# Differential equation spring modelling

A $1$ kg body is attached on a spring. The initial displacement is $x_0=0.5$m then we drop it. The equation which describes movement of mass is:

$\ddot x+3\dot x + 2x=0$.

I have managed to solve the differential equation and got:

$x=-0.5e^{-2t}+e^{-t}$

Question is: will mass oscillate and what will be it's maximum displacement from equilibrium point?

What I've done:

I calculated motion amplitude ($A^2$=$c_1^2+c_2^2$ $c_1$=-0.5 and $c_2=1$) and my answer is that it will oscillate since it's not critical damping.

I don't think that it's allowed what I've done (I'm not sure if equation for amplitude holds for damped vibration and if amplitude really is maximum displacement). I am also not sure how to find out whether system will oscillate.

• Exponentials in real variables are monotonic functions. For oscillation you would get something like an exponential times a sine or cosine. Jun 13, 2018 at 16:41
• It is overdamped. No oscillation. The original displacement is the maximum value Jun 13, 2018 at 16:43
• But how to find maximum displacement generally? Jun 13, 2018 at 16:45
• No matter what, if you don't have an initial velocity, you cannot go farther than the initial displacement. But, if you have an initial velocity, the idea is to just take the derivative of the position with respect to time, then set it to 0. Calculate the time, then the position. Jun 13, 2018 at 16:48
• I only know that is dropped so I though that initial velocity is 0 so x'(0)=0 Jun 13, 2018 at 16:48

I don't remember all the physics words for this underdamped, overdamped et.c. It was a long time ago I did it. But with Laplace/Fourier transforms you can derive that solving the second order polynomial equation

$$ar^2+br+c=0 \hspace{1cm}(\text{ solve for r })$$

$$ay''+by'+c=0$$
If you know of complex numbers, then the magnitude of the solutions $r_1,r_2$ determines the real exponent and the argument of the solution determines the frequency of oscillation.