# Limiting values of x for particle under force

1. Jun 18, 2016

### MikeN

1. The problem statement, all variables and given/known data
This problem is from Kibble and Berkshire's Classical mechanics (Fifth edition), Chapter 2, Question 20.
A particle of mass $m$ moves in the region $x > 0$ under the force $F = -m\omega^2\left(x-\frac{a^4}{x^3}\right)$ where $\omega$ and $a$ are constants. The particle is initially at the position of equilibrium ($x=a$) and is moving with a velocity $v$, find the limiting values of $x$ in the subsequent motion.

2. Relevant equations
$$F(x) = -m\omega^2\left(x-\frac{a^4}{x^3}\right)$$
$$V(x) = \frac{1}{2}m\omega^2\left(x^2+\frac{a^4}{x^2}\right)$$

3. The attempt at a solution
From conservation of energy, $$\Delta K=-\Delta V$$
The initial kinetic energy is $\frac{1}{2}mv^2$, the particle is at the furthest point when the velocity is zero, therefore the final kinetic energy is zero. The initial potential is $V(a)$ and the final is $V(x)$.
$$0-\frac{1}{2}mv^2=-\left(\frac{1}{2}m\omega^2 \left(x^2+\frac{a^4}{x^2}\right)-m\omega^2 a^2\right)$$
The answer given in the book is:
$$x=\sqrt{a^2+\frac{v^2}{4\omega^2}}\pm\frac{v}{2\omega}$$
Plugging some values into this and comparing with my setup shows that I'm going wrong somewhere.

Solving my equation for $x$ gives:
$$x^2=\frac{v^2}{2\omega^2}+a^2\pm\frac{v}{\omega}\sqrt{\frac{v^2}{4\omega^2}+a^2}$$
which is actually fairly similar to the answer in the book after it's squared, except my answer is missing a term ($\frac{v^2}{4\omega^2}$) but I can't see where this comes from.

2. Jun 18, 2016

### haruspex

More similar than you realise, it seems.

3. Jun 18, 2016

### MikeN

Haha, I feel silly now. When I copied the answer from the book down onto paper I had a 2 inside of the square root instead of the 4. Thank you!