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(B) For a string fixed at one end $(x=0)$ and free at the other $(x=L)$,the length $L$ in the $n$-th overtone is given by $L = (2n+1) \frac{\lambda}{4

Vedclass · Accepted Answer

$\sqrt{2} \ mm$

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(B) For a string fixed at one end $(x=0)$ and free at the other $(x=L)$,the length $L$ in the $n$-th overtone is given by $L = (2n+1) \frac{\lambda}{4}$.For the second overtone,$n=2$,so $L = 5 \frac{\lambda}{4}$.Given $L = 10 \ cm$,we have $10 = \frac{5\lambda}{4} \Rightarrow \lambda = 8 \ cm$.The amplitude of a particle at position $x$ is given by $A(x) = A_{max} \sin(kx)$,where $k = \frac{2\pi}{\lambda}$.Here $A_{max} = 2 \ mm$ and $x = 9 \ cm$ from the fixed end.$A(9) = 2 \sin \left( \frac{2\pi}{8} 	imes 9 ight) = 2 \sin \left( \frac{9\pi}{4} ight) = 2 \sin \left( 2\pi + \frac{\pi}{4} ight) = 2 \sin \left( \frac{\pi}{4} ight) = 2 	imes \frac{1}{\sqrt{2}} = \sqrt{2} \ mm$.

$A$ string fixed at one end and free at the other is vibrating in its second overtone. The length of the string is $10 \ cm$ and the maximum amplitude of vibration of particles of the string is $2 \ mm$. The amplitude of the particle at $9 \ cm$ from the fixed end is

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