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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Once again, those equations are \[\frac{ d \theta }{ dt }=v\]\[\frac{ dv }{ dt }=\frac{ g }{ l }\sin \theta\frac{ b }{ ml }v\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1with 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks irishboy, I'm studying Nonlinear 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.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1hi @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 *nonperiodic* 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/doesdampingofperiodaffecttheperiod.399471/ there's a clue in there too :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0small 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in particular it's the smallangle approximation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like @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 exponentiallymodulated sinusoids: $$\theta(t)=Ae^{\alpha t}\cos(\beta t)+Be^{\alpha t}\sin(\beta t)$$ if \(\alpha\pm\beta i\) are the roots of \(z^2+\frac{b}{ml}z+\frac{g}l\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much  I just solved it!
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