The amplitude of a particle executing $S.H.M.$ is $3 \,cm$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is (in $\,cm$)

$A$ 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

Starting from the mean position,a body oscillates simple harmonically with a period of $2\,s$. After what time will its kinetic energy be $75\%$ of the total energy?

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: $.............cm$.

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

Consider the following statements. The total energy of a particle executing simple harmonic motion depends on its:
$(1)$ Amplitude $(2)$ Period $(3)$ Displacement
Of these statements: