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Amplitude of oscillation

  1. Dec 11, 2016 #1

    yecko

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    1. The problem statement, all variables and given/known data
    A mass of 2.0 kg hangs from a spring with a force constant of 50 N/m. An oscillating force F = (4.8 N) cos[(3.0 rad/s)t] is applied to the mass. What is the amplitude of the resulting oscillations? Neglect damping. Answer: 0.15 m

    2. Relevant equations
    F=kx , the mass only change the equilibrium position, amplitude is when max F thus cos is equal to 1

    3. The attempt at a solution
    F=4.8N, k=50N/m
    x=F/k
    =4.8/50
    =0.096m.

    What's wrong with my calculation? Thanks
     
  2. jcsd
  3. Dec 11, 2016 #2

    ehild

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    The mass experiences three forces, gravity, mg, spring force, (-kx) and the driving force F. Gravity changes the equilibrium position only, but F is not equal to kx.
     
  4. Dec 11, 2016 #3

    yecko

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    Yes, I understand and i have mentioned it in the first post
    So what to calculate?
    Thanks
     
    Last edited: Dec 12, 2016
  5. Dec 11, 2016 #4

    ehild

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    The amplitude of vibration.
     
  6. Dec 12, 2016 #5

    yecko

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    I know, frankly its the question~
    i mean how i can calculate in order to get the amplitude of vibration (the answer!)
    what are the formulae to be used and the steps for getting the answer, if my steps are wrong (whats wrong?? why F is not equal to k*dx ?)
    thanks

    (ps. what I mentioned in the first posts are things all i understand... please tell me what i have misunderstood...)
     
  7. Dec 12, 2016 #6
    Really wish I could help you here, but no matter how I look at this question I get the same answer you do, and the only answer I can find online is formatting that makes no sense;

    - A = ( F m ) ( 2 2 ) 2 ( b 2 m ) 2 = ( 4.8 N 2 kg ) [ ( 3 s 1 ) 2 ( 50 N m 2 kg ) ] 2 + = 0.15 m -

    I'd keep trying but its like 2 am, hope somebody else helps you find your answer.
     
  8. Dec 12, 2016 #7

    yecko

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    well, indeed the format goes wrong by directly copy and paste would make it unreadable

    it seems we are on the another half of the planet:) it is still early here~
    however i am going to have exam tomorrow, hope all problems in my exercises can be settled by today:)
    thank you very much
     
  9. Dec 12, 2016 #8

    ehild

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    Can you write Newton's second law for this object? Remember, you have to take two forces into account.
    The object does SHM -simple harmonic motion. What is its displacement x in terms of time?
    You certainly learnt about driven oscillation. The object will oscillate with the frequency of the diving force after some time. What is the angular frequency of the external force?
    See the x(t) function of the forced oscillator in http://www.physics.louisville.edu/cldavis/phys298/notes/resonance.html, with the damping b neglected.
     
  10. Dec 12, 2016 #9

    yecko

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    F=ma
    x=A cos(ωt+Θ)
    ω=3
     
  11. Dec 12, 2016 #10

    ehild

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    F = ma is valid if F means the net force applied to the object. What is that net force in this case?
    The driving force is FD=4.8 cos(3t) . The elastic force is Fe=-kx. What is the net force?
    If what you said FD=kx, the net force is zero, and the acceleration is zero. The object can not oscillate.
     
  12. Dec 12, 2016 #11

    yecko

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    Fd+Fe?
    Why? I mean when the spring length is amplitude, force is max, accerleration is max.
     
  13. Dec 12, 2016 #12

    ehild

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    Yes, and what is in terms of time and x?
    You speak about the spring force. But there are two forces, applied on the object, and the acceleration is (Fd+Fe)/m.
     
  14. Dec 12, 2016 #13

    yecko

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    Fd=4.8 cos(3t) , Fe=-kx
    F(net)=4.8 cos(3t) +-kx
    are they contradict?
     
  15. Dec 12, 2016 #14

    ehild

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    Yes. So ma=4.8 cos(3t) -kx. And the object performs SHM, with the frequency of the driving force, so x=Bcos(3t+θ). What is the acceleration then?
     
  16. Dec 12, 2016 #15

    yecko

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    nothing
     
  17. Dec 12, 2016 #16

    yecko

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    nothing
     
  18. Dec 12, 2016 #17

    ehild

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    yes. And ma=4.8cos(3t)-kx. What should be A and θ?
     
  19. Dec 12, 2016 #18

    yecko

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    A = 0.096(?) and θ=0
    step:50Acos(3t+θ)=4.8cos(3t)
    whats wrong with my calculation?
     
  20. Dec 12, 2016 #19

    ehild

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    How did you get the equation 50Acos(3t+θ)=4.8cos(3t)? You think that the acceleration is zero? That is wrong.
     
  21. Dec 12, 2016 #20

    yecko

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    substitute a and x into ma=4.8cos(3t)-kx
     
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