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 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:

$A$ particle of mass $m$ oscillates with simple harmonic motion between points $x_1$ and $x_2$,the equilibrium position being $O$. Its potential energy is plotted. It will be as given below in the graph:

Vibrational motion possesses which type of energy?

$A$ 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 kinetic energy of a particle executing simple harmonic motion at a displacement of $3 \ cm$ from the mean position is $4 \ mJ$. If the amplitude of the particle is $5 \ cm$,then the maximum force acting on the particle is (in $N$)