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One More hard Oscillations Problem

  1. Nov 21, 2007 #1
    1. The problem statement, all variables and given/known data
    An oscillator with a mass of 420 g and a period of 1.00 s has an amplitude that decreases by 1.20% during each complete oscillation.

    If the initial amplitude is 8.20 cm, what will be the amplitude after 50.0 oscillations?
    If the initial amplitude is 8.20 , what will be the amplitude after 50.0 oscillations?

    2. Relevant equations

    x(t)= Ae[tex]^{-bt/2m}[/tex]cos([tex]\omega[/tex]t)


    3. The attempt at a solution


    I have no idea :(
     
  2. jcsd
  3. Nov 21, 2007 #2

    G01

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    What term in the equation you gave is always making the amplitude smaller as time increases?
     
  4. Nov 21, 2007 #3
    Time or the dampening coeffecient?
     
  5. Nov 21, 2007 #4

    G01

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    Your part of the way there. It is the damping coefficient term. Read my post in your other thread. It may clear some things up on this problem as well.
     
  6. Nov 21, 2007 #5
    this doesnt make sense i hate physics :(
     
  7. Nov 21, 2007 #6

    G01

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    Don't give up yet!

    Your amplitude starts at a value A and decreases due to the damping. The amplitude at any time then, is given by:

    [tex]A(t)=Ae^{-bt/2m}[/tex]

    How can you use this equation to find what you are looking for?
     
  8. Nov 21, 2007 #7
    i think you can solve for the time but i dont undersand what b would be in this equation
     
  9. Nov 21, 2007 #8

    G01

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    In order to find b you have to use the information for how much A decreases in a certain amount of time. They tell us, in one second, A decreases by 1.2%: We can say this mathematically like so:

    A(1)=.012A

    But from the equation in my above post, what is A(1) in terms of b? Using these two equations then, you should be able to solve for b.
     
  10. Nov 21, 2007 #9
    I got b= -3.7132 but i dont know if that is correct/ makes sense
     
  11. Nov 21, 2007 #10

    G01

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    I get positive 3.7 . I think you just forgot a negative sign somewhere.
     
  12. Nov 21, 2007 #11
    ah i see it now, now with this b i find time?
     
  13. Nov 21, 2007 #12
    repost sorry
     
    Last edited: Nov 21, 2007
  14. Nov 21, 2007 #13

    G01

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    Yep that's the plan:smile:
     
  15. Nov 21, 2007 #14
    im such an idiot... i got the first part already :\ =4.48cm

    i meant this one: At what time will the energy be reduced to 18.0% of its initial value?
     
  16. Nov 21, 2007 #15

    G01

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    Remember that the energy of a wave is proportional to the amplitude squared. I can't seem to find the actual equation describing that, but it should be simple enough to look up. So, HINT: You want to find the time at which the amplitude is such that it makes the energy 18% of its initial value.

    Use the percents like we did above with the amplitude, but this time work with the energy:

    [tex].18E_{initial}= E_{at A(t)}[/tex]

    Then use this energy to find A. Then, you should be able to find t from A.
     
    Last edited: Nov 21, 2007
  17. Feb 8, 2010 #16
    i got b as well, but how do i find time at this point
     
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