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(D) The given equation of the transverse wave is $y = 0.021 \sin(x + 30t)$.Comparing this with the standard wave equation $y = A \sin(kx + \omega t)$,

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$0.12$

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(D) The given equation of the transverse wave is $y = 0.021 \sin(x + 30t)$.Comparing this with the standard wave equation $y = A \sin(kx + \omega t)$,we get the angular frequency $\omega = 30 \ rad/s$ and the 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 in a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Given $\mu = 1.3 	imes 10^{-4} \ kg/m$.Substituting the values: $30 = \sqrt{\frac{T}{1.3 	imes 10^{-4}}}$.Squaring both sides: $900 = \frac{T}{1.3 	imes 10^{-4}}$.$T = 900 	imes 1.3 	imes 10^{-4} = 1170 	imes 10^{-4} = 0.117 \ N$.Rounding to two decimal places,$T \approx 0.12 \ N$.

$A$ string is producing transverse vibration whose equation is $y = 0.021 \sin(x + 30t)$,where $x$ and $y$ are in meters and $t$ is in seconds. If the linear density of the string is $1.3 \times 10^{-4} \ kg/m$,then the tension in the string in $N$ will be:

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