Question

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

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

Answer 1

Answer: (D)

The displacement of a particle executing SHM is given by
$y = 5 \, \sin \, 4t + \frac{x}{3}.$ ......(i)
Velocity of particle
$ \frac{dy}{dt} = \frac{5d}{dt} \sin4t + \frac{\pi}{3} $
$= 5 \cos 4t + \frac{\pi}{3} $
$= 20 \cos 4t + \frac{\pi}{3} $
Velocity at $ t = \frac{T}{4} $
$ \frac{dy}{dt}_{t = \frac{T}{4}} = 20 \cos 4 \times\frac{T}{4} + \frac{\pi}{3} $
Or $u = 20 \cos T + \frac{\pi}{3}$ ....(ii)
Now, putting value of T in Eq. (ii), we get
$ u = 20 \cos \frac{\pi}{2} + \frac{\pi}{3} $
$= - 20 \sin \frac{\pi}{3}$
$ = - 20 \times \frac{ 3}{2} $
$= -10 \times 3$
The kinetic energy of particle,
$ KE = \frac{1}{2} mu^{2} $
$ \because \, \, m = 2g = 2 \times10^{-3} kg$
$ = \frac{1}{2} \times 2 \times 10^{-3}\times -10 \overline{3} ^{2} $
$= 10^{-3} \times 100 \times 3 $
$= 3 \times 10^{-1} $
$KE = 0.3 \,J $