Limiting values of x for particle under force

[MikeN]

Homework Statement

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.

Homework 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)

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.

 

[haruspex]

which is actually fairly similar to the answer in the book

More similar than you realize, it seems.

 

[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!