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Simple harmonic motion of a charged particle in a rod

  1. Dec 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Two points, each of charge Q, are fixed at either end of a frictionless rod of length 2R. Another point charge, of charge q (not Q) is free to move along the rod.
    Show that if charge q is displaced a small distance x (0<x<<R) from the centre of the rod, it will undergo simple harmonic motion.
    Derive an expression for the period of oscillation.


    2. Relevant equations



    3. The attempt at a solution
    Uhh.. I really have no idea how to start this.
    I could use F=(kQq)/R2 (ignoring x due to its small size relative to R) to find the force between the particles, but I would have no idea how to relate this to SHM
     
  2. jcsd
  3. Dec 1, 2013 #2

    WannabeNewton

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    Hi Sam! You can't ignore ##x## like that. When we say ##x << R## what we mean is that we want our equations to be valid up to first order in ##\frac{x}{R}##. So write down the Coulomb force from the fixed charges at the respective ends of the rod on the charged particle that's displaced to ##x## and Taylor expand the expression to first order in ##\frac{x}{R}## (i.e. apply the binomial expansion and ignore terms of 2nd order and higher in ##\frac{x}{R}##).
     
  4. Dec 1, 2013 #3
    So, at any given point the force on charge q would be F1-F2 where
    F1 is the force from the closest end = (kQq)/(R-x) and
    F2 is the force from the closest end = (kQq)/(R+x)
    Would I then combine these to get Fresultant=(2kQqx)/(R2-x2) or am I going about this all wrong?
    Where (and how) would I use the Taylor-series expansion?

    Thanks
     
  5. Dec 1, 2013 #4

    WannabeNewton

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    Don't forget the directions of the forces! And you don't need to combine the two terms, you can just apply the binomial expansion (same as Taylor expansion in this case) to each term separately (again to first order in ##\frac{x}{R}##).
     
  6. Dec 1, 2013 #5
    So the expansion for F1 would leave me with (kQq/R)+(kQqx)/2R, and for F2 would be the same but with a minus sign in the middle.

    What the would I do with this.
    When I combine them I would be left with Fresultant=(kQqx)/R
    Is this correct? How would I relate this to simple harmonic motion?
     
  7. Dec 1, 2013 #6

    WannabeNewton

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    Hey Sam, it looks like you missed an overall negative sign somewhere. Let's fix the origin of our coordinate system to the center of the rod with ##+\hat{x}## pointing to the right and displace the particle ##x## to the right (no loss of generality in doing this for the given system). As you wrote before, the charge ##Q## to the right exerts a force ##\vec{F}_1 = -\frac{kqQ}{(R - x)^2}\hat{x}## whereas the charge ##Q## to the left exerts a force ##\vec{F}_2 = \frac{kqQ}{(R + x)^2}\hat{x}## so the magnitude of the net force on ##q## is ##F = \frac{kqQ}{(R + x)^2} -\frac{kqQ}{(R - x)^2}##. It seems you got all that already so let's move on from there.

    I will rewrite the expression for ##F## as ##F = \frac{kqQ}{R^2}(1 + \frac{x}{R})^{-2} -\frac{kqQ}{R^2}(1 - \frac{x}{R})^{-2}##. Now Taylor expand this to 1st order in ##\frac{x}{R}## and be careful with signs! After that, use Newton's 2nd law.
     
  8. Dec 1, 2013 #7
    So I come out with the net force F=-(2kqQx)/R3 This agrees with Simple Harmonic Motion as the net force (the restoring force) is directed towards the equilibrium position. Am I right in saying that it also agrees with SHM as when used with F=ma, acceleration is at a maximum when x is also at a maximum (the maximum amplitude) and acceleration is 0 at the equilibrium position?

    How would I go about deriving an expression for the time period?
    This is what I think of when it comes to SHM:
    As it is starting from an initial (maximum) displacement of A:
    displacement: x = Acos(ωt)
    Velocity: v = -Aωsin(ωt)
    Acceleration: a = -Aω2cos(ωt)
    Then, from F=ma, I could equate the two expressions for a? And acceleration would be max when t=time period, so cos(ωt)=1
     
  9. Dec 1, 2013 #8

    WannabeNewton

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    You're making this harder for yourself than it needs to be :)

    From Newton's 2nd law we have ##\ddot{x} = -\frac{2kqQ}{mR^3}x##. Recall that the equation of SHM is ##\ddot{x} + \omega^{2}x = 0## where ##\omega## is the angular frequency of the oscillation. Is it clear what to do from here?
     
  10. Dec 1, 2013 #9
    Just substitute in :)
    Use ω=2∏/T and rearrange.
     
  11. Dec 1, 2013 #10

    WannabeNewton

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