$$F = - bV,b$$   depends on all the three i.e, shape and size of he block and viscosity of the medium.

At maximum energy of the particle, velocity resonance takes place, which occurs when frequency of external periodic force is equal to natural frequency of undamped vibrations, i.e. $${\omega _2} = {\omega _0}.$$
Further amplitude resonance takes place at a frequency of external force which is less than the frequency of undamped natural vibrations, i.e. $${\omega _1} \ne {\omega _0}.$$

When a simple pendulum oscillates in air continously then its time period remains same whereas amplitude decreases.

$$\eqalign{ & \frac{T}{2} = 0.5\,\sec \Rightarrow T = 1\,\sec \cr & v = \frac{1}{T} = 1\,Hz \cr} $$

The restoring force
$$\eqalign{ & F = - \left( {PA} \right) \cr & = - \left[ {\rho g\left( {2\,y\sin {{30}^ \circ }} \right)} \right]A \cr & = \rho gA\left( { - y} \right) \cr & \therefore a = \frac{F}{m} = \frac{{\rho gA}}{m}\,\,{\text{Thus}}\,\,T = 2\pi \sqrt {\frac{m}{{\rho gA}}} \cr} $$

The potential energy of a particle executing $$SHM$$  is given by,
$$U = \frac{1}{2}m{\omega ^2}{x^2}$$
$$U$$ is maximum, when $$x = a =$$  amplitude of vibration i.e. the particle is passing from the extreme position and is minimum when $$x = 0,$$  i.e. the particle is passing from the mean position
$$\therefore {U_{\max }} = \frac{1}{2}m{\omega ^2}{a^2}\,......\left( {\text{i}} \right)$$
Also, total energy of the particle at instant $$t$$ is given by
$$E = \frac{1}{2}m{\omega ^2}{a^2}\,......\left( {{\text{ii}}} \right)$$
So, when $$E = 160\,J,$$   then maximum potential energy of particle will also be $$160\,J.$$
Alternative
$${\left( {KE} \right)_{\max }} = {\left( {PE} \right)_{\max }} = {\text{Total Mechanical Energy}}$$
So, Total Mechanical Energy $$= 160\,J$$

$$\because A = {A_0}{e^{ - \frac{{bt}}{{2m}}}}$$     (where, $${A_0}$$ = maximum amplitude)
According to the questions, after 5 second,
$$0.9{A_0} = {A_0}{e^{ - \frac{{b(5)}}{{2m}}}}\,......\left( {\text{i}} \right)$$
After 10 more second,
$$A = {A_0}{e^{ - \frac{{b(15)}}{{2m}}}}\,......\left( {{\text{ii}}} \right)$$
From eqns (i) and (ii)
$$A = 0.729{A_0}\,\therefore \alpha = 0.729$$

Use the equation of motion of a body executing $$SHM.$$
i.e. $$x = a\sin \omega t$$
As we know, the velocity of body executing $$SHM$$  is given by
$$\eqalign{ & v = \frac{{dx}}{{dt}} = a\omega \cos \,\omega t = a\omega \sqrt {1 - {{\sin }^2}\omega t} \cr & = \omega \sqrt {{a^2} - {x^2}} \cr & {\text{Here,}}\,x = \frac{a}{2} \cr & \therefore v = \omega \sqrt {{a^2} - \frac{{{a^2}}}{4}} = \omega \sqrt {\frac{{3{a^2}}}{4}} \cr & = \frac{{2T}}{T}\frac{{a\sqrt 3 }}{2} \cr & = \frac{{\pi a\sqrt 3 }}{T} \cr} $$

Derive an expression from the given values which must be similar to $$a = - {\omega ^2}x.$$   Then calculate time period from the values in place of $$\omega .$$

The situation is shown in figure. Consider two springs of spring constants $${k_1}$$ and $${k_2}.$$ Here, the body of weight $$mg$$  is suspended at the free end of the two springs in series combination. When the body is pulled downwards through a little distance $$y,$$ the two springs suffer different extensions say $${y_1}$$ and $${y_2}.$$ But the restoring force is same in each spring.
$$\eqalign{ & \therefore F = - {k_1}{y_1} \cr & {\text{and}}\,\,F = - {k_2}{y_2} \cr & {\text{or}}\,\,{y_1} = - \frac{F}{{{k_1}}} \cr & {\text{and}}\,\,{y_2} = - \frac{F}{{{k_2}}} \cr} $$
$$\therefore $$ Total extension, $$y = {y_1} + {y_2}$$
$$\eqalign{ & = - \frac{F}{{{k_1}}} - \frac{F}{{{k_2}}} \cr & = - F\left( {\frac{{{k_1} + {k_2}}}{{{k_1}{k_2}}}} \right) \cr & {\text{or}}\,\,F = - \left( {\frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}} \right)y \cr} $$
If $$k$$ is the spring constant of series combination of springs then $$F = - ky$$
$$\therefore k = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}$$
If the body is left free after pulling down, it will execute $$SHM$$  of period
$$\eqalign{ & T = 2\pi \sqrt {\frac{m}{k}} \cr & = 2\pi \sqrt {\frac{{m\left( {{k_1} + {k_2}} \right)}}{{{k_1}{k_2}}}} \cr} $$

Total mechanical energy is constant throughout the motion and equals $$\frac{1}{2}m{\omega ^2}{A^2}.$$