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Frequency of oscillation problem

  1. Jun 17, 2007 #1
    So.. I have an ideal linear spring that streches 20 cm when a 40g mass is hung from it. The spring is then mounted horizontally on a frictionless surface (screams conservation of energy/momentum) and a 60g mass is attached to it. The 60g mass is then displaced 20cm from equilibrum and released.

    Find the following:
    1. frequency of oscillation
    2. equation of motion for the mass
    3. the speed and acceleration of the mass at +10 cm from equilibrium.
    4. What would be the period of an external force that would drive the system at resonance and explain how I know this period is correct.

    So this is what I've chiseled at so far:

    I found the spring constant using hooke's law and got 1.96 N/m for k.

    Then I know that because its a spring it will obey Kfinal +Uspringfinal = Kinitial +Uspringinitial

    So pulling it means .5(1.96N/m)(.2m)^2 = .039 J of potential energy in the system. And since there wont be any other outside forces acting, the total energy for this system is going to be .039 J.

    The problem is, all I feel like I can get out of the problem is the basic puzzle pieces and I dont know how to put it all together. I'm trying to look things up on simple harmonic motion its just not clicking for me.

    Any leads/help?
     
  2. jcsd
  3. Jun 17, 2007 #2
    I thought about finding the speed and acceleration at +10 cm.

    At this point Kinetic + Potential Spring energy has to be equal to .039 J. So I can do .5mv^2 +.5kx^2, plugging in the mass, spring constant, and distance to find the velocity of the mass which for kinetic and potential to equal .039 J would have to be .986 m/s.
     
  4. Jun 17, 2007 #3
    so I found the frequency using T = 2pi*sqrt(m/k), then 1/T. So my frequency came out to being .91 Hz.

    I'm still working on the equation of motion for the mass, the acceleration at +10 cm from equilibrium, and the period of an external force that would drive the system at resonance.

    Any help/leads?
     
  5. Jun 18, 2007 #4
    Haha well so I'm still plugging away.

    Using hookes law of f = -kx, I can then look at f = ma and then have ma = -kx, rearranging it to get a = -kx/m, plugging things in to get an acceleration of -3.27 m/s^2. The negative sign makes sense because since it is past equilibrium but not at the max amplitude yet so its slowing down as it reaches it.

    Now I'm just stuck on the junk of the equation of motion of the mass and the period of the external force to get that bad boy into resonance.
     
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