At any instant the total energy is
$$\frac{1}{2}k{A^2} = {\text{constant, where }}A = {\text{amplitude}}$$
hence total energy is independent of $$x.$$

As seen from figure after one time period the bob return to its equilibrium position, so displacement of the particle is zero, but distance covered by the particle in one time period is $$4\,A$$  (where $$A$$ is amplitude of bob, when it does $$S.H.M.$$  )

At extreme position
$$\eqalign{ & y = A\sin \left( {\omega t + \frac{\pi }{2}} \right) = A\cos \omega t \cr & {\text{Now}}\,y = \frac{A}{2}\,{\text{then}}\,\frac{A}{2} = A\cos \omega t \cr & \Rightarrow \omega t = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\,\,\frac{{2\pi }}{T}t = \frac{\pi }{3} \Rightarrow t = \frac{T}{6} \cr} $$

$${t_0} = 2\pi \sqrt {\frac{\ell }{g}} .$$
The restoring force in a liquid
$$\eqalign{ & F = - \left( {mg - V{\rho _e}g} \right)\sin \theta \cr & = - \left( {mg - \frac{m}{{\left( {\frac{4}{3} \times 1000} \right)}} \times 1000g} \right)\left( {\frac{x}{\ell }} \right) \cr & {\text{or}}\,\,a = \left( {g - \frac{{3g}}{4}} \right)\left( {\frac{{ - x}}{\ell }} \right) = \frac{g}{4}\left( {\frac{{ - x}}{\ell }} \right) \cr & \therefore t = 2\pi \sqrt {\frac{\ell }{{\left( {\frac{g}{4}} \right)}}} = 2{t_0}. \cr} $$

$$y = \frac{1}{2} - \frac{1}{2}\cos 2\omega t$$

In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position.

The variation of $$PE$$  and $$KE$$  with time is shown in figure, by dotted parabolic curve and solid parabolic curve respectively.
Figure indicates that maximum values of total energy, $$KE$$  and $$PE$$  of $$SHM$$  are equal.
$$\eqalign{ & {\text{Now,}}\,\,KE = {K_0}{\cos ^2}\omega t \cr & \therefore K{E_{\max }} = {K_0} \cr & {\text{So,}}\,P{E_{\max }} = {K_0} \cr & {\text{and}}\,\,{\left( E \right)_{{\text{Total}}}} = {K_0} \cr} $$

The resonance wave becomes very sharp when damping force is small.

Two sinusoidal displacements have amplitude $$A$$ each, with a phase difference of $$2\frac{\pi }{3}.$$  It is given that sinusoidal displacement $${x_3}\left( t \right)$$  brings the mass to a complete rest. This is possible when the amplitude of third is $$A$$ and is having a phase difference of $$4\frac{\pi }{3}$$ with respect to $${x_1}\left( t \right)$$  as shown in the figure.

Consider a particle of mass $$m,$$ executing linear $$SHM$$  with amplitude $$a$$ and constant angular frequency $$\omega .$$ Suppose $$t$$ second after starting from the mean position, the displacement of the particle is $$x,$$ which is given by
$$x = a\sin \omega t$$
So, potential energy of particle is
$$U = \frac{1}{2}m{\omega ^2}{x^2}\,......\left( {\text{i}} \right)$$
and kinetic energy of particle is
$$T = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii)
$$\frac{T}{U} = \frac{{{a^2} - {x^2}}}{{{x^2}}}$$

From the graph it is clear that the amplitude is $$1\,cm$$  and the time period is 8 second. Therefore the equation for the $$S.H.M.$$  is
$$x = a\sin \left( {\frac{{2\pi }}{T}} \right) \times t = 1\sin \left( {\frac{{2\pi }}{8}} \right)t = \sin \frac{\pi }{4}t$$
The velocity $$\left( v \right)$$ of the particle at any instant of time $$'t'$$ is
$$v = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sin \left( {\frac{\pi }{4}} \right)t} \right] = \frac{\pi }{4}\cos \left( {\frac{\pi }{4}} \right)t$$
The acceleration of the particle is
$$\eqalign{ & \frac{{{d^2}x}}{{d{t^2}}} = - {\left( {\frac{\pi }{4}} \right)^2}\sin \left( {\frac{\pi }{4}} \right)t \cr & {\text{At}}\,\,t = \frac{4}{3}s\,\,{\text{we}}\,{\text{get}} \cr & \frac{{{d^2}x}}{{d{t^2}}} = - {\left( {\frac{\pi }{4}} \right)^2}\sin \frac{\pi }{4} \times \frac{4}{3} = \frac{{ - \sqrt 3 {\pi ^2}}}{{32}}cm/{s^2} \cr} $$