$A$ 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 $5 \ cm$?

An object of mass $0.5\, \text{kg}$ is executing simple harmonic motion. Its amplitude is $5\, \text{cm}$ and time period $T$ is $0.2\, \text{s}$. What will be the potential energy of the object at an instant $t = \frac{T}{4}\, \text{s}$ starting from the mean position? Assume that the initial phase of the oscillation is zero. (In $\text{J}$)

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

When the displacement of a simple harmonic oscillator is one third of its amplitude,the ratio of total energy to the kinetic energy is $\frac{x}{8}$,where $x=$ . . . . . . .

Obtain the expressions for kinetic energy,potential energy,and total energy in simple harmonic motion.

The average kinetic energy in one time period of a $SHM$ is :-