1. Nov 21, 2013

### iRaid

I have a homework problem that I need to use the steady periodic oscillation to solve, so instead of having help on the problem I'd rather just understand how they did it then apply it to my homework (I think that's alright?)

I'm kind of wondering where my book gets this from: $$x_{sp}(t)=5cos4t+4sin4t=\sqrt{41}\left( \frac{5}{\sqrt{41}}cos4t+\frac{4}{\sqrt{41}}sin4t \right)=\sqrt{41}cos(4t-\alpha)$$
Honestly, I have no idea where they even get the square root of 41 either..

I feel like it's some trig substitution, identity, etc. that I'm forgetting, but I cant figure it out

Any help is appreciated.

Last edited: Nov 21, 2013
2. Nov 21, 2013

### LCKurtz

It's a standard technique. You divide and multiply by $\sqrt{5^2+4^2}$. That makes the sum of the squares of the new coefficients = 1. So you can think of $\frac{5}{\sqrt{41}}$ as $\cos(\alpha)$ and $\frac{4}{\sqrt{41}}$ as $\sin(\alpha)$ and you have an addition formula. You can draw a little right triangle with legs 4 and 5 to see $\alpha$.

3. Nov 21, 2013

### iRaid

OK that makes sense, but then where do they get the $\sqrt{41}cos(4t-\alpha)$?

4. Nov 21, 2013

### LCKurtz

Put $\cos(\alpha)$ and $\sin(\alpha)$ in for those two fractions and use the addition formula.

5. Nov 21, 2013

### iRaid

Ah I see now.. $cos(\alpha)cos(4t)+sin(\alpha)sin(4t)=cos(\alpha-4t)$ from the trig addition formulas.

One more question, I'm not sure if you'll know it, but might as well ask... For a forced oscillation, the motion is stated as $x(t)=x_{tr}(t)+x_{sp}(t)$. In the differential equation how do you know what is the $x_{sp}(t)$ or $x_{tr}(t)$?
The book states that the transient motion and the steady periodic oscillation of the mass are given by $x_{sp}(t)$, but I don't even understand what they are trying to say.

Thanks!

6. Nov 22, 2013

### Staff: Mentor

If your solution looks something like x(t) = cos(at) + e-3tsin(at), the steady-state part is the first term; the transient part is the decaying exponential term, so called because it drops off to zero after a short time.

7. Nov 22, 2013

### iRaid

Ahh I see now. Thank you!