Fractional energy in a damped harmonic oscillator

[endeavor]

Homework Statement

Show that the fractional energy lost per period is
$$\frac{\Delta E}{E} = \frac{2\pi b}{m\omega_0} = \frac{2\pi}{Q}$$
where $$\omega_0 = \srqt{k/m}$$ and $$Q = m\omega_0 / b$$

Homework Equations

$$E = 1/2 k A^2 e^{-(b/m)t} = E_0 e^{-(b/m)t}$$

The Attempt at a Solution

$$\Delta E = 1/2 k A^2 e^{-(b/m)(t + T)} - 1/2 k A^2 e^{-(b/m)t}$$ where $$T = 2\pi / \omega_0$$
$$\frac{\Delta E}{E} = e^{-(b/m)T} - 1$$
$$\frac{\Delta E}{E} = e^{-\frac{2\pi b}{m\omega_0}} - 1$$
What should I do now?

 

[endeavor]

Using the Taylor series of e^x,
$$e^{-(2\pi b)/(m\omega_0)} = 1 + -\frac{2\pi b}{m\omega_0}$$
Can you explain why I would drop the rest of the terms?
So
$$\frac{\Delta E}{E} = -\frac{2\pi b}{m\omega_0}$$
first of all, is this right?
second, how do i account for the negative sign?

 

[AlephZero]

All this assumes the damping is small, i.e. b << m \omega_0, so you can drop the higher terms of the Taylor series.

Re the negative sign, the question asks for the fractional energy lost.

Energy lost in 1 cycle (a positive number) = Initial energy - final energy.

Your equation found the change in energy as

Change in energy in 1 cycle (a negative number) = final energy - initial energy.