Question

A particle of mass $m$ is located in a one-dimensional field where potential energy is given by:$V\left(x\right)=A\left(1-\cos px\right)$ where $A$and $p$ are constants. Calculate the period of small oscillations of the particle.

Answer 1

Step 1. Given data:

Potential energy:$V\left(x\right)=A\left(1-\cos px\right)$, where $A$and $p$ are constants.

Mass of particle = $m$

Step 2. Formula used:

Force on the particle, $F=-\frac{dV}{dx}$, where $V=$potential energy. ( The force on an object is the negative of the derivative of the potential energy )

Step 3. Calculations:

Putting the value of potential energy in the above equation, we get

$F=-\frac{d}{dx}\left(A\left(1-\cos px\right)\right)=-Ap.\sin px$

For small oscillations, $\sin px=px$,

Hence, $F=-A{p}^2.x$

Now, the acceleration would be given by, $\alpha =\frac{F}{m}\Rightarrow \frac{-A{p}^{2}x}{m}........\left(1\right)$

Also, we know that, $\alpha =\frac{F}{m}=-{\omega }^2.x......\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} \Rightarrow -\frac{A{p}^2}{m}.x=-{\omega }^2.x \\ \Rightarrow \omega =\sqrt{\frac{A{p}^2}{m}} \end{gathered}$$

Now, $\omega =\frac{2\mathrm{\pi }}{T}$.

$\Rightarrow T=\frac{2\mathrm{\pi }}{\omega }=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\mathrm{Ap}}^2}}$

Hence, the period of small oscillations of the particle is $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\mathrm{Ap}}^2}}$