Question 11 Mark
A body of mass m is situated in a potential field $\text{U}(\text{x})=\text{U}_0(1-\cos\text{ax}),$ Where $U_0$ and a are constant. find the time period of small oscillation.
Answer
View full question & answer→$\because\text{dW}=\text{F}.\text{dx}$ if W = U, then$\text{dU}=\text{F}.\text{dx}\ \text{or}\ \text{F}=\frac{-\text{dU}}{\text{dx}}$ (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) $k = a^2U_0$
$\text{m}\omega^2=\text{a}^2\text{U}_0\Rightarrow\omega^2=\text{a}^2\frac{\text{U}_0}{\text{m}}$
$\Big(\frac{2\pi}{\text{T}}\Big)^2=\text{a}^2\frac{\text{U}_0}{\text{m}}\Rightarrow\text{T}^2=4\pi\frac{\text{m}}{\text{U}_0\text{a}^2}\ \text{or}$
$\text{T}=\frac{2\pi}{\text{a}}\sqrt{\frac{\text{m}}{\text{U}_0}}.$
From (i) this time period is valid for small angle ax.
$\text{F}=\frac{-\text{d}}{\text{dx}}[\text{U}_0(1-\cos\text{ax})=\frac{-\text{d}}{\text{dx}}[\text{U}_0+\text{U}_0\cos\text{a}_\text{x}]$
$\text{F}=-[0-\text{U}_0(-\sin\text{ax}).\text{a}]$
$\text{F}=-\text{aU}_0\sin\text{a}\text{x}$
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) $k = a^2U_0$
$\text{m}\omega^2=\text{a}^2\text{U}_0\Rightarrow\omega^2=\text{a}^2\frac{\text{U}_0}{\text{m}}$
$\Big(\frac{2\pi}{\text{T}}\Big)^2=\text{a}^2\frac{\text{U}_0}{\text{m}}\Rightarrow\text{T}^2=4\pi\frac{\text{m}}{\text{U}_0\text{a}^2}\ \text{or}$
$\text{T}=\frac{2\pi}{\text{a}}\sqrt{\frac{\text{m}}{\text{U}_0}}.$
From (i) this time period is valid for small angle ax.
