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 performs linear $S$.$H$.$M$. with amplitude $a$. When it is at a distance $\frac{a}{3}$ from the extreme position,the magnitude of velocity is $\frac{1}{3}$ times the magnitude of acceleration. The period of $S$.$H$.$M$. is:

At $t=0$,a particle executing $SHM$ with a time period $3 \ s$ is in phase with another particle executing $SHM$. The time period of the second particle is $T$ (less than $3 \ s$). If they are again in the same phase for the third time after $45 \ s$,then the value of $T$ is .... . (in $s$)

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

$A$ small block is connected to one end of a massless spring of un-stretched length $4.9 \ m$. The other end of the spring is fixed at $O$. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \ m$ and released from rest at $t = 0$. It then executes simple harmonic motion with angular frequency $\omega = \frac{\pi}{3} \ rad/s$. Simultaneously at $t = 0$,a small pebble is projected with speed $v$ from point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \ m$ from $O$. If the pebble hits the block at $t = 1 \ s$,the value of $v$ is (take $g = 10 \ m/s^2$):

$A$ particle of mass $m$ moves in the potential energy $U(x)$ shown in the figure. The potential energy is given by $U = \frac{1}{2}kx^2$ for $x < 0$ and $U = mgx$ for $x > 0$. The period of the motion when the particle has total energy $E$ is