$$\eqalign{ & {\text{Maximum velocity,}}\,{v_{\max }} = a\omega \cr & {v_{\max }} = a \times \frac{{2\pi }}{T} \cr & \Rightarrow T = \frac{{2\pi a}}{{{v_{\max }}}} = \frac{{2 \times 3.14 \times 7 \times {{10}^{ - 3}}}}{{4.4}} \approx 0.01\,s \cr} $$

Let us plot the graph of the mathematical equation $$U\left( x \right) = K\left[ {1 - {e^ - }{{^x}^{^2}}} \right],F = \frac{{dU}}{{dx}} = 2kx{e^{ - x}}^{^2}$$

From the graph it is clear that the potential energy is minimum at $$x = 0.$$  Therefore, $$x = 0$$  is the state of stable equilibrium. Now if we displace the particle from $$x = 0$$  then for displacements the particle tends to regain the position $$x = 0$$  with a force $$F = \frac{{2kx}}{{{e^x}^{^2}}}.$$   Therefore for small values of $$x$$ we have $$F \propto x.$$

If a particle executing simple harmonic motion, has a displacement $$x$$ from its equilibrium position, at an instant the magnitude of the restoring force $$F$$ acting on the particle at that instant is given by
$$F = - kx$$
where, $$k$$ is known as force constant.
Hence, in given options, option (A) is correct.
Here, $$k = AK$$

As we know, time period,
$$T = 2p\sqrt {\frac{\ell }{g}} $$
When additional mass $$M$$ is added then
$$\eqalign{ & {T_M} = 2p\sqrt {\frac{{\ell + D\ell }}{g}} \cr & \frac{{{T_M}}}{T} = \sqrt {\frac{{\ell + D\ell }}{\ell }} \,{\text{or}}\,{\left( {\frac{{{T_M}}}{T}} \right)^2} = \frac{{\ell + \Delta \ell }}{\ell } \cr & {\text{or,}}\,{\left( {\frac{{{T_M}}}{T}} \right)^2} = 1 + \frac{{Mg}}{{Ay}}\,\,\,\,\,\,\,\,\,\left[ {\because \Delta \ell = \frac{{Mg\ell }}{{Ay}}} \right] \cr & \therefore \frac{1}{y} = \left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{Mg}} \cr} $$

$$\eqalign{ & {\text{Force constant}}\,k = 2 \times {10^6}\,N/m \cr & {\text{Amplitude}}\left( x \right) = 0.01\,m \cr & {\text{Potential Energy}} = \frac{1}{2}k{x^2} \cr & = \frac{1}{2} \times \left( {2 \times {{10}^6}} \right) \times {\left( {0.01} \right)^2} = 100\,J \cr} $$

As we know,
$$\eqalign{ & F = ma \Rightarrow a \propto F\,{\text{or,}}\,\,a \propto \sin t \cr & \Rightarrow \frac{{dv}}{{dt}} \propto \sin t \Rightarrow \int\limits_0^0 {dV} \propto \int\limits_0^t {\sin t\,dt} \cr & V \propto - \cos \,t + 1 \cr & \int\limits_0^x {dx} = \int\limits_0^t {\left( { - \cos t + 1} \right)dt} \cr & x = \sin t - \frac{1}{2}\sin 2t \cr} $$

Here, the phase difference between waves is $$\frac{\pi }{4}.$$
So, the resultant path of particle will be ellipse.

$$\eqalign{ & y = k{t^2} \cr & \therefore \frac{{dy}}{{dt}} = 2kt\,{\text{or}}\,\frac{{{d^2}y}}{{d{t^2}}} = 2k \cr & {\text{or}}\,{a_g} = 2m/{s^2}\,\left( {\because k = 1m/{s^2}\,{\text{given}}} \right) \cr & {\text{We know that }}T = 2\pi \sqrt {\frac{\ell }{g}} \cr & \therefore \frac{{T_1^2}}{{T_2^2}} = \frac{{{g_2}}}{{{g_1}}} \Rightarrow \frac{{T_1^2}}{{T_2^2}} = \frac{{12}}{{10}} = \frac{6}{5} \cr & \left[ {\because {g_1} = 10m/{s^2}{\text{ and }}{g_2} = g + 2 = 12m/{s^2}} \right] \cr} $$

KEY CONCEPT: The time period $$T = 2\pi \sqrt {\frac{\ell }{g}} $$    where $$\ell = $$ distance between the point of suspension and the centre of mass of the child. This distance decreases when the child stands
$$\therefore T' < T$$   i.e., the period decreases.

The two simple harmonic motions can be written as
$$\eqalign{ & x = a\sin \omega t\,......\left( {\text{i}} \right) \cr & {\text{and}}\,y = a\sin \left( {\omega t + \frac{\pi }{2}} \right) \cr & y = a\cos \omega t\,......\left( {{\text{ii}}} \right) \cr} $$
On squaring and adding Eqs. (i) and (ii), we obtain
$${x^2} + {y^2} = {a^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right)\,\,{\text{or}}\,\,{x^2} + {y^2} = {a^2}$$
This is the equation of a circular motion with radius $$a.$$
NOTE
Simple harmonic motion is of two types.
1. Linear simple harmonic motion
2. Angular simple harmonic motion