$A$ particle is executing $SHM$ with amplitude $A,$ time period $T,$ maximum acceleration $a_0$ and maximum velocity $v_0.$ It starts from the mean position at $t=0.$ At time $t,$ it has displacement $A/2,$ acceleration $a,$ and velocity $v.$ Then:

$A$ system is oscillating with undamped simple harmonic motion. Then the

The figure shows a graph between velocity $(v)$ and displacement $(x)$ from the mean position of a particle performing $SHM$.

Match the following physical quantities for a particle executing Simple Harmonic Motion $(SHM)$ given by $y = A \sin(\omega t)$:
$(a)$ Velocity $(v)$
$(b)$ Potential Energy $(PE)$
$(c)$ Total Energy $(TE)$
$(d)$ Acceleration $(a)$
$(i)$ Constant
(ii) $A\omega \cos(\omega t)$
(iii) $\frac{1}{2} k A^2 \sin^2(\omega t)$
(iv) $-\omega^2 y$

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M$. The piston and the cylinder have equal cross-sectional area $A$. When the piston is in equilibrium, the volume of the gas is $V_0$ and its pressure is $P_0$. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surroundings, the piston executes a simple harmonic motion with frequency

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