$A$ system of two identical rods ($L$-shaped) of mass $m$ and length $l$ are resting on a peg $P$ as shown in the figure. If the system is displaced in its plane by a small angle $\theta$,find the period of oscillations:

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along the horizontal axis and momentum is plotted along the vertical axis. The phase space diagram is the $x(t)$ vs. $p(t)$ curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to the right) is positive and downwards (or to the left) is negative.
$1.$ The phase space diagram for a ball thrown vertically up from the ground is:
$2.$ The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and $E_1$ and $E_2$ are the total mechanical energies respectively. Then:
$(A) E_1 = \sqrt{2} E_2$
$(B) E_1 = 2 E_2$
$(C) E_1 = 4 E_2$
$(D) E_1 = 16 E_2$
$3.$ Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is:
Give the answer for questions $1, 2,$ and $3.$

You are holding a shallow circular container of radius $R$,filled with water to a height $h$ $(h \ll R)$. When you walk with speed $v$,it is seen that water starts spilling over. This happens due to the resonance of the periodic impulse given to the container (due to walking) with the oscillation of the water in the container. If the time period of water oscillating in the container is inversely proportional to $\sqrt{h}$,then $v$ is proportional to

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):

In a $SHM$,at which point are the velocity and acceleration both zero?

$A$ 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: