The equation of a transverse wave propagating | vedclass

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(C) The standard equation of a wave is $y = A \sin(kx + \omega t)$.Comparing this with the given equation $y = 1.5 \sin((5 	imes 10^{-3} x) + 20 t)$,

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$6$

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(C) The standard equation of a wave is $y = A \sin(kx + \omega t)$.Comparing this with the given equation $y = 1.5 \sin((5 	imes 10^{-3} x) + 20 t)$,we get:Wave number $k = 5 	imes 10^{-3} \ cm^{-1}$Angular frequency $\omega = 20 \ rad/s$The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{20}{5 	imes 10^{-3}} = 4000 \ cm/s = 40 \ m/s$.The mass per unit length $\mu$ is $\frac{m}{L} = \frac{3 \ g}{80 \ cm} = \frac{3 	imes 10^{-3} \ kg}{0.8 \ m} = 3.75 	imes 10^{-3} \ kg/m$.The wave speed in a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.Therefore,$T = v^2 \mu = (40)^2 	imes (3.75 	imes 10^{-3}) = 1600 	imes 0.00375 = 6 \ N$.

The equation of a transverse wave propagating along a stretched string of length $80 \ cm$ is $y=1.5 \sin \{(5 \times 10^{-3} x) + 20 t\}$,where $x$ and $y$ are in $cm$ and the time $t$ is in seconds. If the mass of the string is $3 \ g$,then the tension in the string is: (in $N$)

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