$$\eqalign{ & {\text{Given,}}\,\,x = \frac{A}{2} \cr & \therefore {\text{from}}\,x = A\sin \omega t \Rightarrow \omega t = {30^ \circ } \cr & {\text{And,}}\,\,\frac{{KE}}{{PE}} = {\cot ^2}\omega t = {\left( {\sqrt 3 } \right)^2} = 3 \cr} $$

Calculate the effective force constant of parallel spring, then by putting the values of time period $$T = 2\pi \sqrt {\frac{M}{K}} ,$$   we get the new time period of spring.
We can write time period for a vertical spring-block system as
$$T = 2\pi \sqrt {\frac{l}{g}} $$
Here, $$l$$ is extension in the spring when the mass $$m$$ is suspended from the spring.
This can be seen as under :
$$\eqalign{ & kl = mg\,\,\left( {{\text{in equilibrium position}}} \right) \cr & \Rightarrow \frac{m}{k} = \frac{l}{g} \cr & \therefore T = 2\pi \sqrt {\frac{m}{k}} \cr & \therefore {T_1} = 2\pi \sqrt {\frac{m}{{{k_1}}}} \cr & \Rightarrow {k_1} = 4{\pi ^2}\frac{m}{{T_1^2}}\,......\left( {\text{i}} \right) \cr & {T_2} = 2\pi \sqrt {\frac{m}{{{k_2}}}} \cr & \Rightarrow {k_2} = 4{\pi ^2}\frac{m}{{T_2^2}}\,......\left( {{\text{ii}}} \right) \cr} $$
Since, springs are in parallel, effective force constant
$$\eqalign{ & k = {k_1} + {k_2} \cr & \therefore T = 2\pi \sqrt {\frac{m}{{{k_1} + {k_2}}}} \cr & \Rightarrow {k_1} + {k_2} = 4{\pi ^2}\frac{m}{{{T^2}}}\,......\left( {{\text{iii}}} \right) \cr} $$
Substituting values of $${k_1}$$ and $${k_2}$$ from Eqs. (i) and (ii) in Eq. (iii), we get
$$\eqalign{ & 4{\pi ^2}\frac{m}{{T_1^2}} + 4{\pi ^2}\frac{m}{{T_2^2}} = 4{\pi ^2}\frac{m}{{{T^2}}} \cr & \Rightarrow \frac{1}{{{T^2}}} = \frac{1}{{T_1^2}} + \frac{1}{{T_2^2}} \cr & {\text{or}}\,\,{T^{ - 2}} = T_1^{ - 2} + T_2^{ - 2} \cr} $$

$$\eqalign{ & 2k\,{\sin ^2}\theta = {k_1}\,\,{\text{or}}\,\,{k_1} = 2k\,{\sin ^2}\theta \cr & {\text{and}}\,\,{k_2} = 2\left( {2k} \right){\sin ^2}\beta \cr} $$

$$\eqalign{ & {\text{Then}}\,\,{k_{{\text{eq}}}} = {k_1} + {k_2} = 2k\left[ {{{\sin }^2}\theta + 2{{\sin }^2}\beta } \right] \cr & = 2k\left[ {{{\sin }^2}{{45}^ \circ } + 2{{\sin }^2}{{30}^ \circ }} \right] \cr & = 2k\left( {\frac{1}{2} + \frac{1}{2}} \right) = 2k \cr & {\text{Then}}\,\,T = 2\pi \sqrt {\frac{m}{{{k_{{\text{eq}}}}}}} = 2\pi \sqrt {\frac{m}{{2k}}} = \pi \sqrt {\frac{{2m}}{k}} \cr} $$

Velocity of the particle executing $$SHM$$  at any instant is defined as the time rate of change of its displacement at that instant.
Let the displacement of the particle at an instant $$t$$ is given by
$$\eqalign{ & x = a\sin \omega t \cr & \therefore {\text{velocity}}\,v = \frac{{dx}}{{dt}} = \frac{{d\left( {a\sin \omega t} \right)}}{{dt}} \cr & = a\omega \cos \omega t = a\omega \sqrt {\left( {1 - {{\sin }^2}\omega t} \right)} \cr & = a\omega \sqrt {\left( {1 - \frac{{{x^2}}}{{{a^2}}}} \right)} \cr & = \omega \sqrt {\left( {{a^2} - {x^2}} \right)} \cr} $$
At mean position, $$x = 0$$
$$\therefore {v_{\max }} = \omega a$$
According to problem, $$v = \frac{{{v_{\max }}}}{2} = \frac{{a\omega }}{2}$$
$$\eqalign{ & {\text{But}}\,v = \omega \sqrt {{a^2} - {x^2}} \cr & \therefore \frac{{a\omega }}{2} = \omega \sqrt {{a^2} - {x^2}} \,\,{\text{or}}\,\,x = \frac{{\sqrt 3 }}{2}a \cr} $$

$$\eqalign{ & T = 2\pi \sqrt {\frac{M}{k}} \cr & T' = 2\pi \sqrt {\frac{{M + m}}{k}} = \frac{{5T}}{3} \cr & \therefore 2\pi \sqrt {\frac{{M + m}}{k}} = \frac{5}{3} \times 2\pi \sqrt {\frac{M}{k}} \Rightarrow \frac{m}{M} = \frac{{16}}{9} \cr} $$

$$\eqalign{ & \therefore \frac{{\frac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right)}}{{\frac{1}{2}m{\omega ^2}{x^2}}} = \frac{1}{4} \Rightarrow \frac{{{A^2} - {x^2}}}{{{x^2}}} = \frac{1}{4} \cr & 4{A^2} - 4{x^2} = {x^2} \Rightarrow {x^2} = \frac{4}{5}{A^2} \Rightarrow x = \frac{2}{{\sqrt 5 }}A. \cr} $$

$$K.E. = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)$$
When $$x = 0,\,K.E.$$   is maximum and is equal to $$\frac{1}{2}m{\omega ^2}{a^2}.$$

Two simple harmonic motions can be written as
$$\eqalign{ & x = a\sin \left( {\omega t + \delta } \right)\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\,y = a\sin \left( {\omega t + \delta + \frac{\pi }{2}} \right) \cr & {\text{or}}\,\,y = a\cos \left( {\omega t + \delta } \right)\,......\left( {{\text{ii}}} \right) \cr} $$
Squaring and adding Eqs. (i) and (ii), we obtain
$$\eqalign{ & {x^2} + {y^2} = {a^2}\left[ {{{\sin }^2}\left( {\omega t + \delta } \right) + {{\cos }^2}\left( {\omega t + \delta } \right)} \right] \cr & {\text{or}}\,\,{x^2} + {y^2} = {a^2}\,\,\left( {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right) \cr} $$
This is the equation of a circle.
$$\eqalign{ & {\text{At}}\,\,\left( {\omega t + \delta } \right) = 0;x = 0,y = a \cr & {\text{At}}\,\,\left( {\omega t + \delta } \right) = \frac{\pi }{2};x = a,y = 0 \cr & {\text{At}}\,\,\left( {\omega t + \delta } \right) = \pi ;x = 0,y = - a \cr & {\text{At}}\,\,\left( {\omega t + \delta } \right) = \frac{{3\pi }}{2};x = - a,y = 0 \cr & {\text{At}}\,\,\left( {\omega t + \delta } \right) = 2\pi ;x = 0,y = a \cr} $$

Thus, it is obvious that motion of particle is traversed in clockwise direction.

Fora particle executing SHM
At mean position; $$t = 0,\omega t = 0,y = 0,V = {V_{\max }} = a\omega $$
$$\therefore K.E. = K{E_{\max }} = \frac{1}{2}m{\omega ^2}{a^2}$$
At extreme position : $$t = \frac{T}{4},\omega t = \frac{\pi }{2},y = A,V = {V_{\min }} = 0$$
$$\therefore K.E. = K{E_{\min }} = 0$$
Kinetic energy in SHM, $$KE = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {y^2}} \right)$$
$$ = \frac{1}{2}m{\omega ^2}{a^2}{\cos ^2}\omega t$$
Hence graph (B) correctly depicts kinetic energy time graph.

The velocity of a body executing S.H.M. is maximum at its centre and decreases as the body proceeds to the extremes. Therefore if the time taken for the body to go from $$O$$ to $$\frac{A}{2}$$ is $${T_1}$$ and to go from $$\frac{A}{2}$$ to $$A$$ is $${T_2}$$ then obviously $${T_1} < {T_2}$$