1. ## Classical Mechanics

I'm starting on classical mechanics in university (second year) and I skipped first year physics. I do have quite a good mathematical knowledge but I'm not getting this question:

Verify by direct substitution that the function $\phi(t) = A \sin (\omega t) + B \sin(\omega t)$ is a solution to the second order differential equation $\ddot{\phi} = -\omega ^2 \phi$
(Since this solution involves two arbitary constants - the coefficients of the sine and cosine functions - it is in fact the general solution)

So I was just thinking of substituting the first equation into the second to get:

$\ddot{\phi} = -\omega ^2 (A \sin (\omega t) + B \sin(\omega t))$

And then distribute and integrate twice to get the differential?
I'm really confused about the question.

2. ## Re: Classical Mechanics

No, substitute on the left side also. In other words differentiate twice and see if the two sides of the equation are really equal.

3. ## Re: Classical Mechanics

Actually I just realised typed in the equation wrong. The first one is supposed to be: $\phi(t) = A \sin (\omega t) + B \cos(\omega t)$ but I don't think it really matters.

So something like this??

$\phi(t) = A \sin (\omega t) + B \cos(\omega t)$ \\EQUATION 1

$\dot{\phi}(t) = A \omega \cos (\omega t) - B \omega \sin(\omega t)$

$\ddot{\phi}(t) = - A \omega ^2 \sin (\omega t) - B \omega^2 \cos(\omega t)$ \\EQUATION 2

So I substitute equation 1 onto the right hand side and equation 2 onto the left hand side of $\ddot{\phi} = -\omega ^2 \phi$?

That'll give me:

$- A \omega ^2 \sin (\omega t) - B \omega^2 \cos(\omega t) = -\omega ^2 (A \sin (\omega t) + B \cos(\omega t))$

And yes if I distribute the righthand side, it is clear that they are equal.