A transverse wave propagating on a stretched | vedclass

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(B) The given equation of the wave is $y=0.2 \sin (1.5 x+60 t)$.Comparing this with the standard wave equation $y=A \sin (kx+\omega t)$,we get:$k=1.5

Vedclass · Accepted Answer

$0.48$

Vedclass · Answer

(B) The given equation of the wave is $y=0.2 \sin (1.5 x+60 t)$.Comparing this with the standard wave equation $y=A \sin (kx+\omega t)$,we get:$k=1.5 ~m^{-1}$ and $\omega=60 ~rad/s$.The velocity of the wave is given by $v = \frac{\omega}{k} = \frac{60}{1.5} = 40 ~m/s$.The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Given $\mu = 3 	imes 10^{-4} ~kg/m$.Squaring both sides,we get $v^2 = \frac{T}{\mu}$,which implies $T = v^2 \mu$.Substituting the values: $T = (40)^2 	imes (3 	imes 10^{-4}) = 1600 	imes 3 	imes 10^{-4} = 4800 	imes 10^{-4} = 0.48 ~N$.

$A$ transverse wave propagating on a stretched string of linear density $3 \times 10^{-4} ~kg ~m^{-1}$ is represented by the equation $y=0.2 \sin (1.5 x+60 t)$,where $x$ is in metres and $t$ is in seconds. The tension in the string (in newton) is

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