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[tex]Q = \frac{\omega_0}{\omega_1 - \omega_2} = \frac{m\omega_0}{b}[/tex]

where Q is the quality factor or Q value of a forced harmonic oscillator, and [itex]\omega_2[/itex] and [itex]\omega_1[/itex] are the frequencies where the square of the amplitude [itex]A_0[/itex] is half its maximum value.

The equation for [itex]A_0[/itex] is:

[tex]A_0 = \frac{F_0}{m\sqrt{(\omega^2 - \omega_0) + (b\omega/m)^2)}}[/tex]

where [itex]\omega[/itex] is the applied frequency and [itex]\omega_0[/itex] is the natural frequency. The amplitude has maximum value when [itex]\omega = \omega_0[/itex] which is [itex]F_0/(b\omega_0)^2[/itex].

So, since the amplitude depends on the applied frequency, all I need to do is solve for [itex]\omega[/itex] in:

[tex]A_0(\omega)^2 = \frac{\mathrm{max}(A_0)^2}{2}[/tex]

which should give me two answers, since it's a quadratic, which correspond to [itex]\omega_2[/itex] and [itex]\omega_1[/itex]. I can then use these to find the Q value. Is this correct? The equations I get are ugly and I can not simply them to obtain the answer.