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Two masses and spring oscillation

  1. May 22, 2014 #1
    1. The problem statement, all variables and given/known data

    The ratio of the time periods of small oscillation of the insulated spring and mass system before and after charging the masses is

    (a) ≥ 1
    (b) > 1
    (c) ≤ 1
    (d) = 1


    2. Relevant equations



    3. The attempt at a solution

    First I calculated the time period of neutral masses .

    Let us consider origin to be at the left of m1 .Take the x axis to be along the length of the spring and call the left mass be m1 at coordinate x1 and right mass be m2 at coordinate x2 .

    x=x2-x1-L

    d2x/dt2 = d2(x2-x1)/dt2

    EOM for mass m1 is m1d2x1/dt2 = kx

    EOM for mass m2 is m2d2x2/dt2 = -kx

    After manipulating the above two equations we arrive at d2/dt2 + (k/μ)x = 0 ,where μ = m1m2/(m1+m2) .

    The time period is 2π√(μ/k) .

    Now when the masses are charged ,

    EOM for mass m1 is m1d2x1/dt2 = kx - keq2 / (x+L)2

    EOM for mass m2 is m2d2x2/dt2 = -kx + keq2 / (x+L)2 .

    Is this the correct way to approach the problem ? If yes ,how should I proceed .
     

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    Last edited: May 22, 2014
  2. jcsd
  3. May 22, 2014 #2

    NascentOxygen

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    This is just one of a few dozen multi-choice questions on an exam paper? If that's the case, there is probably not a lot of calculation to be done.

    You are not told whether in this test the weights are changed from a smaller mass to a larger, or whether it's vice-versa? So even if the period were observed to change, how could you say whether the corresponding ratio is >1 when it might be <1 ?

    So, purely on the absence of information allowing you to decide on any contrary option, it seems you are confined to only one answer.

    spelling: probably "isolated" spring, and after "changing" :smile:
     
    Last edited: May 22, 2014
  4. May 22, 2014 #3

    ehild

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    The potential consists of an elastic part and a Coulomb-part. Expand the potential function into Taylor series about the equilibrium point. Keep up to the second order term. From the equilibrium condition and from the coefficient of the second-order term, you get the equivalent force constant.


    ehild
     
  5. May 22, 2014 #4
    Have I done it correctly in the OP ? Have I correctly written the EOM for the two charged masses ?Could you help me proceed using the force method .
     
  6. May 22, 2014 #5
    No, they really meant insulated and charging. Look at the picture. The question is whether charging the masses with electric charges q affects the osculation period.
     
  7. May 22, 2014 #6

    ehild

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    It looks correct. Expand the force about the equilibrium point.

    ehild
     
  8. May 22, 2014 #7

    NascentOxygen

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    Big oops! I guess I should click on the thumbnail. :redface:
     
  9. May 22, 2014 #8
    I am not sure what do we mean by equilibrium point in this two body problem ?

    At equilibrium keq2/(x2e-x1e)2 = k(x2e-x1e-L) ,where x2eand x1e are respective equilibrium positions .

    Or, keq2/xe2 = k(xe-L) ,where xe is the separation between the masses at equilibrium.

    Is it correct ?


    EOM for mass m1 is m1d2x1/dt2 = kx - keq2 / (x+L)2

    EOM for mass m2 is m2d2x2/dt2 = -kx + keq2 / (x+L)2 .

    Do I need to expand the 2nd term on the RHS about xe ?
     
    Last edited: May 22, 2014
  10. May 22, 2014 #9

    ehild

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    Yes, and write the differential equation for x-xe.

    ehild
     
  11. May 22, 2014 #10

    vela

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    Do you know about reduced mass and converting a two-body problem into a one-body problem?
     
  12. May 22, 2014 #11
    Hello Vela

    Please check my understanding about reduced mass .

    In the two body problem ,just as in case 1 where uncharged masses are present and the two masses are under the action of internal spring force ,we may consider mass 1 to be stationary (or treat it as origin) and replace mass 2 with reduced mass μ = m1m2/(m1+m2) .The relative separation 'x' works as the x coordinate of the 2nd mass .

    Is this correct ?

    I initially intended using the concept of reduced mass ,but there may be something wrong with my implementation . Please have a look at the OP where I have found the time using reduced mass .

    What is the correct way of converting this two body problem in one body problem ?
     
  13. May 22, 2014 #12

    vela

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    Yes, that's correct, and the differential equation you derived reflects just what you said. The point is, you didn't need to rederive this result. You could have gone straight to
    $$\mu \frac{d^2 x}{dt^2} = -kx.$$ For the charged masses, tack on the electric force and expand to first order to find the effective spring constant.
     
  14. May 22, 2014 #13
    $$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{(x+L)^2}$$

    Is this correct ?
     
  15. May 22, 2014 #14

    vela

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    Looks good.
     
  16. May 22, 2014 #15
    When x<<L , $$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{L^2}(1-\frac{2x}{L})$$

    $$\mu \frac{d^2 x}{dt^2} +(k+\frac{2k_eq^2}{L^3})x = k_e\frac{q^2}{L^2}$$

    Here effective spring constant ##K = k+\frac{2k_eq^2}{L^3}## and time period is ## T = 2\pi \sqrt{\frac{\mu}{K}}##

    Is this correct ?

    Edit :Fixed errors
     
    Last edited: May 22, 2014
  17. May 22, 2014 #16

    vela

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    Right idea, but you made some algebra mistakes.
     
  18. May 22, 2014 #17
    OK..I have fixed the errors .Does post#15 make sense ?
     
  19. May 22, 2014 #18

    vela

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    Yup.
     
  20. May 22, 2014 #19
    OK...

    Can I conclude that the time period of oscillation of charged masses is less than that of the uncharged masses i.e the correct option is b) >1 ?
     
    Last edited: May 22, 2014
  21. May 23, 2014 #20

    ehild

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    Yes.

    ehild
     
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