Question

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 $2\, g$, 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(4 t + \frac{\pi}{3} \right). 4 $
$= 20 \cos \left(4 t + \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 + \phi\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} mu^{2} $
$ = \frac{1}{2} \times2 \times10^{-3} \times\left(-10 \sqrt{3}\right)^{2} $
$ = 10^{-3} \times100 \times3 $
$KE = 0.3\, J $