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 of mass $m$ executes simple harmonic motion with amplitude $a$ and frequency $v$. The average kinetic energy during its motion from the position of equilibrium to the end is

$A$ linear harmonic oscillator has a total mechanical energy of $300 \ J$. If its potential energy at the mean position is $100 \ J$,find its kinetic energy at $x = +\frac{A}{\sqrt{2}}$. (in $J$)

$A$ block kept on a frictionless horizontal surface is connected to one end of a horizontal spring of constant $100 \text{ Nm}^{-1}$ whose other end is fixed to a rigid vertical wall. Initially, the block is at its equilibrium position. The block is pulled to a distance of $8 \text{ cm}$ and released. The kinetic energy of the block when it is at a distance of $3 \text{ cm}$ from the mean position is (in $\text{ J}$)

$A$ particle is executing simple harmonic motion with a time period $T$. At time $t = 0$,it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like:

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