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Homework Help: Find time period of SHM

  1. Sep 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the time period of the mass m as shown in the spring-block system. Neglect air resistance and pulley and springs are massless. All the three springs shown have spring constant k .


    2. Relevant equations



    3. The attempt at a solution
    I am not really sure how to start with this. If the block is displaced by x, the springs ##S_1## and ##S_2## also elongate by x. But how do I calculate the displacement of ##S_0##? I don't know if finding out the displacements is the correct approach.

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Sep 1, 2013 #2
    If the block is displaced by x, the springs S1 and S2 are also elongate by x.

    -How? You are getting a total displacement of 2x if both of them elongate by x.
     
  4. Sep 1, 2013 #3
    Does that mean they elongate by x/2? I am still not sure what to do.
     
  5. Sep 1, 2013 #4
    Life would be so much simpler without just one spring...
     
  6. Sep 1, 2013 #5
    Are you hinting at finding the equivalent spring constant? :uhh:
     
  7. Sep 1, 2013 #6
    Warning proceed at your own risk, I'm not completely sure if I'm correct- lets say 89%
    The S1 and S2 can be considered as a serial combination. Find S.H.M eq of that. Then calculate the force on the pulley, find displacement of S3, Find Shm eq of pulley, add the two eq up- get time period.
    Tell me if it works, too lazy to work it out myself.
     
  8. Sep 1, 2013 #7
    The series combination is equivalent to k/2. If the block is displaced by x, the force in the string connected with ##S_0## is kx. Now how do I find the equivalent, I have a pulley in the set-up. :confused:
     
  9. Sep 1, 2013 #8
    Simple- don't, add both displacement-time equations of the two shm and use some trigo.
    BTW what about gravity?
     
  10. Sep 2, 2013 #9
    Let the downward displacement of the lower end of spring 3 = δ3
    Let the upward displacement of the upper end of spring 2 = δ2
    Let the downward displacement of the upper end of spring 1 = δ1
    Let the downward displacement of the mass m = δm
    Kinematically, how is δ1 related to δ2 and δ3? How is the increased tension (relative to equilibrium) in spring 3, T3, related to δ3? How is the increased tension (relative to equilibrium) in spring 2, T2, related to δ2? How is the increased tension (relative to equilibrium) in spring 1, T1, related to δ1 and δm? How is the tension in spring 2 related to the tension in spring 1?

    Chet
     
  11. Sep 2, 2013 #10
    ##\delta_3=2\delta _1=2\delta _2##?
    Is it wrong to say ##\delta_m=\delta_1##? :confused:
     
  12. Sep 2, 2013 #11
    Not really. If the pulley did not move (δ3=0), then δ1 would be equal to δ2. If the pulley moved down, but spring 2 did not stretch (δ2=0), δ1 would be equal to δ3. But, if the pulley moves down and spring 2 stretches, then δ123.

    If spring 1 stretches, then δm does not equal δ1.

    Chet
     
  13. Sep 2, 2013 #12
    Sorry but I still don't get it. :(

    For example, consider a spring attached to the ceiling and a mass is hanging at the end of spring. If mass descends by ##\delta_m##, then the spring also stretches by ##\delta_m##, why doesn't the same apply here? :confused:
     
  14. Sep 2, 2013 #13
    The upper end of spring 1 is not attached to the ceiling (or any other rigid constraint). It is free to move up and down. And it does move up and down. The amount that spring 1 stretches is amount that the bottom end moves downward minus the amount that the upper end moves downward.
     
  15. Sep 2, 2013 #14

    ehild

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    The springs can stretch independently. The length connecting S1 and S2 stays constant. How does the change of length of an individual spring related to the displacement of the mass? Look at the figure. If the upper spring S3 stretches by δ, the pulley moves downward, the piece of string connected to S2 gets shorter by δ, so the right piece has to get longer by δ. What is the resulting displacement of the block?
    If S2 stretches only, the left piece of the string gets shorter, the right piece gets longer. What about the block?
    It is easy with S1.

    ehild
     

    Attached Files:

    Last edited: Sep 2, 2013
  16. Sep 3, 2013 #15
    The amount that spring 1 stretches is equal to the downward displacement of its lower end (δm) minus the downward displacement of its upper end (δ1). Note that both ends of this spring can move.
     
  17. Sep 3, 2013 #16
    2δ?

    I still have trouble understanding it. For instance, if there were no S1 and S2, the displacement of block would have been 2δ. Correct?
     
  18. Sep 3, 2013 #17

    ehild

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    Yes, if neither S1 nor S2 changed length, the displacement of the block would be 2δ.

    ehild
     
  19. Sep 3, 2013 #18
    Well..now what should I do? :uhh:
     
  20. Sep 3, 2013 #19

    ehild

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    How do you get the displacement of the block if all springs stretch, by δ1, δ2, δ3?
     
  21. Sep 3, 2013 #20
    To expand on what echild wrote in response #17, I suggest you follow the set of steps I outlined in response #9. In response #11, I indicated that δ1 would be related to δ2 and δ3 by:
    δ123
    But, in subsequent postings, echild showed this result to be incorrect, and that the correct relationship should be
    δ12+2δ3
    The amount that spring 3 stretches is δ3. The amount that spring 2 stretches is δ2. The amount that spring 1 stretches is δm1. Follow the rest of the steps in my response #9 to get the tensions and to do the force balances on the pulley and on the mass.

    Chet
     
  22. Sep 4, 2013 #21
    Honestly, I cannot see how you both easily reach the conclusion that δ12+2δ3. But I have tried something else. Please review it.

    In the attachment, I have marked the distances from the ceiling. ##x_{S1T}## is the distance of top end of S1 from the ceiling. Similarly, ##x_{S2T}## is defined. I haven't marked two distances i.e the distances of bottom ends of S1 and S2 because there was not enough space. I call them ##x_{S1D}## and ##x_{S2D}## respectively.

    I now use the fact that length between S1 and S2 and between S1 and m stays constant. Let ##l_1## be the length of spring between S1 and S2.
    $$l_{1i}=x_{S1T}-x_P+x_{S2T}-x_P$$
    When the three spring stretches,
    $$l_{1f}=x_{S2T}-\delta_2-(x_P+\delta_3)+x_{S1T}'-(x_P+\delta_3)$$
    where ##x_{S1T}'## is the distance of top end of S1 when the spring stretches.
    ##\because l_{1f}-l_{1i}=0##
    $$x_{S1T}'-x_{S1T}=\delta_2+2\delta_3$$

    Similarly, I apply this constraint for the string between S1 and m. Let the length be ##l_2##.
    $$l_{2i}=x_m-x_{S1D}$$
    $$l_{2f}=x_m+\delta_m-x_{S1D}'$$
    Again, ##l_{2f}-l_{2i}=0 \Rightarrow x_{S1D}'-x_{S1D}=\delta_m##
    The change in length of spring S1 is ##\delta_1##, hence
    $$x_{S1D}'-x_{S1T}'-(x_{S1D}-x_{S1T})=\delta_1$$
    $$\Rightarrow \delta_m=\delta_1+\delta_2+2\delta_3$$

    Okay, this gives me a relation between elongation of springs and the displacement of block but I still don't end up with the relation you both have posted. :(
     

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  23. Sep 4, 2013 #22

    ehild

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    I haven't posted any relation for δm. Your equation is just what I have suggested by my picture. If S3 stretches by δ3 the mass goes down by 2δ3. If S2 stretches by δ2 the mass moves down by δ2. If S1 stretches by δ1 the mass goes down by δ1. The effect of all springs to the displacement of the mass is $$\delta_m=\delta_1+\delta_2+2\delta_3$$

    Make the free-body diagrams both for the mass and the pulley. As the pulley is massless, T1=T2. Use Hook's law for the tensions, T=kδ for all springs.

    ehild
     
  24. Sep 4, 2013 #23
    That's a very nice and neat way to explain it. It is much easier instead of making the constraint equations. Thank you. :smile:
    From T1=T2, ##\delta_1=\delta_2##. Also, T3=T1+T2, ##\Rightarrow \delta_3=2\delta_1##.
    Hence, ##\delta_m=6\delta_1##. The restoring force acting on m is due to spring i.e T1 and its magnitude is ##k\delta_1=k\delta_m/6##.

    Therefore, the time period is $$2\pi\sqrt{\frac{6m}{k}}$$. Looks correct?

    Thanks a lot ehild and Chet! :)
     
  25. Sep 4, 2013 #24
    For the three tensions, I get:

    T2=kδ2
    T3=kδ3
    T1=k(δm1)

    Therefore, I get:
    δ2m1m2-2δ3
    δ3=δ2+δm1m-2δ3

    Therefore, I get:
    δ3m/3
    δ2m/6
    δ1=5δm/6
    δm1m/6

    This gives the same final result for the tension T1 of the wire on the mass, but the intermediate results differ.
     
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