# Derivation of damped SHM velocity expression

This is from Main: Vibrations and Waves in Physics.

This is a (basic) standard treatment, regarding SHM light damping and using a trial solution $x=Ce^{pt}$.

Roots of p are $-\frac {\gamma}{2} \pm i{\omega_f}$

Where $${\omega_f} =({ \omega_0^2}-\frac {\gamma^2}{4})^{1/2}$$$$={\omega_0}[1-(\frac {\gamma}{2 \omega_0})^2]^{1/2}$$

So far, so standard. Main then gives (Page 36), the 4 different varieties of expressing $x(t)$ in terms of trig identities and complex notation.

What is losing me here are the implicit steps in the differentiation of the first of these expressions below (producing the second), necessary to establish the initial conditions, I just can't see it.

$$x=A \cos {\theta} ={A_1}$$

$${\dot x}=-\frac {\gamma}{2}A\cos {\theta}-{\omega_f}A\sin {\theta}=0$$

I have looked through various sources, but what with various notations and trig identities, I have lost the flow.

I have read through pdfs such as French and MIT Open Courseware but I would like to keep the notation and approach used by Main.

I self study, (and I follow 90 percent of the rest of the book) but this stupid brain freeze is irritating me.

• Isn't it just $\frac{dx}{dt} = \frac{dA(t)}{dt}\cos\theta(t) - A(t)\sin\theta(t)\frac{d\theta(t)}{dt}$? – Alfred Centauri Nov 10 '17 at 19:56

You'd normally have something like $$x=A_0\ \text{exp}\left(-\frac{\gamma}{2} t\right)\ \text{cos}\ \omega t$$ So differentiating gives $$\frac{dx}{dt}=A_0\ \text{exp}\left(-\frac{\gamma}{2} t\right) \left[-\frac{\gamma}{2} \text{cos} \omega t \ -\omega\ \text{sin} \omega t \right]$$