A transverse wave is passing through a string | vedclass

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(C) The wave equation is given by $y = 2 \sin(20t - 10x)$. Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$,we get angular f

Vedclass · Accepted Answer

$0.004$

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(C) The wave equation is given by $y = 2 \sin(20t - 10x)$. Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$,we get angular frequency $\omega = 20 \ rad/s$ and wave number $k = 10 \ m^{-1}$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{20}{10} = 2 \ m/s$.The speed of a transverse wave on a string is also given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Linear mass density $\mu = 	ext{density} 	imes 	ext{area} = 1 \ kg/m^3 	imes 0.01 \ m^2 = 0.01 \ kg/m$.The tension $T$ in the string is provided by the weight of the hanging mass $M$,so $T = Mg$.Substituting these into the speed formula: $2 = \sqrt{\frac{Mg}{0.01}}$.Squaring both sides: $4 = \frac{Mg}{0.01} \implies 4 = \frac{M 	imes 10}{0.01} \implies 4 = 1000M$.Therefore,$M = \frac{4}{1000} = 0.004 \ kg$.

$A$ transverse wave is passing through a string as shown in the figure. The mass density of the string is $1 \ kg/m^3$ and the cross-sectional area of the string is $0.01 \ m^2$. The equation of the wave in the string is $y = 2 \sin(20t - 10x)$. The hanging mass is (in $kg$):-

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