# Diff. Eq. related to SHM question

1. Sep 2, 2011

### astropi

1. The problem statement, all variables and given/known data
This problem relates to a mass on a spring, or most any SHM. For an undamped free vibration, I can solve the problem and reduce it to this equation:

$l = Acoswt + Bsinwt$ (1)

I know the equation is correct. I also know that the above equation reduces to:

$Rcos(wt -\delta)$ (2)

Sad to say, I can not figure out how 1 reduces to 2 (though simple it should be)! There is the trig identity

cos(x-y) = cosx*siny + sinx*cosy

but I am uncertain as to whether this helps in this particular case. I also believe that

$R^2 = A^2 + B^2$

and so putting everything together I *should* be able to reduce 1 to 2, but alas have failed. If someone can explain the steps, that would be appreciated :)

2. Sep 2, 2011

### SammyS

Staff Emeritus
Can you reduce (2) TO (1)?

If so, then work it in reverse.

3. Sep 2, 2011

### davo789

Hi astropi,

I'll give you the important info to start you off!

What you can say is that $A = asin(\delta)$, and $B = acos(\delta)$. By doing this, you introduce the variable $\delta$. Take a couple of minutes to make sure that setting A and B equal to these values makes sense in your head.

You can then say: $A^2 + B^2 = a^2 (sin^2(\delta) + cos^2(\delta)) = a^2$ so that $a=sqrt(A^2 + B^2)$.

Hopefully you can see where this is going, so I'll let you have a go at finishing!

4. Sep 2, 2011

### astropi

Hi Sammy, although I should be able to reduce 2 to 1, for some reason I do not see it.

davo: thanks, I know that's how you get "a", or "R" as I called it in my expression. However, I've been tinkering with the formula trying to reduce it to the final expression using a trig identity and this apparently does not work.

By the way, if it makes a difference, I'm not doing this for class. I'm working out various elementary problems (such as SHM) from basic principles, and sadly became stuck in the math here! Rather than skip some steps I'm trying to be as thorough as possible. It's a bit embarrassing to ask for help with something so trivial, but oh well :)

5. Sep 2, 2011

### vela

Staff Emeritus
Use the identity $\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$ where $\alpha = \omega t$ and $\beta = \delta$.

6. Sep 2, 2011

### astropi

Right, I see where I went wrong. Thanks everyone, and sorry to ask such a trivial question :)