The kinetic energy and the potential energy of a particle executing $S.H.M.$ are equal. The ratio of its displacement and amplitude will be

For a particle performing $S.H.M.$,the total energy is '$n$' times the kinetic energy,when the displacement of a particle from the mean position is $\frac{\sqrt{3}}{2} A$,where $A$ is the amplitude of $S.H.M.$ The value of '$n$' is

$A$ particle starts executing simple harmonic motion $(SHM)$ of amplitude $a$ and total energy $E$. At any instant,its kinetic energy is $\frac{3E}{4}$. Then its displacement $y$ is given by:

For a simple pendulum,a graph is plotted between its kinetic energy $(KE)$ and potential energy $(PE)$ against its displacement $d.$ Which one of the following represents these correctly? (graphs are schematic and not drawn to scale)

$A$ particle is executing simple harmonic motion with amplitude $A$. The ratio of the kinetic energies of the particle when it is at displacements of $\frac{A}{4}$ and $\frac{A}{2}$ from the mean position is

$A$ 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?