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 $T=\frac{2 \pi}{\omega}$.

Let us assume that the required displacement be $x$.

$\therefore$ Potential energy of the simple harmonic oscillator $=\frac{1}{2} k x^2$

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

Maximum energy of the oscillator

$$ \begin{aligned} &\mathrm{TE}=\frac{1}{2} m \omega^2 A^2 \quad\left[\because x_{\max }=A\right] \quad\text { ....(ii) } \end{aligned}$$

$$\begin{aligned} \text{where,}\quad A & =\text { amplitude of motion } \\ \text{Given,}\quad\mathrm{PE} & =\frac{1}{2} \mathrm{TE} \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad & \frac{1}{2} m \omega^2 x^2 =\frac{1}{2}\left[\frac{1}{2} m \omega^2 A^2\right] \\ \Rightarrow \quad & x^2 =\frac{A^2}{2} \\ \text { or } \quad & x =\sqrt{\frac{A^2}{2}}= \pm \frac{A}{\sqrt{2}} \end{aligned}$$

Sign $$\pm$$ indicates either side of mean position.

Given potential energy associated with the field

$$U(x)=U_0(1-\cos \alpha x)\quad \text{... (i)}$$

Now, force $F=-\frac{d U(x)}{d x}$

$$\left[\because \text { for conservatine force } f, \text { we can write } f=\frac{-d u}{d x}\right]$$

$$\begin{aligned} &\text { [We have assumed the field to be conservative] }\\ &\begin{aligned} & F=-\frac{d}{d x}\left(U_0-U_0 \cos \alpha x\right)=-U_0 \alpha \sin \alpha x \\ & F=-U_0 \alpha^2 x\quad \text{... (ii)} \end{aligned} \end{aligned}$$

$$[\because \text { for small oscillations } \alpha x \text { is small, } \sin \alpha x \approx \alpha x \text { ] }$$

$$\Rightarrow \quad F \propto(-x)$$

As, $U_0, \alpha$ being constant.

$\therefore$ Motion is SHM for small oscillations.

Standard equation for SHM

$$F=-m \omega^2 x\quad \text{... (iii)}$$

Comparing Eqs. (ii) and (iii), we get

$$\begin{aligned} m \omega^2 & =U_0 \alpha^2 \\ \omega^2 & =\frac{U_0 \alpha^2}{m} \text { or } \omega=\sqrt{\frac{U_0 \alpha^2}{m}} \\ \therefore \quad \text { Time period } T & =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{U_0 \alpha^2}} \end{aligned}$$

Consider the diagram of the spring block system. It is a SHM with amplitude of 5 cm about the mean position shown.

Given, $\quad$ spring constant $k=50 \mathrm{~N} / \mathrm{m}$

$m=$ mass attached $=2 \mathrm{~kg}$

$\therefore \quad$ Angular frequency $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5 \mathrm{rad} / \mathrm{s}$

Assuming the displacement function

$$y(t)=A \sin (\omega t+\phi)$$

where, $\phi=$ initial phase

But given at $t=0, y(t)=+A$

$$y(0)=+A=A \sin (\omega \times 0+\phi)$$

or $$\quad \sin \phi=1 \Rightarrow \phi=\frac{\pi}{2}$$

$\therefore$ The desired equation is $\quad y(t)=A \sin \left(\omega t+\frac{\pi}{2}\right)=A \cos \omega t$

Putting $A=5 \mathrm{~cm}, \omega=5 \mathrm{rad} / \mathrm{s}$

we get, $y(t)=5 \sin 5 t$

where, $t$ is in second and $y$ is in centimetre.