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A mass-spring system subject to an external periodic forcing of angular frequency $H \lt 0$ satisfies the differential equation

$Y''+18Y'+243Y= \cos(Ht)$

$y(t)=R \cos(Ht-p)$ , $p= \phi$

Find $R$ and $p$

The values should be numbers I think

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Welcome to MSE! It really helps readability to format things using MathJax. Please make sure I got your question correct! Regards – Amzoti 1 min ago edit –  Amzoti Feb 1 '13 at 18:55
    
hint:you can use Fourier series to solve it –  Maisam Hedyelloo Feb 1 '13 at 19:04
    
I don't know what the Fourier series is yet. Can you explain how to use it? Or any web links where I can learn? –  user60587 Feb 1 '13 at 19:07
    
    
@MaisamHedyelloo: the OP is only asking for the amplitude and phase of a particular cosine term. The answer is simply an exercise in algebra. –  Ron Gordon Feb 1 '13 at 19:42

1 Answer 1

You simply plug the stated solution into the differential equation and get

$$-R H^2 \cos{(H t-p)}-18 R H \sin{(H t-p)}+243 R \cos{(H t-p)}=\cos{(H t)}$$

Expand the trig functions by the addition theorems and factor out terms in $\cos{(H t)}$ and $\sin{(H t)}$. After some algebra, the result is

$$[R (243- H^2) \cos{p} + 18 R H \sin{p}] \cos{(H t)} + [R (243- H^2) \sin{p} - 18 R H \cos{p}] \sin{(H t)} = \cos{(H t)}$$

Set the coefficient of $\cos{(H t)}$ on the LHS to $1$ and the coefficient of $\sin{(H t)}$ to $0$. The latter will reveal $p$:

$$\tan{p} = \frac{18 H}{243-H^2}$$

Use this in the former equation to reveal $R$:

$$R = [(243-H^2)^2 + 324 H^2]^{-1/2}$$

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