The amplitude of a particle executing $SHM$ about $O$ is $10 \, cm.$ Then:

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):

The energy of a particle executing simple harmonic motion is given by $E = Ax^2 + Bv^2$,where $x$ is the displacement from mean position $x = 0$ and $v$ is the velocity of the particle at $x$. Choose the incorrect statement.

$A$ body describes simple harmonic motion with an amplitude of $5\; cm$ and a period of $0.2\; s$. Find the acceleration and velocity of the body when the displacement is $(a)\; 5\; cm$,$(b)\; 3\; cm$,and $(c)\; 0\; cm$.

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

$A$ particle executing simple harmonic motion has an amplitude of $6\, cm$. Its acceleration at a distance of $2 \,cm$ from the mean position is $8\, cm/s^2$. The maximum speed of the particle is ... $cm/s$.