$A$ particle is executing linear simple harmonic motion of amplitude $A$. At what displacement is the energy of the particle half potential and half kinetic?

$A$ linear harmonic oscillator of force constant $2 \times 10^6\, N/m$ and amplitude $0.01\, m$ has a total mechanical energy of $160\, J$. Which of the following statements is correct?

In $S.H.M.$,the displacement of a particle at an instant is $Y = A \cos 30^{\circ}$,where $A = 40 \ cm$ and kinetic energy is $200 \ J$. If the force constant is $1 \times 10^{x} \ N/m$,then $x$ will be $(\cos 30^{\circ} = \sqrt{3}/2)$.

$A$ particle of mass $m$ is hanging vertically by an ideal spring of force constant $K$. If the mass is made to oscillate vertically,its total energy is

$A$ particle is executing $S.H.M.$ of amplitude $A$. When the potential energy of the particle is half of its maximum value during the oscillation,its displacement from the equilibrium position is

The displacement of a particle of mass $2 \text{ g}$ executing simple harmonic motion is $x = 8 \cos \left(50 t + \frac{\pi}{12}\right) \text{ m}$,where $t$ is time in seconds. The maximum kinetic energy of the particle is (in $\text{ J}$)