(2)

Here, $\omega =\frac{2\pi }{8}=\frac{\pi }{4}{\text{s}}^{-1}$

$a=-A{\omega }^2=-\left(1\right){\left(\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)}^2=-\frac{{\pi }^2}{16}{\text{ms}}^{-2}$

Thus, option (2) is correct.

(1)

Displacement of the particle, $y=a\sin \omega t,$

$v=\frac{dy}{dt}=a\omega$ $\cos$ $\omega t$

Acceleration,  $a=\frac{dv}{dt}=-a{\omega }^2\sin$ $\omega t$

So, phase difference between displacement and acceleration is $\pi$.

(1)

Since the displacement for a complete vibration is zero, therefore the average velocity will be zero.

(2)

Given, $A=3$ cm, $x=2$ cm

The velocity of a particle in simple harmonic motion is given as $v=\omega \sqrt{A^2-x^2}$

and magnitude of its acceleration is $a={\omega }^{2}x$

Given $\left|v\right|$$=$$\left|a\right|$  $\therefore$  $\omega \sqrt{A^2-x^2}={\omega }^{2}x$

$\omega x=\sqrt{A^2-x^2} \text{or} {\omega }^2{x}^2=A^2-x^2$

${\omega }^2=\frac{A^2-x^2}{x^2}=\frac{9-4}{4}=\frac{5}{4} \text{or} \omega =\frac{\sqrt{5}}{2}$

Time period, $T=\frac{2\pi }{\omega }=2\pi .\frac{2}{\sqrt{5}}=\frac{4\pi }{\sqrt{5}}$ $\text{s}$

(2)

If $A$ and $\omega$ be the amplitude and angular frequency of vibration, then

          $\alpha ={\omega }^{2}A$                                                           ...(i)

and    $\beta =\omega A$                                                            ...(ii)

Dividing eqn. (i) by eqn. (ii), we get

         $\frac{\alpha }{\beta }=\frac{{\omega }^{2}A}{\omega A}=\omega$

$\therefore$   $\text{Time period of vibration is }T=\frac{2\pi }{\omega }=\frac{2\pi }{(\alpha /\beta )}=\frac{2\pi \beta }{\alpha }$

(4)

In SHM, velocities of a particle at distances $x_1$ and $x_2$ from mean position are given by

          $V_1^2={\omega }^2(a^2-x_1^2)$                        ...(i)

           $V_2^2={\omega }^2(a^2-x_2^2)$                       ...(ii)

From equations (i) and (ii), we get

         $V_1^2-V_2^2={\omega }^2(x_2^2-x_1^2)$

$\omega =\sqrt{\frac{V_1^2-V_2^2}{x_2^2-x_1^2}}$$\Rightarrow T=2\pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}$

(3)

Here, $X=A\cos \omega t$

$\therefore$    $\text{Velocity, }v=\frac{dX}{dt}=\frac{d}{dt}\left(A\cos \omega t\right)=-A\omega \sin \omega t$

$\text{Acceleration, }a=\frac{dv}{dt}=\frac{d}{dt}(-A\omega \sin \omega t)=-A{\omega }^2\cos \omega t$

Hence the variation of $a$ with $t$ is correctly shown by graph (3).