$A$ linear harmonic oscillator of force constant $6 \times 10^5 \, N/m$ and amplitude $4 \, cm$ has a total energy of $600 \, J$. Select the correct statement.

If the amplitude of a body executing $SHM$ is doubled,what will be its energy?

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t=\frac{T}{12}$,the ratio of the potential energy to kinetic energy of the particle is $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$

The amplitude of a particle executing $SHM$ is $3\,cm$. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is: $.............cm$.

$A$ 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 $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$

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