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Equation of motion of a dampened pendulum

  1. May 8, 2017 #1
    1. The problem statement, all variables and given/known data

    Hi, I have the following question...

    A pendulum consisting of a mass of 1kg is suspended from a string of length L. Air resistance causes a damping force of bv where b = 10-3 N/m

    1) Derive and solve the equation of motion
    2) Calculate the fractional decrease in amplitude of the pendulum oscillations if the pendulum is operated for 2 hours.

    2. Relevant equations

    F=ma
    F=-kx
    x(t)=A0eαt

    3. The attempt at a solution

    I think I have the derivation handled...

    F=ma=-kx-bv
    ma+bv+kx=0
    m(d2x/dt2)+b(dx/dt)+kx=0
    d2x/dt2+(b/m)(dx/dt)+(k/m)x=0

    As this is a pendulum, I know x=Lsinθ and for small θ sinθ≈θ so...

    d2θ/dt2+(b/m)(dθ/dt)+(k/m)θ=0

    I also know that k=mg/L so...

    d2θ/dt2+(b/m)(dθ/dt)+(g/L)θ=0

    Therefore, I get the equation of motion for the dampened pendulum to be:

    d2θ/dt2+(b/m)(dθ/dt)+(g/L)θ=0

    Now to solve the equation, I can turn it into a quadratic:

    α=(-(b/m)±(√(b2/m2)-(4g/L))/2

    I can tidy this up a little, so...

    α=-(b/2m)±(√(b2/2m2)-(4g/L))

    I know that √g/L = ω so and -√g/L = iω which I can take out as a factor so and tidy up the fraction inside the radical, so...

    α=-(b/2m)±iω√1-(b2L/2)

    I also know that I can let ϑ=ω√1-(b2/4m2ω2) so...

    α=-(b/2m)±iϑ

    I am assuming that this is the equation solved as I don't have a value for L and so can't go any further (can I?)

    As for calculating the fractional decrease, if I sub this back into the equation:

    x(t)=A0eαt I get...

    x(t)=A0e(-b/2m)t+eiϑt

    But I really don't know where to go from here, if, if anywhere?

    Do I simply rearrange the first part so....

    x(t)/A0=e-18/5 so...

    x(t)/A0=0.0273

    Any help gratefully received!
     
  2. jcsd
  3. May 8, 2017 #2

    Ray Vickson

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    Homework Helper

    You can write the solution either as
    $$x = A_1 e^{-rt}e^{i st} + A_2 e^{-rt} e^{-ist}, $$
    or as
    $$x = B_1 e^{-rt} \cos(st) + B_2 e^{-rt} \sin(st),$$
    where ##r## and ##s## are related to your constants ##m, k, b##. If you choose the second form you just need to determine the constants ##B_1, B_2##, which you can do from the initial conditions (position and velocity at ##t = 0##).
     
    Last edited: May 9, 2017
  4. May 9, 2017 #3
    thanks for that,

    I kind of get what you're saying, using the second equation, when t=0, x=B1

    I am still confused as to what to do when t=7200 though
     
  5. May 9, 2017 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    You need to use one more piece of information in order to determine the constant ##B_2##. For example, if the pendulum bob is released at zero velocity from initial position ##x_0## then you have ##x(0) = x_0## and ##\dot{x}(0) = 0##. From those two conditions you can get both ##B_1## and ##B_2## in terms of ##x_0, r,s##.

    Then, you just need to plug in ##t = 7200## into your formula for ##x(t)## and evaluate it.
     
    Last edited: May 9, 2017
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