$A$ linear harmonic oscillator of force constant $2 \times 10^6 \, N/m$ and amplitude $0.01 \, m$ has a total mechanical energy of $160 \, J$. Its

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

Consider a simple harmonic motion $(SHM)$. Let $K$ and $U$ be kinetic energy and potential energy when the displacement in $SHM$ is one-half $\left(\frac{1}{2}\right)$ the amplitude. Which of the following statements is correct?

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.

For a body performing simple harmonic motion,its potential energy is $E_x$ at displacement $x$ and $E_y$ at displacement $y$ from the mean position. The potential energy $E_0$ at displacement $(x+y)$ is

The total energy of a particle executing $S.H.M.$ is $80 \, J$. What is the potential energy when the particle is at a distance of $\frac{3}{4}$ of amplitude from the mean position?