The potential energy of a simple harmonic oscillator of mass $2 \,kg$ at its mean position is $5 \,J$. If its total energy is $9 \,J$ and amplitude is $1 \,cm$, then its time period is

The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is

$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

An object of mass $0.2 \,kg$ executes simple harmonic motion along the $x$-axis with a frequency of $(\frac{25}{\pi}) \,Hz$. At the position $x=0.04 \,m$, the object has a kinetic energy of $0.5 \,J$ and a potential energy of $0.4 \,J$. The amplitude of oscillation is ............ $cm$.

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

If the frequency of a simple harmonic motion is $f$,what is the frequency of its kinetic energy?