A hollow pipe of length 0.8 m is closed at o | vedclass

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(B) The fundamental frequency of a pipe closed at one end is given by $f_P = \frac{v}{4L_P}$,where $v = 320 \ ms^{-1}$ and $L_P = 0.8 \ m$.$f_P = \fra

Vedclass · Accepted Answer

$10$

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(B) The fundamental frequency of a pipe closed at one end is given by $f_P = \frac{v}{4L_P}$,where $v = 320 \ ms^{-1}$ and $L_P = 0.8 \ m$.$f_P = \frac{320}{4 	imes 0.8} = \frac{320}{3.2} = 100 \ Hz$.The frequency of a string vibrating in its second harmonic is $f_S = \frac{2}{2L_S} \sqrt{\frac{T}{\mu}} = \frac{1}{L_S} \sqrt{\frac{T}{\mu}}$,where $L_S = 0.5 \ m$ and $T = 50 \ N$.Since the string resonates with the pipe,$f_S = f_P = 100 \ Hz$.$100 = \frac{1}{0.5} \sqrt{\frac{50}{\mu}} \implies 100 = 2 \sqrt{\frac{50}{\mu}} \implies 50 = \sqrt{\frac{50}{\mu}}$.Squaring both sides: $2500 = \frac{50}{\mu} \implies \mu = \frac{50}{2500} = 0.02 \ kg/m$.The mass of the string $M = \mu 	imes L_S = 0.02 	imes 0.5 = 0.01 \ kg = 10 \ g$.

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