$A$ block of mass $(10 \alpha) \text{ g}$,where $\alpha$ is a constant,is moving with velocity $3 \text{ m/s}$ to the right. It collides inelastically with a block on the right of mass $10 \text{ g}$ and sticks to it. The right block is connected to three springs as shown in the figure. The spring constant of each spring is $k = 2 \text{ N/m}$. If the amplitude of the resulting simple harmonic motion is $A = \frac{1}{2\sqrt{2}} \text{ m}$,then the value of $\alpha$ is:

Two pendulums with identical bobs and lengths are suspended from a common support such that in the rest position the two bobs are in contact. After being displaced by $5^o$,the bob $A$ is released from rest at $t = 0$. Subsequently,it collides elastically head-on with the other bob $B$. Identify the graph showing the variation in energy of pendulum $A$ with time for $0 \leqslant t \leqslant T$ (where $T$ is the period of either pendulum).

For a particle executing $S.H.M.$,where $x$ is the displacement from the equilibrium position,$v$ is the velocity at any instant,and $a$ is the acceleration at any instant,then:

$A$ body executing simple harmonic motion has a maximum acceleration equal to $24 \, m/s^2$ and a maximum velocity equal to $16 \, m/s$. The amplitude of the simple harmonic motion is:

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):

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: