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.

Consider the situations shown in the diagram (i) and (ii)

Assuming the two pendulums follow the following functions of their angular displacements

$$\begin{array}{ll} & \theta_1=\theta_0 \sin \left(\omega t+\phi_1\right) \quad \text{... (i)}\\ \text { and } & \theta_2=\theta_0 \sin \left(\omega t+\phi_2\right)\quad \text{... (ii)} \end{array}$$

As it is given that amplitude and time period being equal but phases being different.

Now, for first pendulum at any time $t$

$\theta_1=+\theta_0\quad$ [Right extreme]

From Eq. (i), we get

$$\begin{aligned} \Rightarrow\quad \theta_0 & =\theta_0 \sin \left(\omega t+\phi_1\right) \text { or } 1=\sin \left(\omega t+\phi_1\right) \\ \Rightarrow\quad \sin \frac{\pi}{2} & =\sin \left(\omega t+\phi_1\right) \\ \text{or}\quad \left(\omega t+\phi_1\right) & =\frac{\pi}{2}\quad \text{... (iii)} \end{aligned}$$

Similarly, at the same instant t for pendulum second, we have

$$\theta_2=-\frac{\theta_0}{2}$$

where $\theta_0=2^{\circ}$ is the angular amplitude of first pendulum. For the second pendulum, the angular displacement is one degree ,therefore $\theta_2=\frac{\theta_0}{2}$ and negative sign is taken to show for being left to mean position.

From Eq. (ii), then

$$-\frac{\theta_0}{2}=\theta_0 \sin \left(\omega t+\phi_2\right)$$

$$\begin{aligned} \Rightarrow\quad\sin \left(\omega t+\phi_2\right) & =-\frac{1}{2} \Rightarrow\left(\omega t+\phi_2\right)=-\frac{\pi}{6} \text { or } \frac{7 \pi}{6} \\ \text{or}\quad \left(\omega t+\phi_2\right) & =-\frac{\pi}{6} \text { or } \frac{7 \pi}{6}\quad \text{.... (iv)} \end{aligned}$$

$$\begin{aligned} &\text { From Eqs. (iv) and (iii), the difference in phases }\\ &\begin{gathered} \left(\omega t+\phi_2\right)-\left(\omega t+\phi_1\right)=\frac{7 \pi}{6}-\frac{\pi}{2}=\frac{7 \pi-3 \pi}{6}=\frac{4 \pi}{6} \\ \text { or }\quad \left(\phi_2-\phi_1\right)=\frac{4 \pi}{6}=\frac{2 \pi}{3}=120^{\circ} \end{gathered} \end{aligned}$$