If it gives correct time at equator, it will give correct time at poles also because the time period of spring-mass system is independent of $$g.$$

Equation of displacement is given by
$$x = A\sin \left( {\omega t + \phi } \right)$$
where $$A = \frac{{{F_0}}}{{m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$

$$\eqalign{ & {\text{Given,}}\,\,x = 10\sin \left( {2t - \frac{\pi }{6}} \right) \cr & A = 10\,\,{\text{and}}\,\,\omega = 2\,Hz \cr & \therefore \upsilon = \omega \sqrt {{A^2} - {d^2}} = 2\sqrt {{{\left( {10} \right)}^2} - {{\left( 6 \right)}^2}} \cr & = 2\sqrt {100 - 36} = 2 \times 8 = 16\,m{s^{ - 1}} \cr} $$

In $$x = A\cos \omega t,$$   the particle starts oscillating from extreme position. So at $$t = 0,$$  its potential energy is maximum.

Amplitude of a damped oscillator at any instant $$t$$ is given by
$$A = {A_0}{e^{ - \frac{{bt}}{{2m}}}}$$
Where $${A_0}$$ is the original amplitude
From question,
$$\eqalign{ & {\text{When}}\,\,t = 2\,s,A = \frac{{{A_0}}}{3}\,\,\therefore \frac{{{A_0}}}{3} = {A_0}{e^{ - \frac{{2b}}{{2m}}}} \cr & {\text{or,}}\,\,\frac{1}{3} = {e^{ - \frac{b}{m}}}\,......\left( {\text{i}} \right) \cr & {\text{When}}\,\,t = 6\,s,A = \frac{{{A_0}}}{n}\,\,\therefore \frac{{{A_0}}}{n} = {A_0}{e^{ - \frac{{6b}}{{2m}}}} \cr & {\text{or,}}\,\,\frac{1}{n} = {e^{ - \frac{{3b}}{m}}} = {\left( {{e^{ - \frac{b}{m}}}} \right)^3}\,\,{\text{or,}}\,\,\frac{1}{n} = {\left( {\frac{1}{3}} \right)^3} \cr & \therefore n = {3^3}\,\,\left( {{\text{Using eq}}{\text{.}}\left( {\text{i}} \right)} \right) \cr} $$

Let the minimum amplitude of $$SHM$$  be $$a.$$
Restoring force on spring
$$F = ka$$
Restoring force is balanced by weight $$mg$$  of block.
For mass to execute simple harmonic motion of amplitude $$a.$$
$$\therefore ka = mg\,\,{\text{or}}\,\,a = \frac{{mg}}{k}$$
$$\eqalign{ & {\text{Here,}}\,m = 2\,kg,k = 200\,N/m, \cr & g = 10\,m/{s^2} \cr & \therefore a = \frac{{2 \times 10}}{{200}} = \frac{{10}}{{100}}m \cr & = \frac{{10}}{{100}} \times 100\,cm = 10\,cm \cr} $$
Hence, minimum amplitude of the motion should be $$10\,cm,$$  so that the mass gets detached from the pan.

$${k_{{\text{eq}}}} = k + 2\,k = 3\,k$$

In equilibrium position
$$\eqalign{ & mg = {F_B} = \frac{m}{{4{\rho _0}}}{\rho _0}\left( {1 + a{y_0}} \right)g \Rightarrow 4 = 1 + a{y_0} \cr & {y_0} = \frac{3}{2} = 1.5 \cr} $$
Now displace it downward by $$\Delta y$$
$$\eqalign{ & \therefore F = mg - \frac{m}{{4{\rho _0}}}{\rho _0}\left[ {1 + a\left( {{y_0} + \Delta y} \right)} \right]g \cr & = mg - \frac{{mg}}{4}\left[ {1 + a{y_0} + a\Delta y} \right] \cr & F = - \frac{{mga}}{4}\Delta y;\,{\text{Acceleration}} = - \left( {\frac{{ag}}{4}} \right)\Delta y \cr & \therefore T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{4}{{ag}}} = \frac{{2\pi }}{{\sqrt 5 }}\sec . \cr} $$

When particle starts from mean position,

$$\eqalign{ & x = A\cos \omega t \cr & \therefore \left( {A - a} \right) = A\cos \left( {\omega \times 1} \right) \cr & {\text{or}}\,\,\left( {A - a} \right) = A\cos \omega \,......\left( {\text{i}} \right) \cr & {\text{Also}}\,\,A - \left( {a + b} \right) = A\cos \left( {\omega \times 2} \right) \cr & {\text{or}}\,\,A - \left( {a + b} \right) = A\cos 2\omega \,......\left( {{\text{ii}}} \right) \cr} $$
After simplifying above equations, we get
$$A = \left[ {\frac{{2{a^2}}}{{3a - b}}} \right]$$

$$K = \frac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right)\,\,\therefore {K_{\max }} = \frac{1}{2}m{\omega ^2}{A^2},\,{\text{at}}\,x = 0.$$