0
$\begingroup$

Given that the potential energy of a particle in 2D space is $$V(x, y) = \frac{1}{2}k(x^2 + y^2),$$ find the equations of motion and show they are circular orbits.

Substituting $r^2 = x^2 + y^2$, I was able to get to the general solution of

$\vec{r}(t) = \vec{C}_1\cos{(\omega t)}+\vec{C}_2\sin{(\omega t)}$

via the fact that $\vec{F} = -\frac{\partial{\vec{V}}}{\partial{r}}$ and understand the process. However, the next step, says to rewrite this equation as

$\vec{r}(t) = \vec{b}_1\cos{(\omega t - \theta)} + \vec{b}_2\sin{(\omega t - \theta)}$

where $\vec{b}_1 = \vec{C}_1\cos{\theta} + \vec{C}_2\sin{\theta}$ and

$\vec{b}_2 = \vec{C}_2\cos{\theta} - \vec{C}_1\sin{\theta}$

In order to find the orbits.

I'm not sure where this substitution is coming from. Any hints/help would be greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Aug 27 '18 at 4:23
  • 1
    $\begingroup$ This is for self-study. I'm a 42 year-old police officer reading "The Theoretical Minimum". Not for a college class. Either way, any sort of assistance would be helpful. $\endgroup$ – lhoernle Aug 27 '18 at 5:54
  • 1
    $\begingroup$ ...and this is from the provided solutions manual on the official website. Just trying to make sense of it. $\endgroup$ – lhoernle Aug 27 '18 at 6:02
  • 1
    $\begingroup$ @JohnRennie I would like to write an answer to this question as it stands which explains in simple terms what the solution is rather than go through the mathematical derivation. I believe that once that is done the mathematical derivation will have more meaning. If you look at the solution then perhaps you will agree that it is rather lacking in explanation of the physical principles involved? madscitech.org/tm/slns/l5e2.pdf I am all for helping a self-learner and perhaps lowering the bar a little? $\endgroup$ – Farcher Aug 27 '18 at 23:00
  • $\begingroup$ I figured it had to do with the trigonometric angle addition identities; however, I've attempted to reconstruct this looking at the solution and am missing a step in the process. For instance, why are we adding the phase shift? $\endgroup$ – lhoernle Aug 28 '18 at 12:46
1
$\begingroup$

There are a lot of ways to combine trigonometric functions like sine and cosine to get these kinds of expressions, and it's worth your while learning as much as possible about manipulating these functions as it will keep cropping up.

I'd encourage you to read as much as possible of Wikipedia's page on trigonometric identities and try working with the expressions for practice.

In your case you'll need to get used to the very important angle sum rules :

$$sin(a\pm b)= sin(a)cos(b)\pm cos(a)sin(b)$$

and

$$cos(a\pm b) = cos(a)cos(b)\mp sin(a)sin(b)$$

This should point you in the direction you need. I really can't repeat how important it is to get to grips with the trigonometric functions if you want to press ahead with self study of physics. They crop up over and over again. Basic knowledge of e.g. Taylor series and calculus is also worth acquiring if you don't already have it. In a University course they'll typically make taking supporting mathematics classes (and exams) mandatory so your maths skills keep pace with what the physics theory requires.

This is for self-study. I'm a 42 year-old police officer reading "The Theoretical Minimum". Not for a college class. Either way, any sort of assistance would be helpful.

Note when we say "homework policy" on Physics SE, we're including any question where the poster would be expected to find out the answer themselves and do basic research to find the answer. This is to encourage people to do their own thinking as well as discourage the people who think it's a "do my homework for me" website - we do get those.

$\endgroup$
  • $\begingroup$ I was about to write a comment like this, thanks. Indeed the two expressions are equivalent. I do find it a bit odd that they bother to introduce $\theta$ though, because in the second form of the solution, the $2 \times 2 +1$ constants in $\vec{b}_1, \vec{b}_2$ and $\theta$ are not all independent, in terms of the $2 \times 2$ initial conditions. I suppose there is a reason in the textbook for them to do this but seems a bit unnatural to me. $\endgroup$ – secavara Aug 27 '18 at 11:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.