Explanation

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