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Help on pendulum problem, thanks!

  1. Sep 11, 2007 #1
    1. The problem statement, all variables and given/known data
    In a science museum, a 110 kg brass pendulum bob swings at the end of a 15.0-m-long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.5 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010 kg/s.

    a. At exactly 12:00 noon, how many oscillations will the pendulum have completed?
    b. And what is its amplitude?

    2. Relevant equations
    w=sqrt(k/m-b^2/4m^2)
    tau=m/b
    w=sqrt(g/L)

    3. The attempt at a solution
    i've got nothing this makes no sense to me, if someone could just give me a hint on where to start.....
     
    Last edited: Sep 11, 2007
  2. jcsd
  3. Sep 11, 2007 #2

    learningphysics

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    What is the general equation of a damped oscillator?
     
  4. Sep 11, 2007 #3
    x(t)=Ae^(-bt/2m)cos(wt+phi), but what do i do with it?
     
  5. Sep 11, 2007 #4
    I posted this same question in Advanced Physics.

    I'm stuck on the number of oscillations. I did what I thought was right, but MP doesn't accept it as an answer.

    The second part is straight forward though, if you understand the equation.
     
  6. Sep 11, 2007 #5
    could u explain how u did the second part?
     
  7. Sep 11, 2007 #6
    I'm as lost as you are. Solid information, but what are we supposed to do with it? Is there more to the problem statement?
     
  8. Sep 11, 2007 #7
    Well, how much of the equation do you understand? Do you understand what each part tells you?

    What physics class are you taking?
     
  9. Sep 11, 2007 #8

    learningphysics

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    [tex]\omega_0 = \sqrt{g/l}[/tex]

    [tex]\omega_r = \sqrt{{\omega_0}^2 - b^2/(4m^2)}}[/tex]

    I believe you can set phi to 0 because it is at its maximum amplitude at t = 0. You can calculate the angle of the pendulum at t = 0... that's the amplitude A.

    The period is [tex]2\pi/\omega_r[/tex]

    That'll give you the time per oscillation.
     
  10. Sep 11, 2007 #9
    i know what the equation means, A =amplitude, b=damping constant, etc, what am i using for x, is it the 1.5 that i pull it away, or something else. i know t= the 4 hours, w=sqrt(g/l), and b is given, as is m, so what else am i missing?
     
  11. Sep 11, 2007 #10
    That makes sense. I wonder why none of my textbooks make any mention of the period equation.... not for damped motion, anyway.

    At any rate, I found a different way to look at the breakdown of the problem. I've got a little experience in Diff-Eq, but I like looking at things in more managable chunks so I don't get confused. Here's how I looked at the problem/equation...

    The equation for Damped SHM looks like this (after solving the DEQ):

    Position X = [tex]A_{0}[/tex] e[tex]^{-(b/2m)t}[/tex]cos([tex]\varpi\acute{}[/tex]t+[tex]\phi[/tex])

    Now, you know m, L, b, and T. You also know A[tex]_{0}[/tex]. That is the starting amplitude, which was given as 1.5.

    So you know A, you know b, you know m, and you know t.

    That's important because the first portion of the equation ([tex]A_{0}[/tex]e[tex]^{-(b/2m)t}[/tex]) is the function of the curve that the amplitude follows. So if you graph that function, you'll have a line that a peak of the whole function touches.

    SO, what value t do you want? Well, where do you want to know the amplitude? They say at Noon, right? So noon is 4 hours past the start time. 4 hours is how many seconds? You know you need seconds, because "b" is given in kg/s.

    Note: Amplitude is always positive in this case.

    Now for the number of oscillations you'll do exactly as was stated above.

    [tex]\varpi\acute{}[/tex] = [tex]\sqrt{\frac{K}{m}-\frac{b^{2}}{4m^{2}}}[/tex]

    K = [tex]\frac{mg}{L}[/tex] (where g = the force of gravity)

    Then the period is [tex]\frac{2\pi}{\varpi\acute{}}[/tex]. That gives you the amount of time per cycle. And since you know the amount of time you want to know the cycles IN, it should be obvious what to do here.
     
  12. Sep 11, 2007 #11

    learningphysics

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    But amplitude for the pendulum refers to the angle from the vertical...
     
  13. Sep 11, 2007 #12

    learningphysics

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    Also, the amplitude at a time t is:

    Ae^(-bt/2m), where A is the angle at t=0.
     
  14. Sep 11, 2007 #13
    Ahh, that's where it's tricky. The problems on Mastering Physics give you a little hint.

    They tell you the units of the answer, in this case Meters.

    It's just a matter of evaluating the DSHM equation at t=14400 seconds for the first half of the equation.
     
  15. Sep 11, 2007 #14
    [tex]A_{0}[/tex] is illustrated (verbally) in the problem to be 1.5 meters horizontal.

    Since the answer requires units of the meter, one can assume they are used throughout. In this case, it is, and [tex]A_{0}[/tex] is the initial amplitude.

    Again, if you were to evaluate this graphically, you'd use 1.5 for [tex]A_{0}[/tex], not the angle at t=0.

    I rekon you could solve the problem either way, but I did it using 1.5 for [tex]A_{0}[/tex] and I was rewarded with a correct answer.
     
  16. Sep 11, 2007 #15

    learningphysics

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    Oh ok... cool. So you did [tex]1.5e^{\frac{-b}{2m}(14400)}[/tex] for the amplitude?
     
  17. Sep 11, 2007 #16
    Yep, that's exactly what I did.
     
  18. Sep 12, 2007 #17
    I was under the impression that the period of a pendulum does not change, but, looking at it logically, a force is being applied when it collides with the air molecules, as there is a slight change in momentum, which means a force is applied.

    Could you use the initial potential energy to calculate the answer? That seems a logical way to do it.
     
  19. Sep 12, 2007 #18
    You could, I imagine.

    Calculate the fractional drop in GPE and then extrapolate it to t=14400. And since potential energy is proportional to the square of the amplitude, you could find the amplitude at a given point. It sounds a bit more difficult, but I bet you'd arrive at the same conclusion, or close enough it wouldn't make much difference.

    I don't see why you COULDN'T do it this way. But doesnt the constant of damping include air resistance? I thought it included all forms of damping, from friction at point of suspension, to air resistance, and gravitational influence, is that not right?

    For small values of B, I think doing it the way above is fine. For large values of B, you HAVE to use the energy situation I think.
     
  20. Feb 3, 2008 #19
    I'm having trouble using this information to solve the problem. You offer the DSHM equation for finding the amplitude at a certain time...but how does that help in determining the amount of oscillations that occur between t=0 and t=14400?

    I've tried determining the time per oscillation and dividing it into 14400 but it's still spitting out the completely wrong answer.
     
    Last edited: Feb 3, 2008
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