A pipe closed at one end has a length of 0.8 | vedclass

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(D) For the fundamental frequency of a pipe closed at one end,the frequency is given by $f = \frac{v}{4L}$,where $v = 320 ~m/s$ and $L = 0.8 ~m$.$f =

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$10$

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(D) For the fundamental frequency of a pipe closed at one end,the frequency is given by $f = \frac{v}{4L}$,where $v = 320 ~m/s$ and $L = 0.8 ~m$.$f = \frac{320}{4 	imes 0.8} = \frac{320}{3.2} = 100 ~Hz$.For a string of length $L' = 0.5 ~m$ vibrating in its second harmonic,the frequency is $f' = \frac{2v_s}{2L'} = \frac{v_s}{L'}$,where $v_s = \sqrt{\frac{T}{\mu}}$.Given that the string resonates with the pipe,$f' = f = 100 ~Hz$.Thus,$100 = \frac{1}{0.5} \sqrt{\frac{50}{\mu}}$.$50 = \sqrt{\frac{50}{\mu}} \Rightarrow 2500 = \frac{50}{\mu} \Rightarrow \mu = \frac{50}{2500} = 0.02 ~kg/m$.The total mass of the string is $m = \mu 	imes L' = 0.02 ~kg/m 	imes 0.5 ~m = 0.01 ~kg = 10 ~g$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

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