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

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$\sqrt{2} \ mm$

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

$A$ 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 $2 \ mm$. Then the amplitude of the particle at $9 \ cm$ from the fixed end is

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