$A$ spring-mass system performs $S.H.M.$ If the mass is doubled keeping the amplitude same,then the total energy of $S.H.M.$ will become:

For a particle executing $S.H.M.$,its potential energy is $8$ times its kinetic energy at a certain displacement $x$ from the mean position. If $A$ is the amplitude of $S.H.M.$,the value of $x$ is:

Draw a graph to show the variation of $PE$,$KE$ and total energy of a simple harmonic oscillator with displacement.

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation,the displacement of the particle from the equilibrium position in terms of its amplitude $a$ is

If a particle repeats its motion after a fixed time interval of $8 \,s$,then after how much time will its maximum value of $PE$ be attained after attaining its minimum value?

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