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Derivation of Q Value

  1. Jun 26, 2005 #1
    I'm supposed to derive the following equation:

    [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.
  2. jcsd
  3. Jun 26, 2005 #2


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    Will it help if you correct your equation for the amplitude? I think it should be

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

    Check your source, and make sure you have it right.
  4. Jun 26, 2005 #3
    I'm under the assumption that the threadmaker simply made a typo... regardless that is a very minor error that would cause a huge difference in value.
  5. Jun 26, 2005 #4


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    A hint then:


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

    find the condition the denominator must satisfy to make the squared amplitude half maximum. Call the frequency greater than [tex] \omega_o [/tex] frequency [tex] \omega_1 [/tex] and the frequency lower than [tex] \omega_o [/tex] frequency [tex] \omega_2 [/tex]. Write the two equations that satisfy the condition in the form of differences of squares to the first power, and add them to eliminate [tex] \omega_o [/tex]. The solution will then be close at hand.
  6. Jun 27, 2005 #5
    Ah yes, I did make a typo. Maybe I made typos when I worked out the problem as well...

    The maximum amplitude is [itex]F_0/(b\omega_0)[/itex]. So, I setup the equation to solve as

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

    Is this correct?
  7. Jun 27, 2005 #6


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    You need to square the right hand side. You also need to recognize that your maximum condition is an approximation for small b, so the resonance is sharply peaked. You need to continue in the spirit of that approximation when you set the denominators of both sides equal. I don't think you can get to the answer if you don't make that assumption. I've been trying to generalize to the exact maximum of the amplitude function, but so far it has not worked out.
  8. Jun 27, 2005 #7
    Well, the book does assume that the harmonic oscillator is lightly damped, so there is a sharp peek in which case the maximum is achieved when [itex]\omega \approx \omega_0[/itex]. This is not generally true though.

    So, we have:

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

    [tex]\frac{1}{2(b\omega_0)^2} = \frac{1}{m^2(\omega^2 - \omega_0^2)^2 + (b\omega)^2}[/tex]

    [tex]m^2(\omega^2 - \omega_0^2)^2 + (b\omega)^2 = 2(b\omega_0)^2[/tex]

    and I end up with an ugly quadratic. Is this right?
  9. Jun 27, 2005 #8


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    Now you need to get a bit tricky to avoid the quadratic ugliness. You can't fool around with the difference term, because the difference is small, and subject to large errors if you do an approximation. But the frequencies in the terms with the [tex] b\omega [/tex] products can be set approximately equal, which will let you equate two squares, and take the square roots. Replace the [tex] \omega_0 [/tex] on the right with [tex] \omega [/tex] , simplify, and take the square root. From this write two equations for the two different frequncies, one of which, [tex] \omega_1 [/tex] , is greater than [tex] \omega_0 [/tex] and one of which, [tex] \omega_2 [/tex] , is less than [tex] \omega_0 [/tex]. Add the two equations to get a difference of squares of the two frequencies on the left, and a sum of the two frequencies on the right. Factor and reduce and your done.
  10. Jun 28, 2005 #9
    Wow. No wonder I couldn't derive it. I really don't understand how you can substitute the natural frequency with the applied frequency in one place in the equation but not in another. I mean, if I write the equation like this:

    [tex]m^2(\omega^2 - \omega_0^2)^2 = b^2(2\omega_0^2 - \omega^2)[/tex]

    I have would also a difference of squares on the RHS so wouldn't approximating result in large errors in that part too. Or is it negligible since that part isn't squared?

    The two equations that you suggest I add are:

    [tex]m(\omega_1^2 - \omega_0^2) = b\omega_1[/tex]
    [tex]m(\omega_0^2 - \omega_2^2) = b\omega_2[/tex]

    Why must the order of [itex]\omega_0[/itex] be reversed on the LHS of the second equation? I know that [itex]\omega_2 < \omega_0[/itex], but so what? Why does the equation change?
  11. Jun 28, 2005 #10


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    Your last question is easy. The first can be approached with varying degrees of rigor. The last two equations come from taking square roots. You have to match positive roots with positve roots. The right hand side of both equations is positive, so the left hand sides must be positive. The negative roots correspond to negative frequencies and are extraneous.

    As for the approximation, here's a slightly different approach. Rewrite the quadratic as

    [tex]m^2(\omega^2 - \omega_0^2)^2 + b^2(\omega^2 - \omega_0^2) - b^2\omega_0^2 = 0 [/tex]

    The middle term is much smaller than the other two terms. The approximation that the peak is at the natural frequency comes from

    [tex] m^2\omega^2 >> b^2 [/tex]

    for all frequencies near the peak, so the second term is much less than the first term. Also, since the second term involves the difference of two nearly equal terms, both comparable in magnitude to the last term, the second term can be neglected. If you discard that second term you get the last two equations with the two driving frequencies replaced by the natural frequncy on the right side. When you add the two equations and factor, you then need to recognize that the sum of the two driving frequencies is very nearly the same as twice the natural frequency.

    A more rigorous justification can be done for this if you need it, but I can't do it right now. You could start by finding the actual location of the peak amplitude at

    [tex] \omega ^2 = \omega _{\rm{o}} ^2 - \frac{{\left( {b/m} \right)^2 }}{2} [/tex]

    and from that finding the true value of the maximum amplitude. I'm not sure you need to go down that road.
  12. Jun 28, 2005 #11
    Ah right. Forgot about the whole +/- root business with square roots.

    That sounds more convincing.

    No. I think we can leave it at that. Thank you very much for your help.
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