$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.
$(D)$ the time at which the particle passes througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.
Velocity $v = A \omega \cos \omega t =\frac{\omega A }{2}$
At the time of collision
$\cos \omega t=\frac{1}{2} $
$\omega t=\frac{\pi}{3} \Rightarrow t=\frac{2 \pi}{3}=\frac{\pi}{3} \sqrt{\frac{m}{k}}$
$Image$
for $(C)$ $ \quad time$$ =\frac{2 \pi}{3} \sqrt{\frac{m}{k}}+\frac{\pi}{2} \sqrt{\frac{m}{k}} $
$ =\frac{5 \pi}{6} \sqrt{\frac{m}{k}}$(So it is incorrect)
for $(D)$ $\quad time \quad=\frac{2 \pi}{3} \sqrt{\frac{m}{k}}+\pi \sqrt{\frac{m}{k}} $
$ =\frac{5 \pi}{3} \sqrt{\frac{m}{k}} \text { (So it is correct). }$
$x\left( t \right) = A\,\sin \,\left( {at + \delta } \right)$
$y\left( t \right) = B\,\sin \,\left( {bt} \right)$
Identify the correct match below



$(A)$ The force is zero $t=\frac{3 T}{4}$
$(B)$ The acceleration is maximum at $t=T$
$(C)$ The speed is maximum at $t =\frac{ T }{4}$
$(D)$ The $P.E.$ is equal to $K.E.$ of the oscillation at $t=\frac{T}{2}$