The displacement of a particle executing SHM | vedclass

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(D) The displacement of the particle executing $SHM$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$.The velocity of the particle is $v = \frac{

Vedclass · Accepted Answer

$0.3$

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(D) The displacement of the particle executing $SHM$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}ight)$.The velocity of the particle is $v = \frac{dy}{dt} = 5 \cdot \cos \left(4 t+\frac{\pi}{3}ight) \cdot 4 = 20 \cos \left(4 t+\frac{\pi}{3}ight)$.Comparing the given equation with the standard $SHM$ equation $y=a \sin (\omega t+\phi)$,we get angular frequency $\omega = 4 	ext{ rad/s}$.Since $\omega = \frac{2\pi}{T}$,the time period is $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} 	ext{ s}$.At $t = \frac{T}{4} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8} 	ext{ s}$,the velocity is:$v = 20 \cos \left(4 \cdot \frac{\pi}{8} + \frac{\pi}{3}ight) = 20 \cos \left(\frac{\pi}{2} + \frac{\pi}{3}ight) = -20 \sin \left(\frac{\pi}{3}ight) = -20 \cdot \frac{\sqrt{3}}{2} = -10\sqrt{3} 	ext{ m/s}$.Given mass $m = 2 	ext{ g} = 2 	imes 10^{-3} 	ext{ kg}$.The kinetic energy is $KE = \frac{1}{2}mv^2 = \frac{1}{2} \cdot (2 	imes 10^{-3}) \cdot (-10\sqrt{3})^2 = 10^{-3} \cdot (100 \cdot 3) = 300 	imes 10^{-3} = 0.3 	ext{ J}$.

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

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