Damped spring - general solution of IVP

**CorruptioN** · May 31st 2010, 11:00 PM

Hi,

I'm working on a 2nd order DE which describes the motion of a damped spring

$mx'' + bx' + kx = 0$

I have found the general solution:

$x(t) = A cos(\omega t + \phi)e^{-\alpha t}$

Where:

$\alpha = b/{2 m}$ and $\omega=\sqrt{k/m - b^2/{4 m^2}}$

which I believe is correct. I now have to find the general solution of $A$ and $\phi$ using $x(0) = x0$ and $x'(0) = v0$ . I'm not really sure where to go with this, so could someone point me in the right direction please?

Thanks!

**HallsofIvy** · June 1st 2010, 02:44 AM

Originally Posted by CorruptioN

Hi,

I'm working on a 2nd order DE which describes the motion of a damped spring

$mx'' + bx' + kx = 0$

I have found the general solution:

$x(t) = A cos(\omega t + \phi)e^{-\alpha t}$

Where:

$\alpha = b/{2 m}$ and $\omega=\sqrt{k/m - b^2/{4 m^2}}$

which I believe is correct. I now have to find the general solution of $A$ and $\phi$ using $x(0) = x0$ and $x'(0) = v0$ . I'm not really sure where to go with this, so could someone point me in the right direction please?

Thanks!

Okay, what you have done looks good. Now put in the given values:
$x(0)= A cos(\phi)= x_0$

From $x(t)= A cos(\omega t+ \phi)e^{-\alpha t}$ , $x'(t)= -\omega Asin(\omega t+ \phi)e^{-\alpha t}- \alpha A cos(\omega t+ \phi)e^{-\alpha t}$
so that $x'(0)= \omega A sin(\phi)- \alpha A cos(\phi)= v_0$ .

Solve the two equations $x(0)= A cos(\phi)= x_0$ and $x'(0)= \omega A sin(\phi)- \alpha A cos(\phi)= v_0$ for A and $\phi$ in terms of $\omega$ , $\alpha$ , $x_0$ , and $v_0$ .

You can eliminate A, getting an equation for $\phi$ only, by dividing one equation by the other.

**CorruptioN** · June 1st 2010, 10:53 AM

Thanks for the reply! That's kind of where I got to, except I found $v_0 = \frac{ \omega A sin(\phi)- \alpha A cos(\phi)}{\alpha^2 + \omega^2}$ . Am I missing something?

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