From the graph it is clear that the amplitude is $$1cm$$  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 $$\frac{{{d^2}x}}{{d{t^2}}} = - {\left( {\frac{\pi }{4}} \right)^2}\sin \left( {\frac{\pi }{4}} \right)t$$
At $$t = \frac{4}{3}s$$  we get
$$\eqalign{ & \frac{{{d^2}x}}{{d{t^2}}} = - {\left( {\frac{\pi }{4}} \right)^2}\sin \frac{\pi }{4} \times \frac{4}{3} = \frac{{ - {\pi ^2}}}{{16}}\sin \frac{\pi }{3} \cr & = \frac{{ - \sqrt 3 {\pi ^2}}}{{32}}cm/{s^2} \cr} $$

For a simple harmonic motion,
$$a \propto \frac{{{d^2}y}}{{d{t^2}}} \propto - y$$
Hence, equations $$y = \sin \omega t - \cos \omega t$$    and $$y = 5\cos \left( {\frac{{3\pi }}{4} - 3\omega t} \right)$$    are satisfying this condition and equation $$y = 1 + \omega t + {\omega ^2}{t^2}$$    is not periodic and $$y = {\sin ^3}\omega t$$   is periodic but not simple hormonic motion.

Figure shows the mapping of the two $$SHMs$$  with circular motions having phase difference
$$\phi = \omega t = \frac{{2\pi }}{6} \times 1 = \frac{\pi }{3}rad$$

The maximum separation between the two particles is
$$\eqalign{ & {S_{\max }} = 2A\sin \frac{\pi }{6} \cr & {\text{or}}\,\,{S_{\max }} = 2 \times 5 \times \frac{1}{2} = 5\,cm. \cr} $$

$${V_m} = \frac{{{V_0}}}{{\left( {a{\omega ^2} - b\omega + c} \right)}}$$
If there is a single resonant frequency, then this equation should be satisfied for only one that particular resonant frequency, hence $$a{\omega ^2} - b\omega + c = 0$$    will have equal roots therefore;
$$D = 0 \Rightarrow {\left( { - b} \right)^2} - 4ac = 0 \Rightarrow {b^2} = 4ac$$

At resonance, amplitude of oscillation is maximum
$$\eqalign{ & \Rightarrow 2{\omega ^2} - 36\omega + 9\,{\text{is}}\,{\text{minimum}} \cr & \Rightarrow 4\omega - 36 = 0\left( {{\text{derivative}}\,{\text{is}}\,{\text{zero}}} \right) \Rightarrow \omega = 9 \cr} $$

$$\eqalign{ & x = 4\left( {\cos \pi t + \sin \pi t} \right) = \sqrt 2 \times 4\left( {\frac{{\sin \pi t}}{{\sqrt 2 }} + \frac{{\cos \pi t}}{{\sqrt 2 }}} \right) \cr & x = 4\sqrt 2 \sin \left( {\pi t + {{45}^ \circ }} \right) \cr} $$

The smallest length of the air column is associated with fundamental mode of vibration of the air column as shown in the diagram.

$$\because {L_{\min }} = \frac{\lambda }{4} \Rightarrow 50\,cm = \frac{\lambda }{4} \Rightarrow \lambda = 200\,cm$$
The next higher length of the air column is
$$\eqalign{ & L = \frac{\lambda }{4} + \frac{\lambda }{2} = \frac{{\lambda + 2\lambda }}{4} = \frac{{3\lambda }}{4} \cr & = \frac{3}{4} \times 200 = 150\,cm \cr} $$

$$K.E = \frac{1}{2}k\left( {{A^2} - {d^2}} \right)$$
and $$P.E. = \frac{1}{2}k{d^2}$$
At mean position $$d = 0.$$  At extremes positions $$d = A$$

For block $$A$$ to move in $$S.H.M.$$

$$mg - N = m{\omega ^2}x$$
where $$x$$ is the distance from mean position
For block to leave contact $$N = 0$$
$$ \Rightarrow mg = m{\omega ^2}x \Rightarrow x = \frac{g}{{{\omega ^2}}}$$

$$\eqalign{ & {\text{Let}}\,\,y = A\sin \,\omega t \cr & {v_{{\text{inst}}}} = \frac{{dy}}{{dt}} = A\omega \cos \omega t = A\omega \sin \left( {\omega t + \frac{\pi }{2}} \right) \cr & {\text{Acceleration}} = - A{\omega ^2}\sin \omega t = A{\omega ^2}\sin \left( {\pi + \omega t} \right) \cr & \therefore \phi = \frac{\pi }{2} = 0.5\,\pi \cr} $$