The mass of a particle is $1 \ kg$ and it is moving along the $x$-axis. The period of its oscillation is $\frac{\pi}{2} \ s$. Its potential energy at a displacement of $0.2 \ m$ is (in $J$)

$A$ block is executing $SHM.$ Let the time period of variation of velocity be $T_1$ and time period of variation of kinetic energy be $T_2.$ The relation between $T_1$ and $T_2$ is

The graph shows the variation of displacement of a particle executing $S$.$H$.$M$. with time. We infer from this graph that

The $P.E.$ of a particle executing $SHM$ at a distance $x$ from its equilibrium position is

$A$ particle is executing simple harmonic motion with a time period $T$. At time $t = 0$,it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like:

An object of mass $0.2 \,kg$ executes simple harmonic motion along the $x$-axis with a frequency of $(\frac{25}{\pi}) \,Hz$. At the position $x=0.04 \,m$, the object has a kinetic energy of $0.5 \,J$ and a potential energy of $0.4 \,J$. The amplitude of oscillation is ............ $cm$.