The displacement of a particle of mass 2 ,g e | vedclass

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(D) Given,$y=5 \sin \left(4 t+\frac{\pi}{3}\right)$.Comparing with $y=A \sin (\omega t+\phi)$,we get angular frequency $\omega = 4 \,rad/s$.The mass o

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$0.3$

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(D) Given,$y=5 \sin \left(4 t+\frac{\pi}{3}ight)$.Comparing with $y=A \sin (\omega t+\phi)$,we get angular frequency $\omega = 4 \,rad/s$.The mass of the particle is $m = 2 \,g = 2 	imes 10^{-3} \,kg$.The velocity of the particle is $v = \frac{dy}{dt} = 5 	imes 4 \cos \left(4 t+\frac{\pi}{3}ight) = 20 \cos \left(4 t+\frac{\pi}{3}ight) \,m/s$.At $t = \frac{T}{4}$,where $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \,s$,we have $t = \frac{\pi}{8} \,s$.Substituting $t = \frac{\pi}{8}$ in the velocity equation:$v = 20 \cos \left(4 	imes \frac{\pi}{8} + \frac{\pi}{3}ight) = 20 \cos \left(\frac{\pi}{2} + \frac{\pi}{3}ight) = 20 \cos \left(\frac{5\pi}{6}ight)$.Since $\cos(150^{\circ}) = -\frac{\sqrt{3}}{2}$,we get $v = 20 	imes \left(-\frac{\sqrt{3}}{2}ight) = -10\sqrt{3} \,m/s$.The kinetic energy $K = \frac{1}{2} mv^2 = \frac{1}{2} 	imes (2 	imes 10^{-3}) 	imes (-10\sqrt{3})^2$.$K = 10^{-3} 	imes 100 	imes 3 = 300 	imes 10^{-3} = 0.3 \,J$.

The displacement of a particle of mass $2 \,g$ executing $SHM$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$. Here,$y$ is in metres and $t$ is in seconds. The kinetic energy of the particle,when $t=\frac{T}{4}$ is (in $\,J$)

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