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(B) The given wave equation is $y = 2 \sin (10x + 300t)$.Comparing this with the standard wave equation $y = A \sin (kx + \omega t)$,we get the wave n

Vedclass · Accepted Answer

$0.054$

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(B) The given wave equation is $y = 2 \sin (10x + 300t)$.Comparing this with the standard wave equation $y = A \sin (kx + \omega t)$,we get the wave number $k = 10 \, rad/m$ and the angular frequency $\omega = 300 \, rad/s$.The linear mass density $\mu = 0.6 	imes 10^{-3} \, g/cm$. Converting this to $SI$ units $(kg/m)$:$\mu = \frac{0.6 	imes 10^{-3} 	imes 10^{-3} \, kg}{10^{-2} \, m} = 6 	imes 10^{-5} \, kg/m$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{300}{10} = 30 \, m/s$.The speed of a transverse wave on a string is related to tension $T$ and linear mass density $\mu$ by $v = \sqrt{\frac{T}{\mu}}$.Therefore,$T = v^2 \mu$.Substituting the values: $T = (30)^2 	imes (6 	imes 10^{-5}) = 900 	imes 6 	imes 10^{-5} = 5400 	imes 10^{-5} = 0.054 \, N$.

$A$ transverse wave propagating on a string can be described by the equation $y = 2 \sin (10x + 300t)$,where $x$ and $y$ are in meters and $t$ is in seconds. If the vibrating string has a linear mass density of $0.6 \times 10^{-3} \, g/cm$,then the tension in the string is .............. $N$.

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