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

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):

$A$ particle of mass $m$ moves in the potential energy $U(x)$ shown in the figure. The potential energy is given by $U = \frac{1}{2}kx^2$ for $x < 0$ and $U = mgx$ for $x > 0$. The period of the motion when the particle has total energy $E$ is

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

Due to some force $F_1$,a body oscillates with a period of $4/5 \, s$,and due to another force $F_2$,it oscillates with a period of $3/5 \, s$. If both forces act simultaneously,the new period will be .... $s$.

As shown in the figures,a uniform rod $OO^{\prime}$ of length $l$ is hinged at the point $O$ and held in place vertically between two walls using two massless springs of same spring constant $K$. The springs are connected at the midpoint and at the top-end $(O^{\prime})$ of the rod,as shown in Fig. $1$,and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $f_1$. On the other hand,if both the springs are connected at the midpoint of the rod,as shown in Fig. $2$,and the rod is made to oscillate by a small angular displacement,then the frequency of oscillation is $f_2$. Ignoring gravity and assuming motion only in the plane of the diagram,the value of $\frac{f_1}{f_2}$ is: