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Limiting values of x for particle under force

  • Thread starter MikeN
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Homework Statement


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

Homework Equations


[tex]F(x) = -m\omega^2\left(x-\frac{a^4}{x^3}\right)[/tex]
[tex]V(x) = \frac{1}{2}m\omega^2\left(x^2+\frac{a^4}{x^2}\right)[/tex]

The Attempt at a Solution


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

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

Answers and Replies

  • #2
haruspex
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which is actually fairly similar to the answer in the book
More similar than you realise, it seems.
 
  • #3
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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!
 

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