The transverse displacement in a stretched st | vedclass

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(A) The given equation is $y = 0.06 \sin \left( \frac{2\pi}{3}x \right) \cos (120\pi t)$.Comparing this with the standard wave equation,we have angula

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$648$

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(A) The given equation is $y = 0.06 \sin \left( \frac{2\pi}{3}x ight) \cos (120\pi t)$.Comparing this with the standard wave equation,we have angular frequency $\omega = 120\pi \, rad/s$ and wave number $k = \frac{2\pi}{3} \, m^{-1}$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{120\pi}{2\pi/3} = 180 \, m/s$.The linear mass density $\mu$ is $\mu = \frac{m}{L} = \frac{3.0 	imes 10^{-2} \, kg}{1.5 \, m} = 2.0 	imes 10^{-2} \, kg/m$.The tension $T$ in the string is given by $T = v^2 \mu$.Substituting the values,$T = (180)^2 	imes (2.0 	imes 10^{-2}) = 32400 	imes 0.02 = 648 \, N$.

The transverse displacement in a stretched string is given by $y = 0.06 \sin \left( \frac{2\pi}{3}x \right) \cos (120\pi t)$,where $x$ and $y$ are in $m$ and $t$ is in $s$. The length of the string is $1.5 \, m$ and its mass is $3.0 \times 10^{-2} \, kg$. The tension in the string is ..... $N$.

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