A body is suspended from a string of length 1 | vedclass

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(D) Mass per unit length of the given string is $\mu = \frac{m_{string}}{l} = \frac{2 	imes 10^{-3} \ kg}{1 \ m} = 2 	imes 10^{-3} \ kg \ m^{-1}$.Sp

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$8 \ kg$

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(D) Mass per unit length of the given string is $\mu = \frac{m_{string}}{l} = \frac{2 	imes 10^{-3} \ kg}{1 \ m} = 2 	imes 10^{-3} \ kg \ m^{-1}$.Speed of transverse waves on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the string. Since the body is suspended,$T = M_{body} 	imes g$.Fundamental frequency for a string fixed at both ends is $f_0 = \frac{v}{2l} = \frac{1}{2l} \sqrt{\frac{M_{body} 	imes g}{\mu}}$.Squaring both sides,we get $f_0^2 = \frac{M_{body} 	imes g}{4l^2 \mu}$.Rearranging for the mass of the body: $M_{body} = \frac{4l^2 f_0^2 \mu}{g}$.Substituting the values: $M_{body} = \frac{4 	imes (1)^2 	imes (100)^2 	imes (2 	imes 10^{-3})}{10} = \frac{4 	imes 10000 	imes 0.002}{10} = \frac{80}{10} = 8 \ kg$.

$A$ body is suspended from a string of length $1 \ m$ and mass $2 \ g$. The mass of the body required to produce a fundamental mode of $100 \ Hz$ frequency in the string is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

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