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Spring-Block system executing SHM in a freely falling elevator

  1. Jul 2, 2013 #1
    A block of mass m is suspended from the ceiling of a stationary elevator through a spring of spring constant k.Suddenly the cable breaks and the elevator starts falling.Show that the block executes S.H.M of amplitude mg/k in the elevator.

    I solved the problem applying a pseudo force mg on the block in the opposite direction to that of the elevator. I would appreciate highly if someone can solve without applying pseudo forces.

    My solution:-As in the equilibrium position,the spring is stretched by mg/k,when we apply mg opposite to that of the weight,it instantly cancels the weight and only force left is kx and that makes it execute in S.H.M and mg/k is the extension when the block is at rest w.r.t. the elevator and so the amplitude.Please solve without pseudo force.Many thanks in advance.
     
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  3. Jul 2, 2013 #2

    tiny-tim

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    welcome to pf!

    hi vijaypandey93! welcome to pf! :wink:
    write out the equations of motion for the mass m, and for the top of the spring (where it is attached to the elevator) …

    show us what you get :smile:
     
  4. Jul 2, 2013 #3
    As the elevator is under free fall n spring is connected to it so i think,the top of the spring is also under free fall,and for the block:mg-kx=ma.a is the accleration of the block which is not equal to g because it's constrained by the spring force.what to do now?
     
  5. Jul 2, 2013 #4

    tiny-tim

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    now write them out, using eg x1 and x2 instead of the same x for both!! :smile:
     
  6. Jul 2, 2013 #5
    bt then we can't have free fall of the ceiling because there'd be two forces working,one Mg(M being the mass of d ceiling) and the other kx1,both downward will get added to produce accleration more than g.what abt this?
     
  7. Jul 2, 2013 #6

    tiny-tim

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    i suspect we're expected to ignore that! :wink:

    (didn't you have to ignore it to use your pseudo force anyway?)
     
  8. Jul 2, 2013 #7
    na,i don't think so because then i am in the elevator's frame of reference in which the ceiling is at rest.
     
  9. Jul 2, 2013 #8

    tiny-tim

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    but if the mass m is not negligible, then the acceleration of your frame isn't constant :redface:
     
  10. Jul 2, 2013 #9
    but in weightlessness spring doesn't apply ny force on the ceiling and can't we treat the free fall of all the bodies separately??
     
  11. Jul 2, 2013 #10

    tiny-tim

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    it jolly well does! :smile:

    it's a spring

    it's what springs do! :wink:
     
  12. Jul 2, 2013 #11
    okay can you pls provide me the mathematical solution of it?
     
  13. Jul 2, 2013 #12

    tiny-tim

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    erm :redface:

    noooo! :rolleyes:
     
  14. Jul 2, 2013 #13

    mfb

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    We do not provide full solutions here.

    If you do not assume that the elevator has an "infinite" mass (large relative to m) the solution will be different. This is easy to see if you use a very small mass (<m, for example) and see how the system will look like.
     
  15. Jul 2, 2013 #14

    BruceW

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    hehe. anyway, as tiny-tim was saying, it is very likely that you are supposed to ignore the sinusoidal motion of the elevator. So just assume the elevator is accelerating downward at g.

    If you really want, you can take into account the SHM of the elevator, then take the limit that the mass of elevator is very large. You will get the same answer this way, but it will require more equations.

    Do whichever way you are more comfortable with. In either case, you need to consider the motion of the ceiling and the motion of the block relative to the ceiling.
     
  16. Jul 2, 2013 #15
    ok,can we be sure it executes SHM?i mean why does it execute SHM.when it's falling freely,the block is accelerating downward only because of the weight of it and the spring force is only applicable when it's stretched.so I assume that it's stretched initially,as the ceiling is also coming down with it,so can't that compression compensate the elongation??:P
     
  17. Jul 2, 2013 #16

    mfb

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    Letting the elevator fall is like "switching gravity off" (that is the basis of General Relativity, by the way) - and the equilibrium position with gravity and without are different.
     
  18. Jul 2, 2013 #17
    yeah i agree with you.bt how'd we show it mathematically?writing eqn for the block:mg-kx=mg will bring absurd results.what about this?can you guide me through solving my problem frm ground frame f reference.i'd really be thankful to you.
     
  19. Jul 2, 2013 #18

    WannabeNewton

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    Note that your method of showing SHM in the elevator frame is the correct one and is the most elegant way to do it mathematically. You simply ascribe a pseudo force in the opposite direction to that of gravity and solve Newton's 2nd law in the elevator frame for the mass-spring attached to the ceiling of the elevator. If you are in the elevator frame you must take this pseudo force into account (as you have).

    As for the ground frame, for simplicity assume the ground frame observer is directly above the elevator. For convenience choose the origin of the elevator frame as the relaxed length of the spring (which will be what?). Now write out Newton's 2nd law in the ground frame for the mass-spring. What position vectors will you have to consider?
     
  20. Jul 2, 2013 #19
    i'm sorry,i din't get wht u meant by chhosing the origin of elevator frame as the relaxed length of the spring.
     
  21. Jul 2, 2013 #20

    WannabeNewton

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    Can you write out Newton's 2nd law in the ground frame for me? It will become clear after that.
     
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