A stretched string of 1 m length and mass 5 | vedclass

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(B) The length of the string $l = 1 \ m = 100 \ cm$. The mass $m = 5 	imes 10^{-4} \ kg$. The tension $T = 20 \ N$. When a string is plucked at a dis

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$200$

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(B) The length of the string $l = 1 \ m = 100 \ cm$. The mass $m = 5 	imes 10^{-4} \ kg$. The tension $T = 20 \ N$. When a string is plucked at a distance $x$ from one end,it vibrates in the $p^{th}$ harmonic such that $x = \frac{l}{2p}$. Given $x = 25 \ cm$,we have $25 = \frac{100}{2p}$,which gives $2p = 4$,so $p = 2$. The frequency of vibration $f$ is given by $f = \frac{p}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length. Here,$\mu = \frac{m}{l} = \frac{5 	imes 10^{-4} \ kg}{1 \ m} = 5 	imes 10^{-4} \ kg/m$. Substituting the values: $f = \frac{2}{2 	imes 1} \sqrt{\frac{20}{5 	imes 10^{-4}}} = 1 	imes \sqrt{4 	imes 10^4} = 200 \ Hz$.

$A$ stretched string of $1 \ m$ length and mass $5 \times 10^{-4} \ kg$ is having a tension of $20 \ N$. If it is plucked at $25 \ cm$ from one end,then it will vibrate with a frequency of ... $Hz$.

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