I have a vaguely ODE-related trigonometry problem dealing with the Differential Equation for Simple Harmonic Motion, question posted below momentarily.

Question

mendicant_bias · Answer

I was looking at how to solve the ODE  $$F = m \frac{d^2x}{dt^2}=-kx,$$or$$\frac{d^2x}{dt^2}+kx=0$$ resulting from equating Hooke's Law and Newton's 2nd Law, is solved, and I understanding how the roots are obtained for the char. Eqn and all, giving you $$c_{1}\cos(\omega t)+c_{2}\sin(\omega t),$$where omega is equal to sqrt(k/m).

mendicant_bias · Answer

What I don't get, though, is that they are then equated to, the general equation for SHM, something like $$c_{1}\cos(\omega t)+c_{2}\sin(\omega t)=A sin (\omega t + \zeta)$$

mendicant_bias · Answer

(I used the wrong Greek letter, but I couldn't remember what the other one was in the argument of sine, lol) But yeah, is there a trig identity being employed or something there? It's on the Wiki page for SHM

mendicant_bias · Answer

This is exactly the equality I'm talking about: http://upload.wikimedia.org/math/6/5/6/656fd81e91b7ad38db0c1f263dd5f4af.png

anonymous · Answer

$$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha$$
where $\alpha=\omega t$, $\beta=\zeta$, $\cos\beta=c_1$ and $\sin\beta=c_2$.

anonymous · Answer

(and the Greek letter is phi, or `\varphi` in latex, giving $\varphi$)

mendicant_bias · Answer

Oh, wow, that's a strange way of employing *that trig identity; I've never seen it done like that before. Neat.

mendicant_bias · Answer

That's still very strange, I'm trying to wrap my head around how the arguments of the sine functions equal to the constants would work. Just going to think about it for a little.

anonymous · Answer

Above I suppose I'm fixing $A=1$, but in general you would have $c_1=A\cos\zeta$ and $c_2=A\sin\zeta$.