For an SHM, the acceleration $$a = - {\omega ^2}x$$   where $${\omega ^2}$$ is a, constant. Therefore $$\frac{a}{x}$$ is a constant. The time period $$T$$ is also constant. Therefore $$\frac{{aT}}{x}$$ is a constant.

Time period of simple pendulum
$$T = 2\pi \sqrt {\frac{l}{g}} $$
where,
$$l = $$  length of pendulum
$$g =$$  acceleration due to gravity
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,T \propto \sqrt l \cr & {\text{Hence,}}\,\,\frac{{{T_2}}}{{{T_1}}} = \sqrt {\frac{{{l_2}}}{{{l_1}}}} \,......\left( {\text{i}} \right) \cr & {\text{Given,}}\,\,{l_2} = 4{l_1},{T_1} = 2\,s \cr} $$
Substituting the values in Eq. (i), we get
$${T_2} = \sqrt {\frac{{4{l_1}}}{{{l_1}}}} \times 2 = 2 \times 2 = 4\,s$$

Figure shows the rod at an angle $$\theta $$ with respect to its equilibrium position. Both the springs are stretched by length $$\frac{{\ell \theta }}{2}.$$
The restoring torque due to the springs $$\tau = - 2{\text{(Restoring force)}} \times {\text{perpendicular distance}}$$
$$\tau = - 2k\left( {\frac{{\ell \theta }}{2}} \right) \times \frac{\ell }{2} = - k\frac{{{\ell ^2}}}{2}\theta \,......\left( {\text{i}} \right)$$
If $$I$$ is the moment of inertia of the rod about $$M$$ then
$$\tau = I\alpha = I\frac{{{d^2}\theta }}{{d{t^2}}}\,......\left( {{\text{ii}}} \right)$$
From (i) & (ii) we get
$$\eqalign{ & I\frac{{{d^2}\theta }}{{d{t^2}}} = - k\frac{{{\ell ^2}\theta }}{2} \Rightarrow \frac{{{d^2}\theta }}{{d{t^2}}} = - \frac{k}{I}\frac{{{\ell ^2}}}{2}\theta = \frac{{ - k}}{{\frac{{M{\ell ^2}}}{{12}}}}\frac{{{\ell ^2}}}{2}\theta \cr & \Rightarrow \frac{{{d^2}\theta }}{{d{t^2}}} = - \frac{{6k}}{M}\theta \cr} $$
Comparing it with the standard equation of rotational SHM we get
$$\eqalign{ & \frac{{{d^2}\theta }}{{d{t^2}}} = - {\omega ^2}\theta \Rightarrow {\omega ^2} = \frac{{6k}}{M} \Rightarrow \omega = \sqrt {\frac{{6k}}{M}} \cr & \Rightarrow 2\pi v = \sqrt {\frac{{6k}}{M}} \Rightarrow v = \frac{1}{{2\pi }}\sqrt {\frac{{6k}}{M}} \cr} $$

$$\eqalign{ & \frac{{Mg}}{A} = {P_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{P_0}V_0^\gamma = P{V^\gamma } \cr & Mg = {P_0}A\,......\left( 1 \right)\,\,\,\,\,\,\,\,\,\,\,{P_0}Ax_0^\gamma = PA{\left( {{x_0} - x} \right)^\gamma } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P = \frac{{{P_0}x_0^\gamma }}{{{{\left( {{x_0} - x} \right)}^\gamma }}} \cr} $$
Let piston is displaced by distance $$x$$

$$\eqalign{ & Mg - \left( {\frac{{{P_0}x_0^\gamma }}{{{{\left( {{x_0} - x} \right)}^\gamma }}}} \right)A = {F_{{\text{restoring }}}} \cr & {P_0}A\left( {1 - \frac{{x_0^\gamma }}{{{{\left( {{x_0} - x} \right)}^\gamma }}}} \right) = {F_{{\text{restoring }}}}\,\left[ {{x_0} - x \approx {x_0}} \right] \cr & F = - \frac{{\gamma {P_0}Ax}}{{{x_0}}} \cr} $$
$$\therefore $$ Frequency with which piston executes SHM.
$$f = \frac{1}{{2\pi }}\sqrt {\frac{{\gamma {P_0}A}}{{{x_0}M}}} = \frac{1}{{2\pi }}\sqrt {\frac{{\gamma {P_0}{A^2}}}{{M{V_0}}}} $$

For first spring, $${t_1} = 2\pi \sqrt {\frac{m}{{{k_1}}}} ,$$
For second spring, $${t_2} = 2\pi \sqrt {\frac{m}{{{k_2}}}} $$
when springs are in series then, $${k_{eff}} = \frac{{{k_1}{k_2}}}{{{k_l} + {k_2}}}$$
$$\eqalign{ & \therefore T = 2\pi \sqrt {\frac{{m\left( {{k_l} + {k_2}} \right)}}{{{k_1}{k_2}}}} \cr & \therefore T = 2\pi \sqrt {\frac{m}{{{k_2}}} + \frac{m}{{{k_1}}}} = 2\pi \sqrt {\frac{{t_2^2}}{{{{(2\pi )}^2}}} + \frac{{t_1^2}}{{{{(2\pi )}^2}}}} \cr & \Rightarrow {T^2} = t_1^2 + t_2^2 \cr} $$
where $$x$$ is the displacement from the mean position

$$\eqalign{ & {f_A} = \frac{1}{{2\pi }}\sqrt {\frac{g}{{{L_A}}}} \,\,{\text{and}}\,\,{f_B} = \frac{{{f_A}}}{2} = \frac{1}{{2\pi }}\sqrt {\frac{g}{{{L_B}}}} \cr & \therefore \frac{{{f_A}}}{{\frac{{{f_A}}}{2}}} = \frac{1}{{2\pi }}\sqrt {\frac{g}{{{L_A}}}} \times 2\pi \sqrt {\frac{{{L_B}}}{g}} \Rightarrow 2 = \sqrt {\frac{{{L_B}}}{{{L_A}}}} \cr & \Rightarrow 4 = \frac{{{L_B}}}{{{L_A}}},\,{\text{regardless of mass}}{\text{.}} \cr} $$

Maximum speed of a particle executing $$SHM$$  is given by,
$$\eqalign{ & {v_{\max }} = a\omega = a\left( {2\,\pi n} \right) \cr & \Rightarrow n = \frac{{{v_{\max }}}}{{2\,\pi a}} \cr} $$
where,
$$a =$$  amplitude of oscillation
$$n =$$  frequency of oscillation
$${\text{Here,}}\,{v_{\max }} = 3.14\,cm/s,\,a = 5\,cm$$
Substituting, the given values, we have
$$n = \frac{{31.4}}{{2 \times 3.14 \times 5}} = 1\,Hz$$

The kinetic energy of a particle executing $$S.H.M.$$  is given by
$$\eqalign{ & K = \frac{1}{2}m{a^2}{\omega ^2}{\sin ^2}\omega t \cr & = < \frac{1}{2}m{\omega ^2}{a^2}{\sin ^2}\omega t > = \frac{1}{2}m{\omega ^2}{a^2} < {\sin ^2}\omega t > \cr & = \frac{1}{2}m{\omega ^2}{a^2}\left( {\frac{1}{2}} \right) \cr & = \frac{1}{4}m{\omega ^2}{a^2}\,\,\left( {\because < {{\sin }^2}\theta > = \frac{1}{2}} \right) \cr & = \frac{1}{4}m{a^2}\left( {2\pi {v^2}} \right)\,\,\left( {\because \omega = 2\pi v} \right) \cr & {\text{or,}}\,\, < K > = {\pi ^2}m{a^2}{v^2} \cr} $$

KEY CONCEPT: The instantaneous kinetic energy of a particle executing S.H.M. is given by
$$\eqalign{ & K = \frac{1}{2}m{a^2}{\omega ^2}{\sin ^2}\omega t \cr & \therefore {\text{average}}\,K.E. = < K > = < \frac{1}{2}m{\omega ^2}{a^2}{\sin ^2}\omega t > \cr & = \frac{1}{2}m{\omega ^2}{a^2} < {\sin ^2}\omega t > \cr & = \frac{1}{2}m{\omega ^2}{a^2}\left( {\frac{1}{2}} \right)\,\left( {\because < {{\sin }^2}\theta > = \frac{1}{2}} \right) \cr & = \frac{1}{4}m{\omega ^2}{a^2} = \frac{1}{4}m{a^2}{\left( {2\pi \nu } \right)^2}\left( {\because \omega = 2\pi \nu } \right) \cr & {\text{or,}}\, < K > = {\pi ^2}m{a^2}{\nu ^2} \cr} $$

For $${X_0} + A$$  to be the maximum seperation $$y$$ one body is at the mean position, the other should be at the extreme.