Two particles $P$ and $Q$ describe simple harmonic motions of same period,same amplitude,along the same line about the same equilibrium position $O.$ When $P$ and $Q$ are on opposite sides of $O$ at the same distance from $O,$ they have the same speed of $1.2 \, m/s$ in the same direction. When their displacements are the same,they have the same speed of $1.6 \, m/s$ in opposite directions. The maximum velocity in $m/s$ of either particle is

$A$ particle is executing $SHM$ with amplitude $A,$ time period $T,$ maximum acceleration $a_0$ and maximum velocity $v_0.$ It starts from the mean position at $t=0.$ At time $t,$ it has displacement $A/2,$ acceleration $a,$ and velocity $v.$ Then:

$A$ 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:

Due to some force $F_1$,a body oscillates with a period of $4/5 \, s$,and due to another force $F_2$,it oscillates with a period of $3/5 \, s$. If both forces act simultaneously,the new period will be .... $s$.

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

$A$ mass of $5\, {kg}$ is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. $A$ simple pendulum of length $4\, {m}$ has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed? (In ${m} / {s}^{2}$)