and $L = 83 \ cm = 0.83 \ m$.The fundamental frequency $(k=1)$ is $f_1 = \frac{V}{4L} = \frac{332}{4 	imes 0.83} = \frac{332}{3.32} = 100 \ Hz$.The possible frequencies are odd multiples of the fundamental frequency: $f_k = (2k - 1) 	imes 100 \ Hz$.We need to find the number of frequencies such that $f_k $(2k - 1) 	imes 100 Since $k$ must be a positive integer,$k$ can be $1, 2, 3, 4, 5$.The frequencies are $100 \ Hz, 300 \ Hz, 500 \ Hz, 700 \ Hz, 900 \ Hz$.There are $5$ such frequencies.

$A$ pipe closed at one end has a length of $83 \ cm$. The number of possible natural oscillations of the air column whose frequencies lie below $1000 \ Hz$ are (velocity of sound in air $= 332 \ m/s$).

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