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Homework Help: Simple Harmonic Motion

  1. Dec 27, 2013 #1
    1. The problem statement, all variables and given/known data

    The masses in figure slide on a frictionless table.m1 ,but not m2 ,is fastened to the spring.If now m1 and m2 are pushed to the left,so that the spring is compressed a distance d,what will be the amplitude of the oscillation of m1 after the spring system is released ?

    2. Relevant equations

    3. The attempt at a solution

    I feel if I can find the position and velocity where m1 loses contact with m2 ,i.e N=0 ,then new amplitude of oscillation can be obtained .

    Let the origin be at m1 when the spring is unstretched.

    The equation of motion for mass m1 is -kx-N = m1d2x/dt2 ,where N is the Normal contact force between the blocks and x is the displacement of the block.

    I would be grateful if someone could help me with the problem.

    Attached Files:

    Last edited: Dec 27, 2013
  2. jcsd
  3. Dec 27, 2013 #2
    When a mass is oscillating on a spring, in some positions the spring is pushing, and in some it is pulling. Obviously, as soon as the spring starts pulling, the masses must separate.
  4. Dec 27, 2013 #3
    Does that mean the blocks separate when the spring acquires its natural length ?
  5. Dec 27, 2013 #4
    You explain what that means :)
  6. Dec 27, 2013 #5
    How can i get the result ( i.e when the blocks separate ) mathematically ?

    -kx-N = m1d2x/dt2

    -kx-N = m1vdv/dx

    Integrating,I get m1v2/2 = (k/2)(x+d)(d-x-N)

    Should I put N=0 here ?
    Last edited: Dec 27, 2013
  7. Dec 27, 2013 #6
    I do not think you need to derive anything mathematically here, because the answer is obvious.

    If you must, first consider the motion of the center of mass of the blocks. It is affected only by the external spring force.

    Then, once you know this motion (and you also know that before separation the blocks are in contact), you can have equations for each block, which will have their interreaction force. It should be zero at the separation point, and with that you can locate the separation point.
  8. Dec 27, 2013 #7
    EOM of CM : -kx=(m1+m2)d2x/dt2

    EOM of m1 : -kx-N = m1d2x/dt2

    EOM of m2 : N = m2d2x/dt2

    Are these equations correct ? If yes , what should I do next ?
  9. Dec 27, 2013 #8
    Before and just at the separation, x is the same in all three equations. At separation, N = 0. Solve for x.
  10. Dec 27, 2013 #9
    Could you please give your response to post#5 .
    Last edited: Dec 27, 2013
  11. Dec 27, 2013 #10
    #5 is wrong. N is not constant, so you cannot integrate like that.

    But from the equations in #7, you can find N as a function of x if you want to.
  12. Dec 27, 2013 #11
    Right...Thanks !!!

    Okay...so using EOM for m1 and m2 ,N=-kxm2/(m1+m2)

    Now putting N=0 → x=0

    Is this correct ?
  13. Dec 27, 2013 #12
    To begin with, both masses will be moving together. At what point in time will the contact force be equal to zero. After that, m1 will slow down faster than if they had been stuck together, while m2 will continue with its existing velocity. How does the motion of m1 vary after that time?
  14. Dec 27, 2013 #13
    Yes, x = 0 is correct.
  15. Dec 27, 2013 #14
    Hi Chet :smile:

    How should I find the time at which the contact force goes to zero ? Which equation should I use ?
  16. Dec 27, 2013 #15
    What is the significance of considering the motion of CM ?
  17. Dec 27, 2013 #16
    The contact force goes to zero the first time that d2x/dt2 goes to zero. After that, m1 and m2 separate.
  18. Dec 27, 2013 #17
    So,i need to solve m2d2x/dt2 = 0 .

    Right ?
  19. Dec 27, 2013 #18
    I think you obtained the equation for the inter-block force from considering the CM motion and the motion of one block. That should answer your question.
  20. Dec 27, 2013 #19

    I considered EOM of m1 and m2 .Knowledge of CM motion was not required .
  21. Dec 27, 2013 #20
    Ah, OK. The equation for CM is not strictly required, but I was not sure how you were going to solve the problem. It seemed to me that you wanted to obtain an explicit solution for x, which is very simple if you consider the CM.
  22. Dec 27, 2013 #21
    You already showed that that happens the first time x = 0.
  23. Dec 27, 2013 #22
    Okay...I thought you were referring to some alternative approach by finding the time when N goes to 0.

    I really like the way you approach the problems.Very neat mathematical ways .No guessing games :smile:
  24. Dec 27, 2013 #23
    If I need a solution for displacement(x) as a function of time(t) then I need to solve the second order DE d2x/dt2+[k/(m1+m2)]x = 0

    Is it correct?
  25. Dec 27, 2013 #24
  26. Dec 27, 2013 #25
    voko...Thank you very much for your valuable guidance.
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