When kinetic energy is maximum, potential energy
is zero and vice versa.
$Maximum\, potential\, energy = total\, energy.$
$0+K_{0}=K_{0}(\mathrm{K} . \mathrm{E} .+\mathrm{P.E.}=\text { total energy })$

$(A)$ 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 $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.
$(A)$ 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 $t=\frac{T}{2}$

$y = \frac{1}{{\sqrt a }}\,\sin \,\omega t \pm \frac{1}{{\sqrt b }}\,\cos \,\omega t$ will be