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Amplitude of a Damped, Driven Pendulum

  1. Jun 29, 2013 #1
    1. The problem statement, all variables and given/known data

    A simple pendulum has a length of 1m. In free vibration the amplitude of its swings falls off by a factor of e in 50 swings. The pendulum is set into forced vibration by moving its point of suspension horizontally in SHM with an amplitude of 1 mm.

    a) [... Built Differential ...]

    b) [... Found Amplitude at exact resonance = 0.1576m ...]

    c) At what angular frequencies is the amplitude half of its resonant value?

    2. Relevant equations

    [itex]A_m = 0.1576 m[/itex]

    [itex]F/m = \frac{gζ}{l}[/itex]

    Let ζ = amplitude of driver.

    [itex]Q= 50 π[/itex]

    [itex]A(ω) = \frac{F/m}{((ω_0^2 - ω^2)^2 + (γω)^2)^{0.5}}[/itex]

    3. The attempt at a solution

    Solution is exactly stated as: "ω0 ± 0.017 sec-1"

    Am I solving for ω or ω0?
     
  2. jcsd
  3. Jun 29, 2013 #2
    Definitely ω, ω0 is the undamped natural frequency.
     
  4. Jun 30, 2013 #3
    Can you explain to me how did you solve this question ? Thank you
     
  5. Jun 30, 2013 #4
    I didn't solve any equation. I just told AJKing whether to solve for ω or ω0
     
  6. Jun 30, 2013 #5
    Do you know by any chance how to proceed ?
     
  7. Jun 30, 2013 #6
    Yes i know how to proceed but let OP reply first. Highjacking a thread is not allowed here at PF!
     
  8. Jun 30, 2013 #7
    Please explain to me how to proceed.
     
  9. Jul 2, 2013 #8
    It would seem obvious how to proceed - solve for ω - but it is difficult to do directly.

    Here's Wolfram Alpha's attempt at solving this:

    [itex]A(ω) = \frac{F/m}{((ω_0^2 - ω^2)^2 + (γω)^2)^{0.5}}[/itex]

    Where ω0 = g0.5

    We must keep out eyes for clever approximations.
     
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