Let $x=A \sin \omega t$ is the displacement function of SHM.

Velocity,

$$\begin{aligned} v & =\frac{d x}{d t}=A \omega \cos \omega t \\ v_{\max } & =A \omega|\cos \omega t|_{\max } \\ & =A \omega \times 1=\omega A \quad \left[\because|\cos \omega t|_{\max }=1\right] \ldots \text { (i) } \end{aligned}$$

$$\begin{aligned} \text { Acceleration, } a=\frac{d v}{d t}=-\omega A \cdot \omega \sin \omega t & \\ & =-\omega^2 A \sin \omega t \\ \left|a_{\max }\right| & =\left|\left(-\omega^2 A\right)(+1)\right| \quad \left[\because(\sin \omega t)_{\max }=1\right] \\ \left|a_{\max }\right| & =\omega^2 A\quad \text{... (ii)} \end{aligned}$$

$$\begin{aligned} &\text { From Eqs. (i) and (ii), we get }\\ &\begin{array}{ll} \Rightarrow & \frac{v_{\max }}{a_{\max }}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega} \\ \Rightarrow \quad & \frac{a_{\max }}{v_{\max }}=\omega \end{array} \end{aligned}$$

The diagram represents

the motion of a particle executing SHM between $A$ and $B$.

Total distance travelled while it goes from $A$ to $B$ and returns to $A$ is

$$\begin{aligned} & =A O+O B+B O+O A \\ & =A+A+A+A=4 A \quad [\because O A=A] \end{aligned}$$

Amplitude $=O A=A$

Hence, ratio of distance and amplitude $=\frac{4 A}{A}=4$

As the particle on reference circle moves in anti-iclockwise direction. The projection will move from $P$ 'to $O$ towards left.

Hence, in the position shown the velocity is directed from $P^{\prime} \rightarrow P^{\prime \prime}$ i.e., from right to left, hence sign is negative.

Let us assume the displacement function of SHM

$$\begin{aligned} \text{where,}\quad x & =a \cos \omega t \\ a & =\text { amplitude of motion } \\ \text { velocity } v & =\frac{d x}{d t} \end{aligned}$$

or $\frac{d x}{d t}=a(-\sin \omega t) \omega=-\omega a \sin \omega t$

or $v=-\omega a \sin \omega t$

$$=\omega \operatorname{acos}\left(\frac{\pi}{2}+\omega t\right) \quad\left[\because \sin \omega t=-\cos \left(\frac{\pi}{2}+\omega t\right)\right]$$

Now, phase of displacement $=\omega t$

Phase of velocity $=\frac{\pi}{2}+\omega t$

$\therefore$ Difference in phase of velocity to that of phase of displacement

$$=\frac{\pi}{2}+\omega t-\omega t=\frac{\pi}{2}$$

Potential energy ( PE ) of a simple harmonic oscillator is where,

$$\begin{aligned} & =\frac{1}{2} k x^2=\frac{1}{2} m \omega^2 x^2 \quad \text{.... (i)}\\ k & =\text { force constant }=m \omega^2 \end{aligned}$$

where, $$\quad k=\text { force constant }=m \omega^2$$

When, PE is plotted against displacement $x$, we will obtain a parabola.

When $x=0, \mathrm{PE}=0$

When $x= \pm A, P E=$ maximum

$$=\frac{1}{2} m \omega^2 A^2$$

$$\begin{aligned} &\text { KE of a simple harmonic oscillator }=\frac{1}{2} m v^2\quad \left[\because v=\omega \sqrt{A^2-x^2}\right] \end{aligned}$$

$$\begin{aligned} & =\frac{1}{2} m\left[\omega \sqrt{A^2-x^2}\right]^2 \\ & =\frac{1}{2} m \omega^2\left(A^2-x^2\right)\quad \text{... (ii)} \end{aligned}$$

This is also parabola, if plot KE against displacement $x$.

$$\begin{aligned} \text{i.e.}\quad & \mathrm{KE}=0 \text { at } x= \pm A \\ \text{and}\quad & \mathrm{KE}=\frac{1}{2} m \omega^2 A^2 \text { at } x=0 \end{aligned}$$

$$\begin{aligned} &\text { Now, total energy of the simple harmonic oscillator }=\mathrm{PE}+\mathrm{KE} \quad \text { [using Eqs. (i) and (ii)] }\\ &\begin{aligned} & =\frac{1}{2} m \omega^2 x^2+\frac{1}{2} m \omega^2\left(A^2-x^2\right) \\ & =\frac{1}{2} m \omega^2 x^2+\frac{1}{2} m \omega^2 A^2-\frac{1}{2} m \omega^2 x^2 \\ \mathrm{TE} & =\frac{1}{2} m \omega^2 A^2 \end{aligned} \end{aligned}$$

Which is constant and does not depend on $x$. Plotting under the above guidelines KE, PE and TE versus displacement $x$-graph as follows