$$t = 2\pi \sqrt {\frac{\ell }{{{g_{eff}}}}} ;\,{t_0} = 2\pi \sqrt {\frac{\ell }{g}} $$

$$\eqalign{ & m{g_{eff}} = mg - B = my - V \times 100 \times g \cr & \therefore {g_{eff}} = g - \frac{{100}}{{\left( {\frac{m}{v}} \right)}}g = g - \frac{{1000}}{{\frac{4}{3} \times 1000}}g = \frac{g}{4} \cr & \therefore t = 2\pi \sqrt {\frac{\ell }{{\frac{g}{4}}}} \cr & \,t = 2{t_0} \cr} $$

The maximum of amplitude and energy is obtained when the frequency is equal to the natural frequency (resonance condition)
$$\therefore {\omega _1} = {\omega _2}$$

Potential energy of the body executing $$SHM$$  is given by
$$U = \frac{1}{2}m{\omega ^2}{x^2}$$
where symbols have their usual meaning. Total energy of the body executing $$SHM$$  is
$$E = \frac{1}{2}m{\omega ^2}{a^2}$$
According to problem,
$$\eqalign{ & U = \frac{1}{4}E \cr & \therefore \frac{1}{2}m{\omega ^2}{x^2} = \frac{1}{4} \times \frac{1}{2}m{\omega ^2}{a^2} \cr & {\text{or}}\,\,{x^2} = \frac{{{a^2}}}{4}\,\,{\text{or}}\,\,x = \frac{a}{2} \cr} $$

$$\eqalign{ & \frac{{{d^2}x}}{{d{t^2}}} = - \alpha x = - {\omega ^2}x \cr & \Rightarrow \omega = \sqrt \alpha \,{\text{or}}\,T = \frac{{2\pi }}{\omega } = \frac{{2\pi }}{{\sqrt \alpha }} \cr} $$

Suppose the liquid in left side limb is displaced slightly by $$y,$$ the liquid in right limb will increase by $$\frac{y}{2}.$$ The restoring force
$$\eqalign{ & F = - PA = - \rho g\left( {\frac{{3y}}{2}} \right) \times 2A = 3\rho gA\left( { - y} \right). \cr & a = \frac{F}{m} = \frac{{3\rho gA\left( { - y} \right)}}{m} \cr} $$

On comparing with, $$a = - {\omega ^2}y,$$   we get
$$\omega = \sqrt {\frac{{3\rho gA}}{m}} \,\,{\text{and}}\,\,T = 2\pi \sqrt {\frac{m}{{3\rho gA}}} $$

$$\eqalign{ & {\text{Given,}} \cr & {\text{Damping force}} \propto {\text{velocity}} \cr & F \propto v \cr & F = kv \cr & \Rightarrow k = \frac{F}{v} \cr & {\text{Unit}}\,{\text{of}}\,k = \frac{{{\text{unit}}\,{\text{of}}\,F}}{{{\text{unit}}\,{\text{of}}\,v}} = \frac{{kg{\text{ - }}m{s^{ - 2}}}}{{m{s^{ - 1}}}} = kg\,{s^{ - 1}} \cr} $$

$$\eqalign{ & T = 2\pi \sqrt {\frac{l}{g}} \cr & V\rho \,{g_{eff}} = V\rho g - \frac{{V\rho }}{{16}}g \cr & {g_{eff}} = \left( {g - \frac{g}{{16}}} \right) = \frac{{15g}}{{16}} \cr & {\text{Now}}\,T' = 2\pi \sqrt {\frac{l}{{{g_{eff}}}}} = 2\pi \sqrt {\frac{l}{{\frac{{15g}}{{16}}}}} = \frac{4}{{\sqrt {15} }}T \cr} $$

In simple harmonic motion, starting from rest,
At $$t = 0,x = A$$
$$x = A\cos \omega t\,......\left( {\text{i}} \right)$$
When $$t = \tau ,x = A - a$$
When $$t = 2\tau ,x = A - 3a$$
From equation (i)
$$\eqalign{ & A - a = A\cos \omega \tau \,......\left( {{\text{ii}}} \right) \cr & A - 3a = A\cos 2\omega \tau \,......\left( {{\text{iii}}} \right) \cr & \cos 2\omega \tau = 2{\cos ^2}\omega \tau - 1......\left( {{\text{iv}}} \right) \cr} $$
From equation (ii), (iii) and (iv)
$$\eqalign{ & \frac{{A - 3a}}{A} = 2{\left( {\frac{{A - a}}{A}} \right)^2} - 1 \cr & \Rightarrow \frac{{A - 3a}}{A} = \frac{{2{A^2} + 2{a^2} - 4Aa - {A^2}}}{{{A^2}}} \cr & \Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa \cr & \Rightarrow 2{a^2} = aA \Rightarrow A = 2a \cr & \Rightarrow \frac{a}{A} = \frac{1}{2} \cr & {\text{Now,}}\,A - a = A\cos \omega \tau \cr & \Rightarrow \cos \omega \tau = \frac{{A - a}}{A} \Rightarrow \cos \omega \tau = \frac{1}{2} \cr & {\text{or,}}\,\frac{{2\pi }}{T}\tau = \frac{\pi }{3} \Rightarrow T = 6\tau \cr} $$

$$K.E = \frac{1}{2}k\left( {{A^2} - {x^2}} \right);\,U = \frac{1}{2}k{x^2}$$
At the mean position $$x = 0$$
$$\therefore K.E = \frac{1}{2}k{A^2} = $$    Maximum and $$U = 0$$

$$\eqalign{ & T = 2\pi \sqrt {\frac{\mu }{{2k}}} \,\,\because \mu = \frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}} \cr & T = 2\pi \sqrt {\frac{{10}}{{7 \times 200}}} = \frac{{2 \times 5}}{7} = \frac{{10}}{7} = 2\pi \sqrt {\frac{1}{{140}}} \cr} $$