Quality Factor of Oscillator, Can someone Look at My Work Please

mmmboh

Hi, so I did the question, but apparently my answer is off by a factor of 2.

An accelerated electron radiates energy at the rate Ke²a²/c³...e is the electron charge, a is the instantaneous acceleration, c is the speed of light, K=6x10⁹Nm²/C²...whatever that is.

So first I had to find how much energy an electron oscillating in a straight line would radiate over one cycle, the motion is describe by x=Asin([tex]2\pi[/tex]ft), I did that and according to the answer book I am right, it is Ke²8A²pi⁴f³/c³.

For the next part I have to find the quality factor...I know that is [tex]2\pi[/tex] x energy stored/energy dissipated per cycle, and the energy stored is 1/2 kA², and since k=w₀m the energy stored becomes (1/2)4pi²f²mA²...when I do that times [tex]2\pi[/tex], and then divide that by the energy dissipated over one cycle that I found in the first part, I get as an answer mc³/(2Ke²pif)...but according to the answer book the denominator should be multiplied by a factor of 4, not two...

Am I missing something here? I can't figure out where my mistake is..

mmmboh

can anyone clear this up please?

ulitoo

I am getting same results as you
factor of 2 , not 4.
I used the aproximation of ln(1-x)=-x if x<1 assuming the radiation is not taking too much energy of the electron