Physics 1 Marks Question

$\text{dU}=\text{F}.\text{dx}\ \text{or}\ \text{F}=\frac{-\text{dU}}{\text{dx}}$ (here restoring force is opposite to displacement)

$\text{F}=\frac{-\text{d}}{\text{dx}}[\text{U}_0(1-\cos\text{ax})=\frac{-\text{d}}{\text{dx}}[\text{U}_0+\text{U}_0\cos\text{a}_\text{x}]$

$\text{F}=-[0-\text{U}_0(-\sin\text{ax}).\text{a}]$

$\text{F}=-\text{aU}_0\sin\text{a}\text{x}$

For SHM. ax is small

So sin ax becomes ax ...(i)

$\therefore\text{F}=-\alpha.\text{U}_0\text{ax}=-\text{a}^2\text{U}_0\text{x}\ ...(\text{ii})$

$\alpha_2\text{U}_0$ are constants.

$\therefore\text{F}\propto-\text{x}.$ so motion is SHM.

Here from (ii) k = a²U₀

_{$\text{m}\omega^2=\text{a}^2\text{U}_0\Rightarrow\omega^2=\text{a}^2\frac{\text{U}_0}{\text{m}}$}

_{$\Big(\frac{2\pi}{\text{T}}\Big)^2=\text{a}^2\frac{\text{U}_0}{\text{m}}\Rightarrow\text{T}^2=4\pi\frac{\text{m}}{\text{U}_0\text{a}^2}\ \text{or}$}

_{$\text{T}=\frac{2\pi}{\text{a}}\sqrt{\frac{\text{m}}{\text{U}_0}}.$}

From (i) this time period is valid for small angle ax.

$\Rightarrow\text{v}=\text{a}\omega\cos\Big(\frac{\pi}{2}+\omega\text{t}\Big)$