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

A second's pendulum means a simple pendulum having time period $T=2 \mathrm{~s}$.

For a simple pendulum,

$$T=2 \pi \sqrt{\frac{l}{g}}$$

where, $l=$ length of the pendulum and $g=$ acceleration due to gravity on surface of the earth.

$$T_e=2 \pi \sqrt{\frac{l_e}{g_e}}\quad \text{.... (i)}$$

$$\begin{aligned} &\begin{aligned} \text { On the surface of the moon, }\quad & T_m=2 \pi \sqrt{\frac{l_m}{g_m}} \quad \text{... (ii)}\\ \therefore \quad & \frac{T_e}{T_m}=\frac{2 \pi}{2 \pi} \sqrt{\frac{l_e}{g_e}} \times \sqrt{\frac{g_m}{l_m}} \end{aligned} \end{aligned}$$

$$\begin{aligned} &T_e=T_m \text { to maintain the second's pendulum time period. }\\ &\therefore \quad 1=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}\quad \text{.... (iii)} \end{aligned}$$

But the acceleration due to gravity at moon is $1 / 6$ of the acceleration due to gravity at earth, i.e.,

$$g_m=\frac{g_e}{6}$$

Squaring Eq. (iii) and putting this value,

$$\begin{aligned} 1 & =\frac{l_e}{l_m} \times \frac{g_e / 6}{g_e}=\frac{l_e}{l_m} \times \frac{1}{6} \\ \Rightarrow \quad \frac{l_e}{6 l_m} & =1 \Rightarrow 6 l_m=l_e \\ \Rightarrow \quad l_m & =\frac{1}{6} l_e=\frac{1}{6} \times 1=\frac{1}{6} \mathrm{~m} \end{aligned}$$

For the calculation purpose, in this situation we will neglect gravity because it is constant throughout and will not effect the net restoring force.

Let in the equilibrium position, the spring has extended by an amount $x_0$.

Now, if the mass is given a further displacement downwards by an amount $x$. The string and spring both should increase in length by $x$. But, string is inextensible, hence the spring alone will contribute the total extension $x+x=2 x$, to lower the mass down by $x$ from initial equilibrium mean position $x_0$. So, net extension in the spring $\left(=2 x+x_0\right)$. Now force on the mass before bulling (in the $x_0$ extension case)

$$\begin{aligned} &\begin{array}{ll} & F=2 T \\ \text { But } & T=k x_0 \quad \text { [where } k \text { is spring constant] }\\ \therefore & F=2 k x_0 \quad \text{... (i)} \end{array}\\ &\end{aligned}$$

When the mass is lowered down further by $x$,

$$F^{\prime}=2 T^{\prime}$$

But new spring length $=\left(2 x+x_0\right)$

$$\therefore \quad F^{\prime}=2 k\left(2 x+x_0\right)\quad \text{.... (ii)}$$

Restoring force on the system.

$$F_{\text {restoring }}=-\left[F^{\prime}-F\right]$$

Using Eqs. (i) and (ii), we get

$$ \begin{aligned} F_{\text {restoring }} & =-\left[2 k\left(2 x+x_0\right)-2 k x_0\right] \\ & =-\left[2 \times 2 k x+2 k x_0-2 k x_0\right] \\ & =-4 k x \\ \text{or}\quad M a & =-4 k x \\ \text{where},\quad a & =\text { acceleration } \quad \text{(As, F = ma)}\\ \Rightarrow \quad a & =-\left(\frac{4 k}{M}\right) x \end{aligned}$$

$k, M$ being constant.

$\therefore$ $a \propto-x$

Hence, motion is SHM.

Comparing the above acceleration expression with standard SHM equation $a=-\omega^2 x$, we get

$$\begin{aligned} &\omega^2=\frac{4 k}{M} \Rightarrow \omega=\sqrt{\frac{4 k}{M}}\\ \therefore \quad &\text { Time period } T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{4 k}{M}}}=2 \pi \sqrt{\frac{M}{4 k}} \end{aligned}$$

We have to convert the given combination of two harmonic functions to a single harmonic (sine or cosine) function.

Given, displacement function

$$\begin{aligned} y & =\sin \omega t-\cos \omega t \\ & =\sqrt{2}\left(\frac{1}{\sqrt{2}} \cdot \sin \omega t-\frac{1}{\sqrt{2}} \cdot \cos \omega t\right) \\ & =\sqrt{2}\left[\cos \left(\frac{\pi}{4}\right) \cdot \sin \omega t-\sin \left(\frac{\pi}{4}\right) \cdot \cos \omega t\right] \\ & =\sqrt{2}\left[\sin \left(\omega t-\frac{\pi}{4}\right)\right]=\sqrt{2}\left[\sin \left(\omega t-\frac{\pi}{4}\right)\right] \end{aligned}$$

Comparing with standard equation

$$y=\operatorname{asin}(\omega t+\phi), \text { we get }=\omega=\frac{2 \pi}{T} \Rightarrow T=\frac{2 \pi}{\omega}$$

Clearly, the function represents SHM with a period $T=\frac{2 \pi}{\omega}$.