# Thread: Damped spring - general solution of IVP

1. ## Damped spring - general solution of IVP

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!

2. 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.

3. 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?