$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $6 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$

For a particle executing simple harmonic motion,the kinetic energy $K$ is given by $K = K_0 \cos^2 \omega t$. The maximum value of potential energy is

The total energy of a body executing $S.H.M.$ is $E$. Then the kinetic energy when the displacement is half of the amplitude,is

Explain and draw the graphs of kinetic energy,potential energy,and mechanical energy versus displacement for $SHM$.

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

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):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$