# Where is the particle free and where is it trapped?

This is one of my homework problems:

A particle of mass $m$ is acted on by a one-dimensional force: $\displaystyle\vec{F}=\left(b\sin\frac{2\pi x}{\lambda}\right)\hat{\mathbf{x}}$ where $b$ and $\lambda$ are positive constants.

a) Obtain the potential energy function $V(x)$.

My answer: Let us define $V$ to be zero at $x=0$.

Then $\displaystyle V(x)=\frac{b\lambda}{2\pi}\left(\cos\frac{2\pi x}{\lambda}-1\right)$

b) Graph the function $V(x)$.

My answer: Let $\lambda=2$ and $b=4$. c) Discuss where the particle is free and where it is trapped (state the values of $x$ where it is trapped).

My answer: I'm not sure I understand what the question means. If the total energy $E$ is positive, the particle will be free. If $E$ is negative, it will be trapped between two consecutive hills. But the question asks for the values of $x$ where it is free and trapped. How do I answer this?

• If you remove the constraint $V(0)=0$ you will have a family of curves. And then, you will need to solve the inequality $V(x) = cos(x) - C < 0$ to find your solution. Note that I rescaled $x$ and $V$ for simplicity. – nicoguaro Feb 13 '17 at 0:41