$A$ mass $0.4 \,kg$ performs $S.H.M.$ with a frequency $\frac{16}{\pi} \,Hz$. At a certain displacement, it has kinetic energy $2 \,J$ and potential energy $1.2 \,J$. The amplitude of oscillation is (in $m$)

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

Starting from the mean position,a body oscillates simple harmonically with a period $T$. After what time will its kinetic energy be $75 \%$ of the total energy? $(\sin 30^{\circ} = 0.5)$

The acceleration $A$ and time $t$ of a body in $S.H.M.$ is given by the curve shown below. Then the corresponding graph between kinetic energy $(K.E.)$ and time $t$ is correctly represented by:

An object of mass $0.2 \ kg$ executes simple harmonic oscillations along the $X$-axis with a frequency of $(\frac{25}{\pi}) \ Hz$. At the position $x=0.04 \ m$,the object has kinetic energy $1 \ J$ and potential energy $0.6 \ J$. The amplitude of oscillation is (in $m$)

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