$$KE$$  and $$PE$$  completes two vibration in a time during which $$SHM$$  completes one vibration. Thus frequency of $$PE$$  or $$KE$$  is double than that of $$SHM.$$

Potential energy of a simple harmonic oscillator
$$U = \frac{1}{2}m{\omega ^2}{x^2}$$
Kinetic energy of a simple harmonic oscillator
$$K = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)$$
Here,
$$x =$$  Displacement from mean position
$$a =$$  Maximum displacement
(or amplitude) from mean position
Total energy is
$$\eqalign{ & E = U + K = \frac{1}{2}m{\omega ^2}{x^2} + \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right) \cr & = \frac{1}{2}m{\omega ^2}{a^2} \cr} $$
When the particle is half way to its end point i.e. at half of its amplitude, then $$x = \frac{a}{2}$$
Hence, potential energy
$$\eqalign{ & U = \frac{1}{2}m{\omega ^2}{\left( {\frac{a}{2}} \right)^2} = \frac{1}{4}\left( {\frac{1}{2}m{\omega ^2}{a^2}} \right) \cr & \Rightarrow U = \frac{E}{4} \cr} $$

$$\eqalign{ & {\text{For}}\,\,x = \left( { - A} \right),{\text{we}}\,{\text{have}} \cr & - A = A\sin \left( {\omega \times 0 + {\phi _0}} \right)\,\,{\text{or}}\,\,{\phi _0} = - \frac{\pi }{2}. \cr & {\text{So}}\,{\text{for}}\,\,x < \left( { - A} \right),{\phi _0} < \left( { - \frac{\pi }{2}} \right). \cr} $$

$$\eqalign{ & {v_1} = \frac{{d{y_1}}}{{dt}} = 0.1 \times 100\pi \cos \left( {100\pi t + \frac{\pi }{3}} \right) \cr & {v_2} = \frac{{d{y_2}}}{{dt}} = - 0.1\pi \sin \pi t \cr & = 0.1\pi \cos \left( {\pi t + \frac{\pi }{2}} \right) \cr & \therefore {\text{Phase diff}}{\text{.}} = {\phi _1} - {\phi _2} = \frac{\pi }{3} - \frac{\pi }{2} = \frac{{2\pi - 3\pi }}{6} = \frac{{ - \pi }}{6} \cr} $$

$$\eqalign{ & F = - \frac{{dU}}{{dx}} = - 20\,x \cr & a = \frac{F}{M} = \frac{{ - 20\,x}}{1} = - 20\,x \cr & - 20\,x = - {\omega ^2}\left( x \right) \cr & \therefore \omega = \sqrt {20} = 2\sqrt 5 \cr & f = \frac{\omega }{{2\pi }} = \frac{{2\sqrt 5 }}{{2\pi }} = \left( {\frac{{\sqrt 5 }}{\pi }} \right) \cr} $$

In $$SHM,$$  the total energy = potential energy + kinetic energy
$$\eqalign{ & {\text{or}}\,\,E = U + K = \frac{1}{2}m{\omega ^2}{x^2} + \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right) \cr & = \frac{1}{2}m{\omega ^2}{a^2} \cr & = \frac{1}{2}k{a^2} \cr} $$
where, $$k =$$  force constant $$ = m{\omega ^2}$$
Thus, total energy depends on $$k$$ and $$a.$$

Let displacement equation of particle executing $$SHM$$  is
$$x = a\sin \omega t$$
As particle travels half of the amplitude from the equilibrium position, so
$$x = \frac{a}{2}$$
$$\eqalign{ & {\text{Therefore,}}\,\,\frac{a}{2} = a\sin \omega t \cr & {\text{or}}\,\,\sin \omega t = \frac{1}{2} = \sin \frac{\pi }{6} \cr & {\text{or}}\,\,\omega t = \frac{\pi }{6}\,\,{\text{or}}\,\,t = \frac{\pi }{{6\omega }} \cr & {\text{or}}\,\,t = \frac{\pi }{{6\left( {\frac{{2\pi }}{T}} \right)}}\,\,\left( {{\text{as,}}\,\,\omega = \frac{{2\pi }}{T}} \right) \cr & {\text{or}}\,\,t = \frac{T}{{12}} \cr} $$
Hence, the particle travels half of the amplitude from the equilibrium in $$\frac{T}{{12}}s.$$

Maximum acceleration of body executing $$SHM$$  is given by
$${\alpha _{\max }} = {\omega ^2}a$$
So, for two different cases,
$$\eqalign{ & \frac{{{\alpha _{\max }}_{_1}}}{{{\alpha _{\max }}_{_2}}} = \frac{{\omega _1^2}}{{\omega _2^2}}\,\,\left( {\because a\,{\text{is}}\,{\text{same}}} \right) \cr & = \frac{{{{\left( {100} \right)}^2}}}{{{{\left( {1000} \right)}^2}}} = \frac{1}{{{{10}^2}}} \cr} $$

Here, $$x = 2 \times {10^{ - 2}}\cos \pi t$$
$$\therefore v = \frac{{dx}}{{dt}} = 2 \times {10^{ - 2}}\pi \sin \pi t$$
For the first time, the speed to be maximum,
$$\eqalign{ & \sin \pi t = 1\,{\text{or,}}\,\sin \pi t = \sin \frac{\pi }{2} \cr & \Rightarrow \pi t = \frac{\pi }{2}\,{\text{or}},\,t = \frac{1}{2} = 0.5\sec . \cr} $$

NOTE : In S.H.M., at extreme position, $$P.E.$$  is maximum when $$t = 0,x = A. $$   i.e., at time $$t = 0,$$  the particle executing S.H.M. is at its extreme position. Therefore $$P.E.$$  is max. The graph I and III represent the above characteristics.