In simple harmonic motion,the total mechanical energy of a given system is $E$. If the mass of the oscillating particle $P$ is doubled,then the new energy of the system for the same amplitude is

$A$ particle oscillates along the $x$-axis according to the law,$x(t) = x_0 \sin^2\left(\frac{t}{2}\right)$,where $x_0 = 1 \text{ m}$. The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph.

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $6 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$

For a body executing $S.H.M. :$
$(a)$ Potential energy is always equal to its $K.E.$
$(b)$ Average potential and kinetic energy over any given time interval are always equal.
$(c)$ Sum of the kinetic and potential energy at any point of time is constant.
$(d)$ Average $K.E.$ in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below:

The variations of potential energy $(U)$ with position $x$ for three simple harmonic oscillators $A, B$ and $C$ are shown in the figure. The oscillators have the same mass. The time period of oscillation is greatest for

$A$ particle performs $S.H.M.$ Its potential energies are $U_{1}$ and $U_{2}$ at displacements $x_{1}$ and $x_{2}$ respectively. At displacement $(x_{1} + x_{2})$,its potential energy $U$ is: