The displacement of a particle of mass $2 \mathrm{~g}$ executing SHM is given by $y=5 \sin \left(4…
The displacement of a particle of mass $2 \mathrm{~g}$ 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
$0.4 \mathrm{~J}$
$0.5 \mathrm{~J}$
$3 \mathrm{~J}$
$0.3 \mathrm{~J}$
Solution
$
\begin{gathered}
\text { Given, } y=5 \sin \left(4 t+\frac{\pi}{3}\right) \\
\omega=4 \text { and so } T=\frac{2 \pi}{4}=\frac{\pi}{2} \mathrm{~s} \\
t=\left[\frac{T}{4}=\frac{\pi}{8} \mathrm{~s}\right. \\
\text { Velocity }=\frac{d y}{d t}=20 \cos \left(4 t+\frac{\pi}{3}\right)
\end{gathered}
$
Velocity at $t=\frac{\pi}{8}$ s is
$
\begin{aligned}
v & =20 \cos \left(4 \times \frac{\pi}{8}+\frac{\pi}{3}\right)=20 \cos 150^{\circ} \\
& =-20 \cos 30^{\circ}=-\frac{20 \sqrt{3}}{2} \mathrm{~ms}^{-1}
\end{aligned}
$
KE of particle is
$
K=\frac{1}{2} m v^2=\frac{1}{2} \times \frac{2}{1000} \times 100 \times 3=0.3 \mathrm{~J}
$