The equation of a travelling wave on a stretc | vedclass

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(C) The given wave equation is $y = 0.03 \sin(450t - 9x)$.Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$,we get angular fr

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$12.5$

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(C) The given wave equation is $y = 0.03 \sin(450t - 9x)$.Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$,we get angular frequency $\omega = 450 \, rad/s$ and wave number $k = 9 \, m^{-1}$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{450}{9} = 50 \, m/s$.The linear mass density $\mu = 5 \, g/m = 5 	imes 10^{-3} \, kg/m$.The speed of a wave on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.Squaring both sides,we get $v^2 = \frac{T}{\mu}$,so $T = \mu v^2$.Substituting the values,$T = (5 	imes 10^{-3} \, kg/m) 	imes (50 \, m/s)^2$.$T = 5 	imes 10^{-3} 	imes 2500 = 5 	imes 2.5 = 12.5 \, N$.

The equation of a travelling wave on a stretched string of linear mass density $5 \, g/m$ is $y = 0.03 \sin(450t - 9x)$,where distance and time are measured in $SI$ units. The tension in the string is ... $N$.

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