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Homework Help: Calculating time period of oscillation

  1. Jun 20, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=59723&stc=1&d=1371745489.png


    2. Relevant equations



    3. The attempt at a solution
    (see attachment 3)
    If the middle charge is moved a y distance, then the other two move a distance y/2 in opposite direction. Similarly, the velocity in y direction of other two can be also calculated. As the rods are rigid, the component of velocities along the rod should be equal i.e
    [tex]v_x\cos \theta-\frac{v}{2}\sin\theta=v\sin\theta \Rightarrow v_x=\frac{3v}{2}\tan\theta[/tex]
    Calculating the energy of system and differentiating it w.r.t time and setting the derivative equal to zero gives the time period.
    [tex]E=\frac{kq^2}{l}+\frac{kq^2}{l}+\frac{kq^2}{2l\cos\theta}+\frac{1}{2}mv^2+2 \times \frac{1}{2}m(v_x^2+\frac{v^2}{4})[/tex]
    [tex]\Rightarrow E=\frac{kq^2}{l}+\frac{kq^2}{l}+\frac{kq^2}{2l\cos\theta}+\frac{3}{4}mv^2+\frac{9mv^2}{4}\tan^2\theta[/tex]
    Since y<<<l,
    [tex]\frac{1}{\cos\theta}=1/(\sqrt{1-\frac{9y^2}{4l^2}})=1+\frac{9y^2}{8l^2}[/tex]
    I don't understand how should I write ##\tan^2\theta##
    [tex]\tan\theta=\cfrac{\cfrac{3y}{2}}{\sqrt{1-\frac{9y^2}{4l^2}}}[/tex]
    [tex]\tan^2\theta=\frac{9x^2}{4}\left(1-\frac{9y^2}{4l^2}\right)^{-1}[/tex]
    [tex]\Rightarrow \tan^2\theta = \frac{9x^2}{4}\left(1+\frac{9y^2}{4l^2}\right)[/tex]
    But I don't think substituting this in the energy expression is a good idea because I will end up with terms consisting ##v^2x^2## and ##v^2x^4##.

    Any help is appreciated. Thanks!
     

    Attached Files:

    Last edited: Jun 20, 2013
  2. jcsd
  3. Jun 20, 2013 #2
    Oscillations are found for small displacementas. Thake the displacement of middle charge dy. Assume that other two charges don't move.
    Then proove that it oscilates in Simple Harmonic motion.
     
  4. Jun 20, 2013 #3
    What? I had to solve it this way? LOL, I was over thinking on the problem.

    Okay, for this type of oscillation, I get:
    [tex]T=2\pi\sqrt{\frac{ml^3}{2kq^2}}[/tex]

    Is this what you get?
     
  5. Jun 20, 2013 #4
    Well, if i don't know whether there are other ways in which it can osscilates. Your diagrams do not seem to agree with this type of osscilation. But if it oscilates like this then i am also getting same answer as yours.
     
  6. Jun 20, 2013 #5
    I plugged in the values and got T=0.0444 but this is wrong.
     
  7. Jun 20, 2013 #6
    I think taking yhe other two charges fixed was wrong. Take them movable, it still is the SHM.

    Is the answer pi/10?
     
  8. Jun 20, 2013 #7
    How do you get ##\pi/10## (I don't know about the answer)? :confused:

    If middle charge moves a distance dy, the other two moves a distance dy/2. That will result in a time period of
    [tex]T=2\pi\sqrt{\frac{ml^3}{3kq^2}}[/tex]
     
  9. Jun 21, 2013 #8
    Look at the diagram given in question, according to that if middle charge moves a distance dy upwards the other two charges mobe a distance dy downwards(as l is constant). Why are you getting dy/2?
     
  10. Jun 21, 2013 #9
    Conservation of linear momentum.

    Is it wrong to use it here? :confused:
     
    Last edited: Jun 21, 2013
  11. Jun 21, 2013 #10
    I am also confused about taking conservation of linear of linear momentum. It violates symmetry and the diagram as there can be other interactions as well I think the the better method would be taking the potential energy as function of 'y' and the finding force. But it is more or less the same.

    I think we need help from senior members. I have asked ehild!
     
  12. Jun 21, 2013 #11
    I have not watched the entire discussion, but: have you been able to express the potential and kinetic energies solely in terms of theta and the given constants?

    When that is done, linearize the equation expressing conservation of energy assuming theta is small.

    It should then be fairly simple to proceed.
     
  13. Jun 21, 2013 #12
    voko, please check the OP, I have already stated why I am not able to proceed. I have used the same method you are stating (I think). Thanks! :)
     
  14. Jun 21, 2013 #13
    I do not see any attempt or any problem in #1 in expressing ##v## via ##\theta## and ##\dot{\theta}##.
     
  15. Jun 21, 2013 #14
    [tex]\sin\theta=\frac{3y}{2l}[/tex]
    [tex]x=\frac{2l\sin\theta}{3}[/tex]
    Differentiating w.r.t time
    [tex]v=\frac{2l\cos\theta}{3} \dot{\theta}[/tex]

    Is this what you ask me?
     
  16. Jun 21, 2013 #15
    Assuming it is correct, yes (sorry, no time to check that now).

    Now you have energy in terms of the angle and the angular velocity. Assuming the angle is small, derive an equation quadratic (at most) in them, and that should give you what you want.
     
  17. Jun 21, 2013 #16
    For small angle, ##tan^2\theta=\theta^2## and ##\cos\theta=1-\theta^2/2##, right?

    Please check my energy equation in the first post when you have time, thank you.
     
  18. Jun 21, 2013 #17

    TSny

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    Can you argue that the last term on the right is of higher order in the small quantities θ and v and can therefore be neglected?

    Everything looks good to me.
     
  19. Jun 21, 2013 #18
    I initially thought of neglecting them but I wasn't too sure. I think its okay to do that.

    Differentiating the energy equation w.r.t time and setting the derivative equal to zero, I get
    [tex]T=2\pi\sqrt{\frac{4ml^3}{3kq^2}}[/tex]
    Is this what you get?
     
  20. Jun 21, 2013 #19

    TSny

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    Yes.
     
  21. Jun 21, 2013 #20
    Thanks a lot TSny! :smile:
     
  22. Jun 21, 2013 #21

    ehild

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    A slightly different approach:

    The problem can be written entirely in terms of θ:

    x1=-Lcosθ, y1=Lsinθ/3
    x2=Lcosθ, y2=Lsinθ/3
    x3=0, y3=-2Lsinθ/3 (middle atom)

    These displacements leave the CM stationary.

    The velocity components are: x1'=Lsinθθ', y1'=L/3 cosθθ' ...

    KE= mL2θ'2(sin2(θ)+cos2(θ)/3)

    PE=2kq2/L+kq2/(2Lcosθ)

    Supposing maximum angular speed θ' was given initially at θ=0. The KE is zero when the angle is θmax=A

    Initial energy: mL2θ'2(1/3)+2kq2/L+kq2/(2L)

    Final energy: 2kq2/L+kq2/(2LcosA)

    cosA≈1-A2/2 if A is small. 1/(1-A2/2)≈1+A2/2


    So we get : mL2θ'2(1/3)=kq2/(2L)A2/2
    θ'2=3/4 kq2/(L3m)A2 where θ' is the maximum angular speed.

    For an SHM, maximum speed = amplitude * ω.

    ehild
     
  23. Jun 21, 2013 #22
    Thank you ehild! This is much better. :smile:
     
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