Question on general solution to harmonic EoM

[Vitani11]

Homework Statement

An equation of motion for a pendulum:

(-g/L)sinΦ = Φ(double dot)

Homework Equations

L = length
g = gravity
ω = angular velocity
Φ_ο = initial Φ

The Attempt at a Solution

The solution is Φ=Asinωt+Bcosωt

solving for A and B by setting Φ and Φ(dot) equal to zero respectively gives:

Φ= Φ_οcosωt

My question is how can you just go ahead and write down Φ=Asinωt+Bcosωt? My professor said it was because this was the only function that when differentiated twice give -sinωt but I can't just take that for what it is because I need to know how to arrive there.

 

[Ray Vickson]

Homework Statement

An equation of motion for a pendulum:

(-g/L)sinΦ = Φ(double dot)

Homework Equations

L = length
g = gravity
ω = angular velocity
Φ_ο = initial Φ

The Attempt at a Solution

The solution is Φ=Asinωt+Bcosωt

solving for A and B by setting Φ and Φ(dot) equal to zero respectively gives:

Φ= Φ_οcosωt

My question is how can you just go ahead and write down Φ=Asinωt+Bcosωt? My professor said it was because this was the only function that when differentiated twice give -sinωt but I can't just take that for what it is because I need to know how to arrive there.

Your solution is incorrect: the differential equation $\ddot{\Phi} = -k \sin(\Phi)$ involves Elliptical functions. It is NOT of the form $\Phi = A \sin \omega t + B \cos \omega t$. See, eg.,
https://en.wikipedia.org/wiki/Pendulum_(mathematics) .

For very small angles $|\Phi| \ll 1$ we have $\sin(\Phi) \approx \Phi$, so for small angles the solution is of the form you want.

As for why that form applies: you can take several points of view.
(1) Two hundred years ago somebody discovered that form of solution, and we have been taught it ever since.
(2) You can recognize that the equation $\ddot{x} = k x$ has solution $x = e^{\sqrt{k} t}.$ When $k = -c^2$ that gives $e^{\pm i c t}$ and Euler's equation gives that as $\cos ct \pm i \sin ct.$ Taking the real and imaginary parts gives us the $\sin$ and $\cos$ forms.
(3) You can try to solve $\ddot{x} = -c^2 x$ as a power series $x = a_0 + a_1 t + a_2 t^2 + \cdots$ and use standard DE methods to determine the coefficients $a_0, a_1, a_2, \ldots$. You will find that you get a combination of the series expansions of $\sin ct$ and $\cos ct.$
(4) You can use the method of Laplace transforms.

There are other ways as well, but realisically, (1) is the easiest explanation. Your professor's explanation is 100% correct (except that both $\sin$ and $\cos$ will work).

 

[Vitani11]

Good explanation, thank you.

 

[Vitani11]

When you say solve x(double dot) = -c²x as a power series in (3) are you referring to perturbation theory? .

 

[Ray Vickson]

When you say solve x(double dot) = -c²x as a power series in (3) are you referring to perturbation theory? .

No, it has nothing to do with perturbation theory. Development of series solutions to DEs is one of the topics that gets covered in just about any DE course. Or, you can Google "series solution to differential equation".