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Simple pendulum with friction ODE

  • Thread starter S_Flaherty
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  • #1
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I'm trying to figure out how to find the general solution for a simple pendulum with friction.


y'' + ky' + (g/L)y = 0


I know how to find the solution for a simple pendulum without friction:
y'' = -(g/L)y ... which leads to ... y = Acos((g/L)x)

So far I have: y'' + ky' + (g/L)y = 0, substituting m = y' and w^2 = g/L,
m^2 + km + w^2 = 0 so m = [-k ± sqrt(k^2 - 4(1)(w^2))]/2 ... I'm stuck here because
I can't remember how to reduce this so I can find the homogeneous equation for y.
 

Answers and Replies

  • #2
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I don't think there is an exact solution, that's why people tend to numerically solve its motion via computer.

You could try making the magnitude of the frictionless pendulum (A) changing over time but not the period and see if it works.
 
  • #3
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This is a linear ODE with constant coefficients. You solve it by first solving its characteristic equation, which in this case is ## \lambda^2 + k\lambda + g/L = 0 ##, giving ## \lambda = \frac {-k \pm \sqrt{k^2 - 4g/L} } {2} ##. If the roots are different, then the solutions are given by ##y = Ae^{\frac {-k + \sqrt{k^2 - 4g/L} } {2}t} + Be^{\frac {-k - \sqrt{k^2 - 4g/L} } {2}t} ##. Note that the roots are typically complex, so A and B are also complex. By choosing them properly, you can transform the equation to ## y = e^{-(k/2)t} (A'\cos (\sqrt{(k/2)^2 - g/L})t + B'\sin (\sqrt{(k/2)^2 - g/L})t)##, where the constants are real.
 
  • #4
HallsofIvy
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I confused as to what you mean by "homogeneous equation". In differential equations there are two uses of the word "homogeneous" one of which applies only to first order equations and so does not apply here. The other is that a linear differential of higher order is "homogeneous" if and only if every term involves y or a derivative of y- in other words there is no function of x only. What you have here is a homogeneous equation!

The 'characteristic equation' of this differential equation* is [itex]r^2+ kr+ (g/L)= 0[/itex]. By the quadratic formula, that has solution [itex]r= (-k\pm\sqrt{k^2- 4g/L})/2[/itex].
We can write the general solution as [itex]Ce^{[(-k+\sqrt{k^2-4g/L})/2]x}+ De^{[-k-\sqrt{k^2-4g/L})/2]x}= e^{-kx}(Ce^{\sqrt{k^2- 4g/L}x}+ De^{-\sqrt{k^2- 4g/L}x})[/itex]

What that "really" is depends on whether [itex]k^2- 4g/L[/itex] is positive, negative or 0. If it is positive then we simply have two exponentials. If it is 0, we have "resonance"- only a single root to the characteristic equation and will have one of the solutions multiplied by x. If it is negative, that square root gives imaginary roots so we will have [itex]e^{-kx}[/itex] times sine and cosine.



* We get that 'characteristic equation' by "assuming" a solution of the form [itex]y= e^{rx}[/itex] so that [itex]y'= re^{rx}[/itex] and [itex]y''= r^2e^{rx}[/itex]. Then [itex]y''+ ky'+ (g/L)y= r^2e^{rx}+ rke^{rx}+ (g/L)e^{rx}= (r^2+ kr+ (g/L))e^{rx}= 0[/itex]. Since [itex]e^{rx}[/itex] is never 0, we must have [itex]r^2+ kr+ (g/L)= 0[/itex].
 
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