The total energy of a particle executing $SHM$ at an amplitude of $4 \, cm$ is $20 \, J$. What will be its total energy at a displacement of $x = 2 \, cm$ (in $J$)?

The kinetic energy of a particle executing simple harmonic motion in a straight line is $pv^2$ and the potential energy is $qx^2$,where $v$ is the speed at a distance $x$ from the mean position. Its time period is given by the expression:

$A$ 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 $t = \frac{T}{6}$.

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 $N$)

Starting from the mean position,a body oscillates simple harmonically with a period $T$. After what time will its kinetic energy be $75 \%$ of the total energy? $(\sin 30^{\circ} = 0.5)$

$A$ linear harmonic oscillator of force constant $2 \times 10^6 \, N/m$ and amplitude $0.01 \, m$ has a total mechanical energy of $160 \, J$. Its