Question

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.

Answer 1

Once again, those equations are

\[\frac{ d \theta }{ dt }=v\]\[\frac{ dv }{ dt }=-\frac{ g }{ l }\sin \theta-\frac{ b }{ ml }v\]

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

in particular it's the small-angle approximation

Answer 7

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 \(z^2+\frac{b}{ml}z+\frac{g}l\)

Answer 8

Thank you so much - I just solved it!