$Assertion :$ In simple harmonic motion,the velocity is maximum when the acceleration is minimum.
$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$.

Two particles,$1$ and $2$,each of mass $m$,are connected by a massless spring and are on a horizontal frictionless plane,as shown in the figure. Initially,the two particles,with their center of mass at $x_0$,are oscillating with amplitude $a$ and angular frequency $\omega$. Thus,their positions at time $t$ are given by $x_1(t) = (x_0 + d) + a \sin \omega t$ and $x_2(t) = (x_0 - d) - a \sin \omega t$,respectively,where $d > 2a$. Particle $3$ of mass $m$ moves towards this system with speed $u_0 = a \omega / 2$ and undergoes an instantaneous elastic collision with particle $2$ at time $t_0$. Finally,particles $1$ and $2$ acquire a center of mass speed $v_{cm}$ and oscillate with amplitude $b$ and the same angular frequency.
$(1)$ If the collision occurs at time $t_0 = 0$,the value of $v_{cm} / (a \omega)$ will be
$(2)$ If the collision occurs at time $t_0 = \pi / (2 \omega)$,then the value of $4b^2 / a^2$ will be

$A$ clock $S$ is based on the oscillation of a spring,and a clock $P$ is based on pendulum motion. Both clocks run at the same rate on Earth. On a planet having the same density as Earth but twice the radius,which of the following is true?

An oscillator of mass $M$ is at rest in its equilibrium position in a potential $V = \frac{1}{2}k(x - X)^2$. $A$ particle of mass $m$ comes from the right with speed $u$ and collides completely inelastically with $M$ and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after $13$ collisions is: $(M = 10, m = 5, u = 1, k = 1)$.

$A$ graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion is

In a $SHM$,at which point are the velocity and acceleration both zero?