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Time Period of SHM

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Two Springs are present (one just infront of the other). The Spring towards the left has +Q charge and towards the right -Q charge (at their ends).The distance between the two charges is d. The Springs are of length l. Find the Time Period of the Simple Harmonic Motion if the charges are of same mass. ( l > d )

    (Wall)-->(Spring)-->+Q -Q<--(Spring)<--(Wall)

    2. Relevant equations

    F(elec)=(k Q^2) / (d^2) where k=(1/4)∏ε
    F(Spring)=( Kl )

    3. The attempt at a solution

    I know the above two equations, but cant proceed. Is there any other force too? I cant figure out why will the charges move back again? I'm having two confusions

    1) The charges are opposite so they will attract each other. When they reach a certain point they will collide (as l > d ) and move back. Is this the reason why they move back? What other equation do i have to use?

    2) Is it the Spring force will pulls the charges back before they collide. But it shouldn't be true as ( l > d ) and electrostatic forces are very strong and spring force cannot overcome it. Am i right? so how should i proceed

    Pls help me....
  2. jcsd
  3. Oct 14, 2011 #2


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    Homework Helper

    You can assume that the system is in equilibrium initially, and then you give a little push to the masses. For a simple harmonic motion, the displacement of the charged masses from their equilibrium positions must be small with respect to the distance between them. Find the time period of small oscillations with this assumption. Do not forget that the springs are connected to two opposite walls, so the sum of the spring lengths and the distance between the masses is constant.

  4. Oct 15, 2011 #3
    But what about the electrstatic force. It constantly changes as the distance between the charges change. how to incorporate that. I know by intergrating the force, but then what to do?
  5. Oct 15, 2011 #4


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    Homework Helper

    Do Taylor expansion of the Coulomb force around the equilibrium position and keep the constant and first-order terms.

  6. Oct 15, 2011 #5
    I havent yet learnt Taylor expansion. So is there any other method ?
    I actually found this question in a magazine. The answer page was torn. So i dont know the answer too.
    I have tried solving it 8-10 times but with no success !! (cant reach the final expansion)
    I found this sum interesting so I picked it up
    I would be helpful to me if you can solve the sum or give me the equations to be solved
    I would be learning things both ways.
  7. Oct 15, 2011 #6


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    Homework Helper

    Sorry, I am not allowed to solve problems. I can only help.

    It is useful to learn how to calculate with small quantities.
    Suppose you have to calculate (1+a)2, where a<<1. Decomposing the square, (1+a)2=1+2a+a2. If a<<1 you can ignore the square of it at approximate (1+a)2≈1+2a. Calculate 1.0012and compare it with 1+2*0.001.

    Suppose you have a fraction, 1/(1+q) and q << 1.

    1/(1+q) is equal to the sum of the geometric series 1-q+q2-q3+... =1/(1+q)
    If q<<1 you can ignore the terms with second or higher power, and use the approximation 1/(1+q)=1-q.
    Try to calculate 1/1.001 and compare it with 1-0.001.

    Here you have the Coulomb force of form A/(R+Δr)2.
    Factor out R: You get ( A/(R2) (1/(1+Δr/R)2.
    Assume that Δr/R<<1. Try to apply the approximations above.

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