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

  • Thread starter mmmboh
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  • #1
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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..
 
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Answers and Replies

  • #2
407
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can anyone clear this up please?
 
  • #3
1
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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
 

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