Hi, so I did the question, but apparently my answer is off by a factor of 2.(adsbygoogle = window.adsbygoogle || []).push({});

An accelerated electron radiates energy at the rate Ke^{2}a^{2}/c^{3}...e is the electron charge, a is the instantaneous acceleration, c is the speed of light, K=6x10^{9}Nm^{2}/C^{2}...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 Ke^{2}8A^{2}pi^{4}f^{3}/c^{3}.

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 kA^{2}, and since k=w_{0}m the energy stored becomes (1/2)4pi^{2}f^{2}mA^{2}...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 mc^{3}/(2Ke^{2}pif)...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..

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# Homework Help: Quality Factor of Oscillator, Can someone Look at My Work Please

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