For a particle executing simple harmonic motion, match the following statements (conditions) from Column-$I$ to statements (shapes of graph) in Column-$II$.

Column-$I$	Column-$II$
$(A)$ Velocity-displacement graph $(\omega \neq 1)$	$(i)$ Straight line
$(B)$ Acceleration-displacement graph	$(ii)$ Sinusoidal
$(C)$ Acceleration-time graph	$(iii)$ Circle
$(D)$ Acceleration-velocity graph $(\omega \neq 1)$	$(iv)$ Ellipse

$A$ small block is connected to one end of a massless spring of un-stretched length $4.9 \ m$. The other end of the spring is fixed at $O$. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \ m$ and released from rest at $t = 0$. It then executes simple harmonic motion with angular frequency $\omega = \frac{\pi}{3} \ rad/s$. Simultaneously at $t = 0$,a small pebble is projected with speed $v$ from point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \ m$ from $O$. If the pebble hits the block at $t = 1 \ s$,the value of $v$ is (take $g = 10 \ m/s^2$):

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

The center of a disk of radius $r$ and mass $m$ is attached to a spring of spring constant $k$,inside a ring of radius $R > r$ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring,without slipping. The spring can only be stretched or compressed along the periphery of the ring,following Hooke's law. In equilibrium,the disk is at the bottom of the ring. Assuming small displacement of the disc,the time period of oscillation of the center of mass of the disk is written as $T = \frac{2 \pi}{\omega}$. The correct expression for $\omega$ is ($g$ is the acceleration due to gravity):

Obtain the ratio of maximum acceleration and maximum velocity of a $SHM$ particle.

$Assertion :$ 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 $\frac{\pi }{2}$.