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anonymous

  • one year ago

Consider the damped pendulum system dθ/dt=v dv/dt=−(g/l)*sinθ−(b/ml)*v where b is the damping coefficient, m is the mass of the bob, l is the length of the arm, and g is the acceleration of gravity (g≈9.8m/s^2). What relationship must hold between the parameters b, m, and l for the period of a small swing back and forth of the damped pendulum to be one second? The answer must be equal to 1/2*pi.

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  1. anonymous
    • one year ago
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    Once again, those equations are \[\frac{ d \theta }{ dt }=v\]\[\frac{ dv }{ dt }=-\frac{ g }{ l }\sin \theta-\frac{ b }{ ml }v\]

  2. IrishBoy123
    • one year ago
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    with given input, this is the DE you need to solve: \(\large \ddot \theta + \frac{b}{ml} \dot \theta + \frac{g}{l} \theta = 0\) but can you explain: a) the assumption i made that \(sin \theta \approx \theta\) ?? and b) how i know the pendulum has unit length ... which means what for l ?? that is the remaining physics. i imagine you can :p loads of people here can help you with the DE. it's a sinusoid with exponential decay. and it's linear homogeneous. me too, if you are stuck. BTW you could probably find a polished formula for this on Hyperphysics. i don't know what you are studying.

  3. anonymous
    • one year ago
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    Thanks irishboy, I'm studying Non-linear Diff Eq. As for your questions, a) For small theta (close to 0) since we're assuming a small swing, the sine of theta would be close to 0 too. b) l is unknown, and must be part of the answer along with b and m.

  4. IrishBoy123
    • one year ago
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    hi @sleung if \(\dot v = \dot \theta\) , it must follow that l = 1? [not that it matters, keep it in as l, if you prefer] so then you have a DE to solve. and a load of messy algebra. so solve it in which ever way you wish knowing that you will get a solution involving something liked \(\theta(t) = A \ sin B t\ e^{-whatever} + C \ sin D t\ e^{-whatever} \) And you know that \(B = D \equiv \omega = 2\pi f = 2\pi \frac{1}{T} \) with: T = period ALSO, from the net: "over damping and critical damping of the system produce *non-periodic* motion. Under damping will produce a frequency that is less than the natural frequency by an amount that depends on the "damping ratio"." Reference https://www.physicsforums.com/threads/does-damping-of-period-affect-the-period.399471/ there's a clue in there too :p

  5. anonymous
    • one year ago
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    small swing seems to suggest \(\theta\ll1\) so \(\sin\theta\approx\theta\) as the higher order terms \(O(\theta^k)\) become negligible; this is called linearization

  6. anonymous
    • one year ago
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    in particular it's the small-angle approximation

  7. anonymous
    • one year ago
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    like @IrishBoy123 said, we get a linear ODE as an approximation: $$\ddot\theta+\frac{b}{ml}\dot\theta+\frac{g}l\theta=0$$which gives a solution as a linear combination of the exponentially-modulated sinusoids: $$\theta(t)=Ae^{\alpha t}\cos(\beta t)+Be^{\alpha t}\sin(\beta t)$$ if \(\alpha\pm\beta i\) are the roots of \(z^2+\frac{b}{ml}z+\frac{g}l\)

  8. anonymous
    • one year ago
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    Thank you so much - I just solved it!

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