Question

The potential energy for a particle that can move along the x-axis is $U=A{x}^2+B\sin \left(\frac{\pi x}{L}\right)$where A, B, and L are constants. What is the force on the particle at

(a) $X=0,$

(b) $X=L/2$and

(c) $X=L$?

Short answer

$$\begin{gathered} \left(a\right)F_x=-B\mathrm{\pi }/\mathrm{L} \\ \left(\mathrm{b}\right){\mathrm{F}}_{\mathrm{x}}=-\mathrm{AL} \\ \left(\mathrm{c}\right){\mathrm{F}}_{\mathrm{x}}=-2\mathrm{AL}+\frac{\mathrm{B\pi }}{\mathrm{L}} \end{gathered}$$

Step-by-step solution

Given information (part a)

The potential energy of the particle along the x-axis is given as

$U=A{x}^2+B\sin \left(\frac{\pi x}{L}\right)$

Explanation (part a)

The force F_X can be found by taking the negative derivative of potential energy with respect to position.

$$\begin{gathered} U=A{x}^2+B\sin \left(\frac{\pi x}{L}\right) \\ {F}_x=-\frac{dU}{dx} \\ =-\frac{d}{dx}\left(A{x}^2+B\sin \left(\frac{\pi x}{L}\right)\right) \\ =-\frac{d}{dx}\left(A{x}^2\right)-\frac{d}{dx}\left(B\sin \left(\frac{\pi x}{L}\right)\right) \\ {F}_x=-2Ax-B\cos \left(\frac{\pi x}{L}\right)\times \left(\frac{\pi }{L}\right) \end{gathered}$$

At $X=0$

$$\begin{gathered} F_x=-2A\times 0-B\cos \left(\frac{\pi \left(0\right)}{L}\right)\times \left(\frac{\pi }{L}\right) \\ =0-B\cos \left(0\right)\times \left(\frac{\pi }{L}\right) \\ =-B\times 1\times \left(\frac{\pi }{L}\right) \\ =-B\left(\frac{\mathrm{\pi }}{\mathrm{L}}\right) \end{gathered}$$

Given information (part b)

The potential energy of the particle along the x-axis is given as

$U=A{x}^2+B\sin \left(\frac{\pi x}{L}\right)$

Explanation (part b)

$AtX=L/2$

localid="1649413461206" $$\begin{gathered} F_x=-2A\times \frac{L}{2}-B\cos \left(\frac{\pi \times L}{2L}\right)\times \left(\frac{\pi }{L}\right) \\ =\frac{-2AL}{2}-B\cos \left(\frac{\pi }{2}\right)\times \left(\frac{\pi }{L}\right) \\ =-AL-B\left(0\right)\left(\frac{\pi }{L}\right) \\ =-AL \end{gathered}$$

Given information (part c)

The potential energy of the particle along the x-axis is given as

$U=A{x}^2+B\sin \left(\frac{\pi x}{L}\right)$

Explanation (part c)

$$\begin{gathered} AtX=L \\ {F}_x=-2A\times L-B\cos \left(\left(\frac{\pi L}{L}\right)\right)\times \left(\frac{\pi }{L}\right) \\ =-2AL-B\cos \left(\pi \right)\times \left(\frac{\pi }{L}\right) \\ =-2AL-B\left(-1\right)\left(\frac{\pi }{L}\right) \\ =-2AL+\frac{B\mathrm{\pi }}{L} \end{gathered}$$