How is the steady periodic oscillation used to solve problems?

[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 can't figure it outAny help is appreciated.

 

[LCKurtz]

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: $$x_{sp}(t)=5cos4t+4sin4t=\sqrt{41}\left( \frac{5}{\sqrt{41}}cos4t+\frac{4}{\sqrt{41}}sin4t \right)=\sqrt{41}cos(4t-\alpha)$$

I feel like it's some trig substitution, identity, etc. that I'm forgetting, but I can't figure it outAny help is appreciated.

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##.

 

[iRaid]

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##.

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

 

[LCKurtz]

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

 

[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!

 

[Mark44]

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.

 

[iRaid]

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.

Ahh I see now. Thank you!