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(C) Given mass per unit length $\mu = 0.135 \, g/cm = 0.135 	imes 10^{-3} \, kg / 10^{-2} \, m = 0.0135 \, kg/m$.The wave equation is $y = -0.21 \sin

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$1215$

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(C) Given mass per unit length $\mu = 0.135 \, g/cm = 0.135 	imes 10^{-3} \, kg / 10^{-2} \, m = 0.0135 \, kg/m$.The wave equation is $y = -0.21 \sin(x + 30t)$.Comparing this with the standard wave equation $y = A \sin(kx + \omega t)$,we get angular frequency $\omega = 30 \, rad/s$ and wave number $k = 1 \, m^{-1}$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{30}{1} = 30 \, m/s$.The speed of a transverse wave on a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.Therefore,$T = v^2 \mu = (30)^2 	imes 0.0135 = 900 	imes 0.0135 = 12.15 \, N$.We are given $T = x 	imes 10^{-2} \, N$,so $12.15 = x 	imes 10^{-2}$.Thus,$x = 12.15 	imes 100 = 1215$.

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