Solving SHM: Converting x(t)^2 to cos^2(wt+φ)

[wil3]

Hello. I am studying second-order differential equations, and I am currently studying simple harmonic oscillations, like those in a spring-block system. When solving the differential equation, I get answers in the form of y^2=

x(t)^2=(c_1^2+c_2^2)(cos^2(\omega t)-(c_2^2/(c_1^2+c_2^2))sin^2(\pi/\omega))

How can I convert this to the general form of the SHM equation:

x(t)^2=(c_1^2+c_2^2)cos^2(\omega t+\phi)

where cos(\phi)=c_1/\sqrt{c_1^2+c_2^2} and sin(\phi)=-c_2/\sqrt{c_1^2+c_2^2}

I'm really close here, but I do not know how to use trig identities to convert the squared constant plus squared cosine function into just a single cosine function containing a shift. Thank you.(PS- I am also trying to use Latex, so check back to see if I've edited the post to make it more readable. Hopefully, you can at least see what I'm going for)

 

[Gerenuk]

In general you can use
\cos(x)=\Re e^{ix}
and
\cos^2 x=\Re e^{ix}\cdot\frac{e^{ix}+e^{-ix}}{2}
You get the idea?

Hmm, in your equation the /\omega looks funny. And are you sure that you don't have too many squares in there?

 

[wil3]

No, that's the correct equation. This is how far I got solving back from:

x(t)=c_1cos(\omega t)+c_2sin(\omega t)

I already used Euler's formula on the solution to the differential equation to get the above equation, which I then manipulated using trig identities to get the equation I printed above. How would I use the formula again to make progress on the problem? Basically, I am looking for a way to show with trig identities that a cosine squared function plus a constant is the same as a cosine squared function with a shifted argument.

 

[Gerenuk]

Then start from the latter equation you quote and use the idea I propose. It's pretty easy then.

 

[uart]

wil3, if your equation in post #3 is correct then your other equation (post #1) is definitely incorrect.