$$\eqalign{ & {\text{Given,}}\,\,V = 36 \times \frac{5}{{18}}\,m/s = 10\,m/s, \cr & g = 10\,m/{s^2},r = 10\,m,l = 1\,m \cr & T = 2\pi \sqrt {\frac{\ell }{{{g^2} + {{\left( {\frac{{{v^2}}}{r}} \right)}^2}}}} = 2\pi \sqrt {\frac{1}{{{{10}^2} + {{\left( {\frac{{100}}{{10}}} \right)}^2}}}} \cr & = \frac{\pi }{{10}}\sqrt 2 \,\sec \cr} $$

Compare the time period of two different oscillation.
Time period of simple pendulum
$$T = 2\pi \sqrt {\left( {\frac{l}{g}} \right)} \propto \sqrt l $$
where, $$l$$ is effective length.
(i.e. distance between centre of suspension and centre of gravity of bob)

Initially, centre of gravity is at the centre of sphere [Fig. (a)]. When water leaks the centre of gravity goes down until it is half-filled [Fig. (b)], then it begins to go up and finally it again goes at the centre [Fig. (c)]. That is effective length first increases and then decreases. As $$T \propto \sqrt l ,$$   so time period first increases and then decreases.

Velocity in $$SHM$$  is given by
$$\eqalign{ & v = \omega \sqrt {{a^2} - {y^2}} \cr & {\text{At}}\,\,y = 4\,cm = 0.04\,m,v = 3\,m/s \cr & \therefore 3 = \omega \sqrt {{a^2} - {{\left( {0.04} \right)}^2}} \,......\left( 1 \right) \cr & {\text{At}}\,\,y = 3\,cm = 0.03\,m,v = 4\,m/s \cr & \therefore 4 = \omega \sqrt {{a^2} - {{\left( {0.03} \right)}^2}} \,......\left( 2 \right) \cr} $$
Dividing (2) by (1),
we get $$a = 0.05 = 5\,cm$$

$$\eqalign{ & {v^2} = {\omega ^2}\left( {{A^2} - {x^2}} \right)\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\,{a^2} = \left( {{\omega ^2}{x^2}} \right) = {\omega ^4}{x^2}\,......\left( {{\text{ii}}} \right) \cr} $$
From above equations, we have
$${v^2} = - \frac{{{a^2}}}{{{\omega ^2}}} + {\omega ^2}{A^2} \Rightarrow y = mx + c$$
It represents straight line with negative slope.

Given, $${y_1} = a\sin \omega t$$
$${y_2} = b\cos \omega t = b\sin \left( {\omega t + \frac{\pi }{2}} \right)$$
The resultant displacement is given by
$$y = {y_1} + {y_2} = \sqrt {{a^2} + {b^2}} \sin \left( {\omega t + \phi } \right)$$
Hence, the motion of superimposed wave is simple harmonic with amplitude $$\sqrt {{a^2} + {b^2}} .$$

Let block be displaced through $$x$$ $$m,$$ then weight of displaced water or upthrust, (upwards) is given by Archimedes principle
$${F_b} = - Ax\rho g$$
where, $$A$$ is the area of cross-section of the block and $$\rho $$ is its density. This must be equal to force $$\left( { = ma} \right)$$  applied, where, $$m$$ is the mass of the block and $$a$$ is the acceleration.
$$\therefore ma = - Ax\rho g\,\,{\text{or}}\,\,a = - \frac{{A\rho g}}{m}x = - {\omega ^2}x$$
This is the equation of simple harmonic motion. Time period of oscillation,
$$\eqalign{ & T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{m}{{A\rho g}}} \cr & \Rightarrow T \propto \frac{1}{{\sqrt A }} \cr} $$

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

The two springs are in parallel.
$$\eqalign{ & f = \frac{1}{{2\pi }}\sqrt {\frac{{{K_1} + {K_2}}}{m}} \,......({\text{i}}) \cr & f' = \frac{1}{{2\pi }}\sqrt {\frac{{4{K_1} + 4{K_2}}}{m}} \cr & = \frac{1}{{2\pi }}\sqrt {\frac{{4\left( {{K_1} + 4{K_2}} \right)}}{m}} = 2\left( {\frac{1}{{2\pi }}\sqrt {\frac{{{K_1} + {K_2}}}{m}} } \right) \cr & = 2f\,\,\,{\text{from}}\,{\text{eqn}}{\text{.(i)}} \cr} $$

For a particle executing $$SHM$$
$$\eqalign{ & {\text{Acceleration}}\left( a \right) \propto - {\omega ^2}{\text{displacement}}\left( x \right)\,......\left( {\text{i}} \right) \cr & {\text{Given}}\,x = a{\sin ^2}\omega t\,......\left( {{\text{ii}}} \right) \cr} $$
Differentiating the above equation w.r.t, we get
$$\frac{{dx}}{{dt}} = 2a\omega \left( {\sin \omega t} \right)\left( {\cos \omega t} \right)$$
Again differentiating, we get
$$\eqalign{ & \frac{{{d^2}x}}{{d{t^2}}} = a = 2a{\omega ^2}\left[ {{{\cos }^2}\omega t - {{\sin }^2}\omega t} \right] \cr & = 2a{\omega ^2}\cos 2\omega t \cr} $$
The given equation does not satisfy the condition for $$SHM$$  (Eq. (i)]. Therefore, motion is not simple harmonic.

For the block is about to slip, $$\mu g = {\omega ^2}a$$
$$\omega = \sqrt {\frac{{\mu g}}{a}} \Rightarrow v = \frac{1}{{2\pi }}\sqrt {\frac{{\mu g}}{a}} $$