A transverse displacement of a vibrating stri | vedclass

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(A) The given equation of the standing wave is $y = 0.06 \sin \left( \frac{2 \pi}{3} x \right) \cos (120 \pi t)$.Comparing this with the standard equa

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$1296$

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(A) The given equation of the standing wave is $y = 0.06 \sin \left( \frac{2 \pi}{3} x ight) \cos (120 \pi t)$.Comparing this with the standard equation $y = A \sin(kx) \cos(\omega t)$,we get the wave number $k = \frac{2 \pi}{3} \ m^{-1}$ and the angular frequency $\omega = 120 \pi \ rad/s$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{120 \pi}{2 \pi / 3} = 120 \pi 	imes \frac{3}{2 \pi} = 180 \ m/s$.The wave speed in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.Given $\mu = 4 	imes 10^{-2} \ kg/m$,we have $180 = \sqrt{\frac{T}{4 	imes 10^{-2}}}$.Squaring both sides: $180^2 = \frac{T}{4 	imes 10^{-2}}$.$32400 = \frac{T}{0.04}$.$T = 32400 	imes 0.04 = 1296 \ N$.

$A$ transverse displacement of a vibrating string is given by $y = 0.06 \sin \left( \frac{2 \pi}{3} x \right) \cos (120 \pi t)$. If the mass per unit length of the string is $4 \times 10^{-2} \ kg/m$,then the tension in the string will be: (in $N$)

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