# Proof of angular frequency in damped oscillator [duplicate]

I was given the equation of motion for a damped oscillator is

$$\frac{{\rm d}^2x}{{\rm d}t^2} + \frac{b}{m}\frac{{\rm d}x}{{\rm d}t} + \frac{k}{m}x= 0$$ and the solution of the motion equation is

$$x(t)=A \exp\left(-(b/2m)t\right)\cos(\omega t+\phi).$$ Now how do I go from the solution of the motion equation to this, while using the first and second derivative to solve: $$ω=\sqrt{\frac{k}{m} - \frac{b^2}{4 m^2}}.$$ I feel like I can do it just that I get confused on what is the derivative for A, $\omega$, and $\phi$. Any help will be appreciated thank you.

• farside.ph.utexas.edu/teaching/315/Waves/node10.html . This is a quite banal problem with loads of relevant pages on it. Google is your friend!
– Gert
Commented Nov 14, 2017 at 20:40
• Possible duplicate of Finding the steady state solution of a damped oscillator
– Gert
Commented Nov 14, 2017 at 20:44
• Plug your solution into the differential equation, do the differentiations, and you get an equation involving $\omega$. Solve for $\omega$.
– Mike
Commented Nov 14, 2017 at 21:07
• @Gert Thanks! And On the webpage you linked the equation they used was x(t)=Ae^(b/2m)tcos⁡(ωt-ϕ). Since the signs are different would the answer still be the same? Commented Nov 15, 2017 at 0:18
• $A, \omega, \phi$ are constants. They do not depend on time. Commented Nov 15, 2017 at 12:04

In general, for an equation of the form $\ddot x + a \dot x + bx = 0$, propose a solution $e^{Rt}$ for $R \in \mathbb C$. Then you will find a quadratic equation for $R$.
Upon solving it, you will find $R = \mu \pm i \nu$ with $\mu,\nu \in \mathbb R$. Using Euler's formula and the fact that linear combinations of solutions are themselves solutions, it can be shown that,
$$x(t) = e^{\mu t}\left( c_1 \cos \nu t + c_2 \sin \nu t\right)$$
assuming no double root ($\nu \neq 0$). You can then apply initial conditions to solve for $c_1$ and $c_2$. This outlines the basic procedure - hope you can take it from here!