$A$ particle starts oscillating simple harmonically from its equilibrium position. The ratio of kinetic energy and potential energy of the particle at time $t = T/12$ is: ($T =$ time period)

The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\, J$ and its amplitude is $0.01\, m,$ its time period would be

$A$ particle of mass $10 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$ and a time period of $2 \ s$. What is its kinetic energy at a displacement of $5 \ cm$?

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

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $4 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)?

$A$ body performs $S.H.M.$ Its kinetic energy $K$ varies with time $t$ as indicated by which graph?