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Equation of motion from given 2D Potential

  1. Aug 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves in two dimensions under the following potential energy
    function:
    V(##\vec{r}##) = ½ k (x2 + 4y2)

    Find the resulting motion, given the initial condition at t=0:
    x = a, y = 0, x' = 0, y' = vo

    2. Relevant equations
    F = ma = -dV/dr

    3. The attempt at a solution
    This will obviously involve a 2nd order diff eq, and there are enough initial conditions to solve for the unknown constants. If the potential were given with the r variable instead of x and y, it would be simpler. As such, I'm not sure how to take dV/dr when V is V(x,y) not V(r)...

    If I use x = r Cosθ, y = r Sinθ, I can put it as:
    V(##\vec{r}##) = (3/2) k r2 Sin2θ

    But now that I have θ in the formula, is it ok to take dV/dr as such and set it in F = -dV/dr?
     
  2. jcsd
  3. Aug 21, 2015 #2
    You are doing it the wrong way, there are vector involved and ##\frac{d}{d \vec r} = \nabla ## and everyone already says it: force is the gradient of the potential, can you work it out now ?
    [Edit: In case I wasn't clear, ##\nabla = \lt \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \gt ## ]
     
  4. Aug 21, 2015 #3
    *facepalm* Thank you!

    ##\vec{F}## = -k( x ##\hat{x}## + 4y ##\hat{y}##)

    Then set to m##\vec{a}## = m(##\ddot{x}## ##\hat{x}## + ##\ddot{y}## ##\hat{y}##), and compare associated vector components... yes?
     
  5. Aug 21, 2015 #4
    Just tried that, for anyone who wants to check:

    X(t) = a ##\cos##(##\sqrt{k/m} t##)

    Y(t) =##\frac{v_{o}}{\sqrt{4k/m}}\sin(\sqrt{4k/m} t)##
     
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