One end of a taut string of length 3 m along | vedclass

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(C) Given: Length of string $L = 3 \ m$,wave speed $v = 100 \ m/s$.The string is fixed at $x=0$ (node) and the other end at $x=3 \ m$ is vibrating (an

Vedclass · Accepted Answer

$(A, C, D)$

Vedclass · Answer

(C) Given: Length of string $L = 3 \ m$,wave speed $v = 100 \ m/s$.The string is fixed at $x=0$ (node) and the other end at $x=3 \ m$ is vibrating (antinode).For a string fixed at one end and free at the other,the possible wavelengths are given by $L = (2n+1) \frac{\lambda}{4}$,where $n = 0, 1, 2, ...$.Thus,$\lambda = \frac{4L}{2n+1} = \frac{12}{2n+1} \ m$.The wave number is $k = \frac{2\pi}{\lambda} = \frac{(2n+1)\pi}{6}$.The angular frequency is $\omega = vk = 100 	imes \frac{(2n+1)\pi}{6} = \frac{(2n+1)50\pi}{3}$.For $n=0$: $k = \frac{\pi}{6}$,$\omega = \frac{50\pi}{3}$. This matches option $(A)$.For $n=1$: $k = \frac{3\pi}{6} = \frac{\pi}{2}$ (Not in options).For $n=2$: $k = \frac{5\pi}{6}$,$\omega = \frac{250\pi}{3}$. This matches option $(C)$.For $n=7$: $k = \frac{15\pi}{6} = \frac{5\pi}{2}$,$\omega = \frac{15 	imes 50\pi}{3} = 250\pi$. This matches option $(D)$.Therefore,options $(A)$,$(C)$,and $(D)$ are correct.

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