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Forced and damped oscillators

  1. Nov 23, 2004 #1
    I have two problems, the second of which I think I might be solving right. The web program we use to do our homework isn't accepting my answer. It might be the program's fault, but I'm not sure, so I'd like to check.
    Here's my first problem:

    Damping is negligible for a 0.131-kg object hanging from a light 6.50-N/m spring. A sinusoidal force with an amplitude of 1.70 N drives the system. At what frequency will the force make the object vibrate with an amplitude of 0.440 m? There are two possible solutions.

    I would think I know how to answer the problem if there weren't two possible solutions. That completely throws me. It also confuses me that amplitude is given in Newtons and in meters. The only thing I can think of to solve the problem is to use the equation for amplitude of a driven oscillator:
    A = (F0/m)/((w2-w02)2 + ((bw)/m)2)1/2
    w02 = k/m
    F0 = 1.70 N?
    bw/m = 0, because damping is negligible
    I got 8.79 Hz when I solved for w. That's still only one answer, so I guess it's not right.

    Problem two:
    Consider a damped oscillator that is an object hanging vertically from a spring and submersed in a viscous liquid. Assume the mass is 382 g, the spring constant is 105 N/m, and b = 0.119 N*s/m. (a) How long does it take for the amplitude to drop to half its initial value? (b) How long does it take for the mechanical energy to drop to half its initial value?

    I used the equation for position of a damped oscillator:
    x = Ae(-b/(2m))tcos(wt + theta)
    I used .5A as x, took the natural long of both sides and solved for t. I ignored cos(wt + theta) because the cos at the maximum amplitude is equal to one. I got 4.45 s. The time it takes for the mechanical energy to drop to half its initial value would be 8.90 s (twice as much), because E = .5kA2. What am I doing wrong?

    Thanks for your help!
     
    Last edited: Nov 23, 2004
  2. jcsd
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