Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system (see figure). Because of the torsional constant $k$,the restoring torque is $\tau = k\theta$ for an angular displacement $\theta$. If the rod is rotated by $\theta_0$ and released,the tension in it when it passes through its mean position will be

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:

State whether the following statements are True or False:
$1.$ If a spring is cut into two equal pieces,the spring constant of each piece decreases.
$2.$ As the displacement of a Simple Harmonic Oscillator $(SHO)$ increases,its acceleration decreases.
$3.$ $A$ system can oscillate with more than one natural frequency.
$4.$ The periodic time of Simple Harmonic Motion $(SHM)$ depends on amplitude,energy,or phase constant.

$A$ $1\,kg$ mass is attached to a spring of force constant $600\,N/m$ and rests on a smooth horizontal surface with the other end of the spring tied to a wall as shown in the figure. $A$ second mass of $0.5\,kg$ slides along the surface towards the first at $3\,m/s$. If the masses make a perfectly inelastic collision,find the amplitude and time period of oscillation of the combined mass.

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

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: $(M = 10, m = 5, u = 1, k = 1)$.