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Homework Help: Harmonic motion of the dipole?

  1. Sep 10, 2008 #1
    the conservation of energy formula for a dipole

    E-field going parallel to the y-axis. if the positive end of the dipole was in the first quad with angle theta from the y axis and the negative end of the dipole was in the third quad, what would be the angular velocity of the dipole at the y-axis? using theta0= theta or the angle the dipole is released.U(pi/2)=0. And what is the period of the harmonic motion of the dipole?

    The answer is w=[tex]\sqrt{(2qED/I) * (1-cos(theta0)}[/tex]
    1/2Iw^=qEDcos(theta). then to find the answer the change in potential energy has to go to theta0 to pi/2?? Do this work?? How do you prove the U(pi/2)=0 since at pi/2 the potential energy from theta0 has turned to kinetic energy? qED(cos(pi/2)-cos(theta0))= 0???

    For the period, The harmonic formula d^2x/dt^2=-w^2x replacing x with theta d^2theta/dt^2 = w*theta
    and since I*angular acceleration= torque and torque=qEdsin(theta) but my teacher changed torque to qED*theta. Why??
    Then I just used I*d^2theta/dt^2 = torque or I*-w^2*theta=torque to find w. And w=2pi/T to find T.

    Why was the torque changed from qEdsin(theta) to qEd*theta????? And is the harmonic formula d^2x/dt^2=-w^2x the same no matter what?????

    Thanks for the help.
    Last edited: Sep 10, 2008
  2. jcsd
  3. Sep 10, 2008 #2
    Re: Dipoles

    I have a test on this tomorrow, so could someone please look at the equations and the questions about how to prove U(pi/2)= 0 and why qEDsin(theta) turned into qED*theta and if the second derivative is always equal to -w^2*x for harmonic motion?

    Thank you.
  4. Sep 10, 2008 #3
    Re: Dipoles

    bump.. Can anyone answer my questions?

    Please! I do not understand why qEDsin(theta) turned into qED*theta
  5. Sep 11, 2008 #4


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    Homework Helper

    Re: Dipoles

    Hi StephenDoty,

    The approximation [tex]\sin\theta\to \theta[/tex] is used when the angle [itex]\theta[/itex] is assumed to be small.

    I don't understand what you are asking about [itex]U(\pi/2)[/itex]. The formula for [itex]U[/itex] shows that it is zero at the angle [itex]\pi/2[/itex]. However, the dipole never gets to that angle, so I'm not sure what you are asking.
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