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Thread: Seems like the quadratic formula but I dont understand the steps

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    Seems like the quadratic formula but I dont understand the steps

    Seems like the quadratic formula but I dont understand the steps-capture.jpg
    Last edited by cxz7410123; Jul 23rd 2017 at 03:13 PM.
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    Re: Seems like the quadratic formula but I dont understand the steps

    Seems like the quadratic formula but I dont understand the steps


    ... the equation is quadratic in $\omega^2$ ...

    $a = m^2$, $b = -4km$, $c = 3k^2$

    $\omega^2 = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{4km \pm \sqrt{(-4km)^2 - 4(m^2)(3k^2)}}{2(m^2)}$

    $\omega^2 = \dfrac{4km \pm \sqrt{16k^2m^2 - 12k^2m^2}}{2(m^2)}$

    $\omega^2 = \dfrac{4km \pm 2km}{2(m^2)}$

    $\omega_1 = \sqrt{\dfrac{k}{m}}$

    $\omega_2 = \sqrt{\dfrac{3k}{m}}$
    Last edited by skeeter; Jul 23rd 2017 at 04:24 PM. Reason: typo
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    Re: Seems like the quadratic formula but I dont understand the steps

    Strictly speaking a quartic (fourth degree) equation will have four solutions. The other two would be -\omega_1 and -\omega_2. But a "frequency" is always a positive number.
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    Re: Seems like the quadratic formula but I dont understand the steps

    Thank you for your replies.
    I have one more question: How do you convert it to quadratic and equalize it to ω^2 ?
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    Re: Seems like the quadratic formula but I dont understand the steps

    Actually, I would say that the equation is quadratic in $m \omega^2.$

    We start with $m^2 \omega^4 - 4km \omega^2 + 3k^2 = 0.$

    $\text {Let } u = m \omega^2 \implies u^2 = m^2 \omega^4 \implies u^2 - 4uk + 3k^2 = 0 \implies$

    $(u - 3k)(u - k) = 0 \implies u = 3k \text { or } u = k \implies $

    $m \omega^2 = 3k \text { or } m \omega^2 = k \implies \omega = \pm \sqrt{\dfrac{3k}{m}} \text { or } \omega = \pm \sqrt{\dfrac{k}{m}}.$

    The mechanics are easy. What may be hard is to see that the substitution

    $u = m \omega^2$ will turn the whole thing into a simple quadratic.

    EDIT: Halls points out that a quartic may have four different roots as this shows. But he also says that the negative solutions can be ignored.
    Last edited by JeffM; Jul 24th 2017 at 12:24 PM.
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    Re: Seems like the quadratic formula but I dont understand the steps

    The given equation is m^2\omega^2- 4km\omega^2+ 3k^2= 0. I would substitute x for \omega^2 (JeffM would use x= m\omega^2- either works). In terms of x, the equation becomes m^2x^2- 4kmx+ 3k^2= 0, a quadratic equation in x. By the quadratic formula, x= \frac{4km\pm \sqrt{16k^2m^2- 4(m^2)(3k^2)}}{2m^2}= \frac{4km\pm \sqrt{4k^2m^2}}{2m^2}= \frac{4km\pm 2km}{2\m^2}.

    The two values for x are x= \frac{4km+ 2km}{2m^2}= \frac{6km}{2m^2}= \frac{3k}{m} and x= \frac{4km- 2km}{2m^2}= \frac{2km}{2m^2}= \frac{k}{m}.

    Since x= \omega^2, \omega= \pm\sqrt{x}= \pm x^{1/2} giving
    \omega= \pm \sqrt{\frac{3k}{m}} and \omega= \pm\sqrt{\frac{k}{m}}.

    Those are the four solutions to the quartic equation. As I said before, since a "frequency" is always positive, to solve this problem we drop the negative solutions.
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    Re: Seems like the quadratic formula but I dont understand the steps

    The given equation is m^2\omega^2- 4km\omega^2+ 3k^2= 0. I would substitute x for \omega^2 (JeffM would use x= m\omega^2- either works). In terms of x, the equation becomes m^2x^2- 4kmx+ 3k^2= 0, a quadratic equation in x. By the quadratic formula, x= \frac{4km\pm \sqrt{16k^2m^2- 4(m^2)(3k^2)}}{2m^2} = \frac{4km\pm \sqrt{4k^2m^2}}{2m^2} = \frac{4km\pm 2km}{2m^2}.

    The two values for x are x= \frac{4km+ 2km}{2m^2}= \frac{6km}{2m^2}= \frac{3k}{m} and x= \frac{4km- 2km}{2m^2}= \frac{2km}{2m^2}= \frac{k}{m}.

    Since x= \omega^2, \omega= \pm\sqrt{x} giving
    \omega= \pm \sqrt{\frac{3k}{m}} and \omega= \pm\sqrt{\frac{k}{m}}.

    Those are the four solutions to the quartic equation. As I said before, since a "frequency" is always positive, to solve this problem we drop the negative solutions.
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    Re: Seems like the quadratic formula but I dont understand the steps

    I did not mean to imply in any way, shape, or form that I disagreed with Hall's method of solution. His method and mine achieve exactly the same result and in fairly similar ways. I merely found my way to be a bit more intuitive, but intuition is in the mind's eye of the beholder.

    As I said, the mechanics are not hard. What may be hard is to see that a substitution may turn the quartic into a quadratic. For some reason, substitution as a technique for getting standard forms is not usually taught until calculus.
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