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

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(C) $1$. The fundamental frequency of a pipe closed at one end is given by $f_p = \frac{v}{4L}$,where $v = 320 \ m/s$ and $L = 0.8 \ m$.$f_p = \frac{3

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$10$

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(C) $1$. The fundamental frequency of a pipe closed at one end is given by $f_p = \frac{v}{4L}$,where $v = 320 \ m/s$ and $L = 0.8 \ m$.$f_p = \frac{320}{4 	imes 0.8} = \frac{320}{3.2} = 100 \ Hz$.$2$. The string is vibrating in its second harmonic. The frequency of the $n$-th harmonic for a string fixed at both ends is $f_s = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $n=2$,$l = 0.5 \ m$,$T = 50 \ N$,and $\mu = \frac{m}{l}$.$3$. Since the string resonates with the pipe,$f_s = f_p = 100 \ Hz$.$100 = \frac{2}{2 	imes 0.5} \sqrt{\frac{50}{m/0.5}} = 2 \sqrt{\frac{25}{m}} = 2 	imes 5 \sqrt{\frac{1}{m}} = \frac{10}{\sqrt{m}}$.$4$. Squaring both sides: $100 = \frac{100}{m}$,which gives $m = 1 \ kg = 1000 \ g$. Wait,re-evaluating the calculation: $100 = \frac{10}{\sqrt{m}} \implies \sqrt{m} = 0.1 \implies m = 0.01 \ kg = 10 \ g$.

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