What is the displacement of a body in $SHM$ when the potential energy becomes three times its kinetic energy?

The total energy of a particle executing $S.H.M.$ is $80 \, J$. What is the potential energy when the particle is at a distance of $\frac{3}{4}$ of amplitude from the mean position?

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 $N$)

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $\omega_1$ and $\omega_2$ and have total energies $E_1$ and $E_2$,respectively. The variations of their momenta $p$ with positions $x$ are shown in the figures. If $\frac{a}{b}= n^2$ and $\frac{a}{R}= n$,then the correct equation$(s)$ is(are):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

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

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 $\,cm$)