The transverse displacement of a string of a | vedclass

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(D) The given equation for the standing wave is $y_{(x, t)} = 0.03 \sin \left(\frac{2 \pi x}{3}\right) \cos (60 \pi t)$.Comparing this with the standa

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$81$

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(D) The given equation for the standing wave is $y_{(x, t)} = 0.03 \sin \left(\frac{2 \pi x}{3}ight) \cos (60 \pi t)$.Comparing this with the standard standing wave equation $y = a \sin(kx) \cos(\omega t)$,we get:Angular frequency $\omega = 60 \pi 	ext{ rad s}^{-1}$.Wave number $k = \frac{2 \pi}{3} 	ext{ m}^{-1}$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{60 \pi}{2 \pi / 3} = 30 	imes 3 = 90 	ext{ m s}^{-1}$.The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Given $\mu = 0.01 	ext{ kg m}^{-1}$.Substituting the values: $90 = \sqrt{\frac{T}{0.01}}$.Squaring both sides: $8100 = \frac{T}{0.01}$.$T = 8100 	imes 0.01 = 81 	ext{ N}$.

The transverse displacement of a string of a linear density $0.01 \text{ kg m}^{-1}$,clamped at its ends is given by $Y_{(x, t)} = 0.03 \sin \left(\frac{2 \pi x}{3}\right) \cos (60 \pi t)$,where $x$ and $y$ are in metres and time $t$ is in seconds. The tension in the string is: (in $\text{ N}$)

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