Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Periodic Motion

  1. Feb 22, 2008 #1
    ----abs( ) refers to absolute value

    The potential energy for a function is given as U = k*abs(x). The motion is not SMH because U is non-differentiable at some point, but is periodic due there being a stable equilibrium. I was asked to come up with a velocity vs. time graph for this...but I dont know how the hell they got it (I dont even know how to derive the sinousidal functions for SMH)...the graph is not a trigonometric function and looks a little like this:

    \/\/\/\/\/\/\ Its sort of like a sine function, except with alternating line segments.

    The t-axis runs horizontally through the middle and the v-axis runs vertically so that the velocity is 0 at t = 0. How do I get this sort of graph.
  2. jcsd
  3. Feb 22, 2008 #2
    Force is the (negative) derivative of potential with respect to position. So, you'll be able to find acceleration as a function of position. Then, do some chain rule stuff and you should be able to find velocity as a function of time.

    edit: oh hey, you're the guy that helped me with my (other) gauss's law post.

    edit2: oh wait, that just rederives K = -U.
    Last edited: Feb 22, 2008
  4. Feb 23, 2008 #3
    I did what you said with a little bit of calc, but I ended up getting a linear function for velocity. The linear function does, however, portray the velocity function over selective intervals.
    Guess I'll shift the question to something else: How would I mathematically derive the equations for the sinousidal functions of Simple Harmonic Motion?
  5. Feb 23, 2008 #4
    It's easy to find velocity as a function of position, but I'm not sure how to find it as a function of time. But, SHM (for springs) is derived like this:

    [tex]F = ma => \frac{d^2x}{dt^2} = \frac{-k}{m} x[/tex]

    In other words, the second derivative of position is proportional to position. The only functions that satisfy this property are the sine and cosine functions. Solving a second-order differential equation is not simple, so introductory physics books derive it this way.
  6. Feb 23, 2008 #5
    oh....I dont know how to solve for second-order differentials.
    Last edited: Feb 23, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook