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How do I prove that the width of resonance in an RLC circuit is the inverse of the proper time of the circuit?

I know that the amplitude of charge in the condensator is $$Q(\omega)=\frac{V_\text{max}/L}{\sqrt{(\omega_0^2-\omega^2)^2+(\gamma\omega)^2}}$$ So I tried to show from the equality $Q(\omega_0\pm\Delta\omega)=\frac{1}{\sqrt{2}}Q(\omega_0)$ that $\Delta\omega = 2 / \gamma$ but the expression is to hard. Do I need to do approximations?

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  • $\begingroup$ What have you done to try to figure out the answer yourself? $\endgroup$ – David Z Aug 31 '15 at 2:27
  • $\begingroup$ Yes, you need the approximation $\Delta \omega \ll \omega_0$. Also, you should ignore everything in the expression except for what's inside the square root, since that's the important part. $\endgroup$ – knzhou Aug 31 '15 at 2:51
  • $\begingroup$ What do you mean by "proper time"? $\endgroup$ – DanielSank Aug 31 '15 at 3:02
  • $\begingroup$ Please define $\gamma$. Is this a series circuit or parallel circuit? Is it voltage or current driven? $\endgroup$ – DanielSank Aug 31 '15 at 3:22
  • $\begingroup$ Something is wrong with your denominator. First term under the radical has units of $\omega^4$ and the second term has $\omega^2$. Unless $\gamma$ has units of inverse time, too. $\endgroup$ – Bill N Aug 31 '15 at 3:25

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