## 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

Once again, those equations are $\frac{ d \theta }{ dt }=v$$\frac{ dv }{ dt }=-\frac{ g }{ l }\sin \theta-\frac{ b }{ ml }v$

2. IrishBoy123

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

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

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

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

in particular it's the small-angle approximation

7. anonymous

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

Thank you so much - I just solved it!