$A$ 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 ... $cm/s$.

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

$A$ $1\,kg$ 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 oscillating systems,a simple pendulum and a vertical spring-mass system,have the same time period of motion on the surface of the Earth. If both are taken to the moon,then-

The figure shows a graph between velocity $(v)$ and displacement $(x)$ from the mean position of a particle performing $SHM$.

$A$ horizontal platform with an object placed on it is executing $S.H.M.$ in the vertical direction. The amplitude of oscillation is $3.92 \times 10^{-3} \, m$. What must be the least period of these oscillations,so that the object is not detached from the platform (in $, s$)?