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 $2 \mathrm{~g}$, the kinetic energy of the particle when $t=\frac{T}{4}$ is given by

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Aicharya · Accepted Answer

The displacement of particle, executing SHM

Velocity of particle $\begin{aligned} \left(\frac{d y}{d t}\right) & =\frac{5 d}{d t} \sin \left(4 t+\frac{\pi}{3}\right) \\ & =5 \cos \left(4 t+\frac{\pi}{3}\right) \cdot 4 \\ & =20 \cos \left(4 t+\frac{\pi}{3}\right) \end{aligned}$ Velocity at $t=\left(\frac{T}{4}\right)$ $\left(\frac{d y}{d t}\right)_{t=\frac{T}{4}}=20 \cos \left(4 \times \frac{T}{4}+\frac{\pi}{3}\right)$

Comparing the given equation with standard equation of SHM, given by We get $\begin{aligned} & y=a \sin (\omega t+\phi) \\ & \omega=4 \end{aligned}$ As $\omega=\frac{2 \pi}{T}$ $\begin{aligned} & \Rightarrow \quad T=\frac{2 \pi}{\omega} \\ & \text { or } & T=\frac{2 \pi}{4} \\ & \text { or } & T=\left(\frac{\pi}{2}\right) \\ & \end{aligned}$ Now, putting value of $T$ in Eq. (ii), we get $\begin{aligned} u & =20 \cos \left(\frac{\pi}{2}+\frac{\pi}{3}\right) \\ & =-20 \sin \frac{\pi}{3} \\ & =-20 \times \frac{\sqrt{3}}{2} \\ & =-10 \times \sqrt{3} \end{aligned}$ The kinetic energy of particle, $\begin{aligned} & \mathrm{KE}=\frac{1}{2} m u^2 \\ & \because \quad m=2 \mathrm{~g}=2 \times 10^{-3} \mathrm{~kg} \\ & =\frac{1}{2} \times 2 \times 10^{-3} \times(-10 \sqrt{3})^2 \\ & =10^{-3} \times 100 \times 3 \\ & =3 \times 10^{-1} \\ & \mathrm{KE}=0.3 \mathrm{~J} \\ & \end{aligned}$

The displacement of a particle executing SHM is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right) .$ If $T$…

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