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 $\,cm$)

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 $N$)

$A$ particle of mass $m$ is executing simple harmonic motion about its mean position. If $A$ is the amplitude and $T$ is the period of $S$.$H$.$M$.,then the total energy of the particle is

$A$ body starting at $t=0$ from the origin oscillates simple harmonically with a period of $4 \ s$. After what time will its kinetic energy be $75 \%$ of its total energy?

The total energy of a simple harmonic oscillator is $E$. What is its kinetic energy at half the amplitude?

$A$ particle starts its oscillation from the equilibrium position with time period $T$. Find the ratio of kinetic energy to potential energy of the particle at time $t = \frac{T}{6}$.