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Spring amplitude problem

  1. Dec 6, 2009 #1
    I attach a 2 kg block to a spring that obeys Hooke's Law and supply 16 J of energy to stretch the spring. I release the block; it oscillates with period .3 s. The amplitude is:


    * 38 cm
    * 19 cm
    * 9.5 cm
    * 4.3 cm


    Can anybody help me with this? I can't figure out what to do or where to start. It seems so easy, but I can't get it. Thanks.
     
  2. jcsd
  3. Dec 6, 2009 #2
    Re: Amplitude

    Do you recall seeing an eqn T=2*pi*sqrt[k/m] . I think that is it. So solve for K by squaring both sides. Recall that energy is 1/2 k x^2 and is equal to 16J. Solve for x. This will be the initial stretch. In SHM, this constant represents 1/2 amplitude.
     
  4. Dec 6, 2009 #3

    Delphi51

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    Homework Helper

    Re: Amplitude

    Could you list the spring formulas you have to work with? There should be one with force, one with energy and one with period or frequency.
     
  5. Dec 6, 2009 #4
    Re: Amplitude

    To Denverdoc: Yes, but on my notes, I have:

    T = 2[tex]\pi[/tex]*[tex]\sqrt{}m/k[/tex]



    To Delphi51: I will list the ones I see here with me.

    Potential Energy: Usp = 1/2(k)(xs)2
    (k is spring constant, xs is compression/extension of spring from relaxed position)

    Fs = -kxs

    Wby sp = 1/2(k)(xs,i)2 - 1/2(k)(xs,f)2

    Those are the only ones I see at the moment.
     
  6. Dec 6, 2009 #5
    Re: Amplitude

    Yes you're right I forgot to flip it in going from w to T. So you have enough to solve it. Just follow the instructions in my last post.
     
  7. Dec 6, 2009 #6
    Re: Amplitude

    It's all good. I do have a question: You said in your last post that to solve for K, I had to square both sides. It seems to me that I have to do much more than that. Am I right?
     
  8. Dec 6, 2009 #7
    Re: Amplitude

    Sure, there is some simple algebra as well. Remember that is to just get K. You still have the other things to do as outlined in my post.
     
  9. Dec 7, 2009 #8
    Re: Amplitude

    Well, when solving for k, I got:

    k = m/(t/2[tex]\pi[/tex])2

    is that right?
     
  10. Dec 7, 2009 #9
    Re: Amplitude

    Yes (but by convention capital T is used for the period to avoid confusion with time.)
     
  11. Dec 7, 2009 #10
    Re: Amplitude

    He's right make sure T=t in your terms or you could end up with anything.
     
  12. Dec 7, 2009 #11
    Re: Amplitude

    Exactly right.
     
  13. Dec 7, 2009 #12
    Re: Amplitude

    Yeah, I just missed hitting the shift key when typing, or I typed it too fast. I do know the difference, and wouldn't get confused.

    Well, I did that and got

    k = 2 kg/(.3 s/2[tex]\pi[/tex])2
    = 9.006
     
  14. Dec 7, 2009 #13
    Re: Amplitude

    Very well. I like that it is round. Now it's perfectly ok to go back and reread my earlier posts as to how to proceed, remember we are looking for x, the initial displacement which is 1/2A.
     
  15. Dec 7, 2009 #14
    Re: Amplitude

    Again he's right more mistakes come from mixing up terms than everything else.

    [tex]a\neq{x}[/itex]

    [tex]c\neq{C}[/itex]

    And so on.

    Looks good to me?

    What expectation value were you expecting for k?
     
  16. Dec 7, 2009 #15
    Re: Amplitude

    Okay. Now I solved for x:

    E = 1/2(k)(x)2
    16 J = 1/2(9.006)(x)2

    x = [tex]\sqrt{}(32 J/9.006)[/tex]
    = 1.885
     
  17. Dec 7, 2009 #16
    Re: Amplitude

    So that means that 2x is my amplitude? so all I need to do is double my x and get my final amplitude value?
     
  18. Dec 7, 2009 #17
    Re: Amplitude

    Indeed. Sounds like you have this all figured out gj.
     
  19. Dec 7, 2009 #18
    Re: Amplitude

    If the question is looking for peak to peak amplitude, yes. Remember your cos or sin function will vary between 1 and -1.

    The initial displacement becomes the A on the outside of the SHM type of eqns you have been writing this weekend. Since it is simple harmonic motion, the initial displacement determines the subsequent behavior. Had we stretched the string more initially, the A would be different. Imagine a pendulum. If there is no friction, the ball swings back and forth to and from ther release point--same principle here.
     
  20. Dec 7, 2009 #19
    Re: Amplitude

    For simple harmonic operators this might help? Don't know?

    https://www.physicsforums.com/showthread.php?t=219445&highlight=Wilson-Sommerfield

    Helped me though.
     
  21. Dec 7, 2009 #20
    Re: Amplitude

    I think I understand. My only issue now is, when I got 3.77 m as my A, how is the answer 38 cm? Did I mess up somewhere with my math?
     
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