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Dipole in electric field

  1. Feb 10, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A molecule with electric dipole moment ##\underline{p}## is initially aligned in an electric field ##\underline{E}## . If this molecule is perturbated from its equilbrium position by a small angle, show that it will perform simple harmonic motion.

    Calculate the frequency of this motion in terms of p, E and I

    3. The attempt at a solution
    What I did first was write the angular EOM for the dipole: Consider it perturbed at some angle ##\theta##. This gives a torque due to the force by the electric field about the centre of the dipole. I could also consider the torque due to gravity, however, I took gravity to be acting through the centre of mass of the dipole and since I take the torques about the centre, I can ignore it. My final expression is $$I \alpha = -pE\sin \theta\,\Rightarrow\,I \alpha = -pE \theta,$$ if I take the dipole moment p = dq, sinθ ≈ θ for small θ and the -ve because the torque acts to lower θ.

    Is it enough to say from here that since this is in the form ##\tau = -k \theta,## with ##k = pE##, then the motion is simple harmonic?

    If so, I can say that $$T = 2\pi \sqrt{\frac{I}{k}}\,\Rightarrow\,f = \frac{1}{2 \pi}\sqrt{\frac{k}{I}}\,\Rightarrow\,f = \frac{1}{2\pi} \sqrt{\frac{pE}{I}}.$$ The dimensions check but have I made appropriate assumptions etc and is it okay to state we have the form ##\tau = -k\theta## so SHM applies?
    Many thanks
     
  2. jcsd
  3. Feb 11, 2013 #2

    SammyS

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    The equation, [itex]\displaystyle \ \ I \alpha = -pE \theta\,, \ [/itex] is enough to show Simple Harmonic Motion.

    It's equivalent to the differential equation, [itex]\displaystyle \ \ I \frac{d^2\theta}{dt^2} = -pE \theta\ .[/itex]
     
  4. Feb 11, 2013 #3

    CAF123

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    Thanks SammyS.
    I realised I could express my eqn in the form $$\ddot{\theta} + \frac{pE}{I} \theta = 0,$$ which obviously has sin and cos as solutions.

    One further question I have is that when I solve this eqn I get $$\theta = A\cos \left(\sqrt{\frac{pE}{I}}\right)t + B\sin \left(\sqrt{\frac{pE}{I}}\right)t $$,

    How do I know that ##\omega## (angular freq)is necessarily the argument of sin and cos? I seem to be taking it for granted.
     
  5. Feb 11, 2013 #4

    SammyS

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    What is the period of [itex]\displaystyle \ \cos \left(\sqrt{\frac{pE}{I}}\ t\right)\ ? [/itex]
     
  6. Feb 11, 2013 #5

    CAF123

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    It has period $$\frac{2\pi}{\left(\sqrt{\frac{pE}{I}}\right)}$$ I see how the result follows. Thanks again.
     
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