The $K.E.$ and $P.E.$ of a particle executing $SHM$ with amplitude $A$ will be equal when its displacement is

$A$ particle of mass $m$ oscillates with simple harmonic motion between points $X_1$ and $X_2$,the equilibrium position being $O$. Its potential energy will be as shown in the following graph:

Draw a graph to show the variation of $PE$,$KE$ and total energy of a simple harmonic oscillator with displacement.

For a simple pendulum, having time period $T$, the variation of kinetic energy $(K.E)$ with time $(t)$ is represented by:

The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = k x^{2} / 2$,where $k$ is the force constant of the oscillator. For $k = 0.5 \; N m^{-1}$,the graph of $V(x)$ versus $x$ is shown in the figure. Show that a particle of total energy $1 \; J$ moving under this potential must 'turn back' when it reaches $x = \pm 2 \; m$.

The maximum potential energy of a block executing simple harmonic motion is $25 \ J$. $A$ is the amplitude of oscillation. At $x = A / 2$,the kinetic energy of the block is $...............$ (in $J$)