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: 2D Harmonic Motion

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle undergoes simple harmonic motion in both the x and y directions simultaneously. Its x and y coordinates are given by
    x=asin(wt)
    y=bcos(wt)

    a. Eliminate t from these equations to show that the path of the particle in the x-y plane is an ellipse.
    b. Calculate the kinetic and potential energy at a point on the particle’s trajectory. Show that the ellipse is a path of constant total energy, and show that the total energy is given by the sum of the separate energies of the x and y oscillations.

    2. Relevant equations
    Equation of an ellipse: (x/a)^2+(y/b)^2=1
    KE = 1/2=mv^2
    PE = ??

    3. The attempt at a solution
    a) I rearranged the parametric eqns. like so: x/a=sinwt and y/b=coswt, squared both sides and using trig identities eliminated the term (sin^2wt+cos^2wt) leaving 1 on the right side and therefore getting the equation of an ellipse: (x/a)^2+(y/b)^2=1
    b)This is where I'm having some troubles... I am pretty sure that I can calculate the KE of the particle by looking at:
    KE(x)= 1/2*mv(t)^2
    = (1/2)mw^2(acoswt)^2

    KE(y)= -(1/2)mw^2(bsinwt)^2

    but I don't know how to come up with the equations for the potential energy.

    Regrading the fact that the ellipse is a path of constant total energy, I am thinking of showing that when the particle is on the x axis it means that y=0 thus eliminating the y component of the energy, same goes for when its on the y axis. which essentially can be modelled with a cosine function where energy is always constant but changes from component to component.

    I'd appreciate any help or some input on my attempt at the solution.
    Thanks!
     
  2. jcsd
  3. Sep 26, 2010 #2

    ehild

    User Avatar
    Homework Helper

    You can find the acceleration and the force. And remember the definition of potential energy.


    ehild
     
  4. Sep 26, 2010 #3

    Delphi51

    User Avatar
    Homework Helper

    Your Ek calc looks good, except the Eky should be positive. I combined Vx and Vy with the Pythagorean theorem, then found the energy and got the sum of your Ekx and Eky - so between us we have the last part of (b). Time should be eliminated to give Ek as a function of x and y:
    Ek = ½mw²(a²y²/b² + b²x²/a²)

    The potential energy is mysterious. What causes it? There apparently is no charge so must it be gravitational potential energy?
    If so, any calculation of it will have a factor of "g" in it. So it will not combine with the Ek expression to make a constant. Looks like a dead end to me.
     
  5. Sep 26, 2010 #4
    @ehild: you mind expanding a bit about the "acceleration and the force" you talk of?
    @Delphi51: Yea I got to the same conclusion regarding the Ek. It seems that the Ep is due to gravity but I'm really not sure how to write the equation.. been trying to work it out with a circle, which seems to work (U=mgl(theta)^2/2) but it's not really related to an ellipse or what the question asks for...

    Any more ideas?
     
  6. Sep 26, 2010 #5

    Delphi51

    User Avatar
    Homework Helper

    Ehild has a brilliant insight! Forget gravity, you're given the acceleration so you can find the force directly without knowing what caused it.
    I have it almost working, just out by a sign from getting a constant Ek + Ep. Find ax and ay, Fx and Fy. Integrate each to get work in each direction, add them to get total Ep. I got an expression with constants and x², y² just like the Ek. Totalling them and then using the original expressions for x and y as functions of time to eliminate x and y, I get a function of time that is some constants times cos²(wt) - sin²(wt). If you can do all that and get rid of that minus sign, it adds up to 1 times the constants.
     
  7. Sep 26, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You know the particle is undergoing simple harmonic motion. What type of force causes simple harmonic motion, i.e., can an arbitrary force do it, or does it have to vary in a specific way?
     
  8. Sep 26, 2010 #7
    I completely forgot to look at the relation between work done and the potential energy!
    so I found Fx= m*ax= -maw^2sin(wt) and Fy= m*ay=-mbw^2cos(wt).
    But how do I do the integration?
     
  9. Sep 26, 2010 #8

    ehild

    User Avatar
    Homework Helper

    You can follow two ways: Make a vector equation:

    [tex]\vec F = m \vec a =-m\omega^2\vec r[/tex]

    How do you get the potential energy if you know the force in terms of the position vector?

    ehild
     
  10. Sep 26, 2010 #9
    so I make my acceleration vector a = sqrt((ax)^2+(ay)^2) and after some rearranging and factoring I get a = w^2 sqrt(x^2+y^2). But what do I do with that result now? How do I reach the expression of PE?
     
    Last edited: Sep 26, 2010
  11. Sep 26, 2010 #10

    Delphi51

    User Avatar
    Homework Helper

    I think it is better to keep your x and y accelerations separate.
    Figure out the work done in each direction, then add them.
    To find work you must integrate dW = Fx*dx and dW = Fy*dy.
    If Fx is a function of x only, it is easy integration. Strictly speaking, you should integrate from x = 0 to some value of x, but the evaluation at x = 0 is zero, so the integral of
    Fx*dx = -maw²*sin(wt)*dx = -mw²x*dx is -mw²x²/2
    Check this very carefully; remember I lost a minus sign at least.
    The PE is the total of the work done in the x direction and the work done in the y direction.
     
  12. Sep 27, 2010 #11

    ehild

    User Avatar
    Homework Helper

    The vector of acceleration is

    [tex]
    \vec a =-\omega^2(a\sin(\omega t) \vec i +b\cos(\omega t )\vec j ) = -\omega^2\vec r.
    [/tex]

    The potential energy is equal to the work done by the force when the body moves from the selected point (x,y) to the one where the potential energy is 0. Let be this point at r=0, and the integration path a radius from point P(x,y) to (0,0). The work done is the integral of (-w^2 m rdr)=w^2 m r^2/2.

    I wanted to write it the formulae in tex, but I did not succeed...

    ehild
     
    Last edited: Sep 27, 2010
  13. Sep 27, 2010 #12

    ehild

    User Avatar
    Homework Helper

    There is an other way to handle the problem. The motion is the superposition of two independent oscillation, one along x, the other along y. The energy in this case is the sum of the energies assigned to both motions. What is the energy of a body performing simple harmonic motion, in terms of its amplitude and angular frequency?

    ehild
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook