Question

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

• A. 0.4J
• B. 0.5 J
• C. 3 J
• D. 0.3 J

Answer 1

Answer: (D)

Particle executing SHM.
Displacement $y=5\sin \left(4t+\frac{\pi }{3}\right)$ ....(i)
Velocity of particle
$\left(\frac{dy}{dt}\right)=\frac{5d}{dt}\sin \left(4t+\frac{\pi }{3}\right)$
$=5\cos \left(4t+\frac{\pi }{3}\right).4$
$=20\cos \left(4t+\frac{\pi }{3}\right)$
Velocity at $t=\left(\frac{T}{4}\right)$
${\left(\frac{dy}{dt}\right)}_{t=\frac{T}{4}}=20\cos \left(4\times \frac{T}{4}+\frac{\pi }{3}\right)$
or $u=20\cos \left(T+\frac{\pi }{3}\right)$ ....(ii)
Comparing the given equation with standard equation of SHM.
$y=a\sin \left(\omega t+\varphi \right)$
We get, $\omega =4$
As $\omega =\frac{2\pi }{T}$
$\Rightarrow T=\frac{2\pi }{\omega }$
or $T=\frac{2\pi }{4}$
$=\left(\frac{\pi }{2}\right)$
Now, putting value of T in Eq. (ii), we get
$u=20\cos \left(\frac{\pi }{2}+\frac{\pi }{3}\right)$
$=-20\sin \frac{\pi }{3}$
$=-10\times \sqrt{3}$
The kinetic energy of particle,
$KE=\frac{1}{2}m{u}^2$
$=\frac{1}{2}\times 2\times 10^{-3}\times {\left(-10\sqrt{3}\right)}^2$
$=10^{-3}\times 100\times 3$
$KE=0.3J$