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Speed of an object on a spring

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A .25kg object suspended on a light spring is released form a postition of .15m above the stretched equilibrium position. THe spring has a spring constatnt of 80N/m. What is the speed of the object when the object is .05m above its equilibrium position? What is the maximum speed and where does it occur?


    2. Relevant equations
    I'm not sure what equation i need for this


    3. The attempt at a solution
    I would belive that the speed of the object would be the same if it were .05m above or below equilibrium. But any help in pointing me in the right direction would be helpful. THanks
     
  2. jcsd
  3. Dec 8, 2007 #2

    Dick

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    I think your belief is correct, which is a good sign. So you are probably picturing the object oscillating around the equilibrium position. There's an explicit formula for this motion y=A*cos(omega*t) where y is deviation from equilibrium. Can you figure out how to get A and omega? (Hint: omega involves the spring constant and the mass.) Now velocity is dy/dt.
     
  4. Dec 8, 2007 #3
    Ok i got omega by omega = sqrt(80n/m/.25kg) = 17.889 but I can't figure out how to get A.
     
  5. Dec 8, 2007 #4

    Dick

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    'A' is the amplitude of oscillation. y goes from +A to -A. When the object is released it's at rest. That's A. BTW the units of omega are 1/sec. Don't leave out units.
     
  6. Dec 8, 2007 #5
    I think Dick made things too hard. (no pun intended)

    Try conservation of energy!
     
  7. Dec 8, 2007 #6
    bah, i dont understand. is the A = to 1 since its between .05 and -.05? and also how do you solve for t?
     
  8. Dec 8, 2007 #7
    this is one of my last questions and i've been trying ot figure it out for like 2 hours now, please help
     
  9. Dec 8, 2007 #8

    Dick

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    The oscillation is between 0.15m and -0.15m from equilibrium. A=0.15m.
     
  10. Dec 8, 2007 #9
    ok so v = sqrt(80n/m/.25kg)*(.15^2-.05^2) = 2.5 m/s and it would be that same for -.05 becuase -.05^2 = .05^2. But how would u find the maximum speed and where is it occuring?
     
  11. Dec 8, 2007 #10
    So it has maximum velolcity at the equilibrium so just plug in "0" instead of .05 and i get 2.7 m/s, this sounds reasonable. Is this correct?
     
  12. Dec 9, 2007 #11

    Dick

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    Sounds correct.
     
  13. Dec 9, 2007 #12
    still way to hard.

    PE of a spring is equal to 1/2kx^2 where x is distance from equilibrium and KE is 1/2 mv^2. An oscilator is just a system that conserves energy. At one extreme (your initial condition) There is only potential energy of the spring because the object is at rest (it is changing directions) at full stretch. You corectly noticed that this is true symetrically on both sides of equilibrium. At the other extreme the spring is at equilibrium (exerting no force) but the mass is moving and has Kinetic Energy. Therefor at maximum speed the KE(final)=PE(initial) and the 1/2 cancel out leaving mv^2=kx^2 and the spring was orignally stretched. This leaves you with v = sqrt(80n/m/.25kg)*(.15^2) and as you said correctly before the change is the spring in part a) is .10 and you therefor did this correctly.
    I think that although your math is correct and Dick helped you with this you don't really have a sense of the physics that is occuring. Try to get a mental picture before you resort to complicated formula's and you life will be easier.
     
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