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Thread: algebraic manipulation

  1. #1
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    algebraic manipulation

    Hello,

    I'm trying to follow the rearranging of a formula from:

    $\displaystyle M=\frac {m\Omega^2}{\sqrt{r^2 \Omega^2+(k-m \Omega^2)^2}}$

    to:

    $\displaystyle M=\frac {\beta^2}{\sqrt{(1-\beta^2)^2+4\alpha^2\beta^2}}$

    Using the following;

    $\displaystyle \alpha=\frac{r}{2\sqrt{mk}}$

    $\displaystyle \omega = \sqrt{k/m}$

    $\displaystyle \beta=\Omega/\omega$

    I have got as far as;

    $\displaystyle M=\frac {m \Omega^2}{\sqrt{4 \alpha^2km \Omega^2+k^2-m^2 \Omega^4}}$

    but always have unwanted variables leftover... any help please.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by robocombot View Post
    Hello,

    I'm trying to follow the rearranging of a formula from:

    $\displaystyle M=\frac {m\Omega^2}{\sqrt{r^2 \Omega^2+(k-m \Omega^2)^2}}$

    to:

    $\displaystyle M=\frac {\beta^2}{\sqrt{(1-\beta^2)^2+4\alpha^2\beta^2}}$

    Using the following;

    $\displaystyle \alpha=\frac{r}{2\sqrt{mk}}$

    $\displaystyle \omega = \sqrt{k/m}$

    $\displaystyle \beta=\Omega/\omega$

    I have got as far as;

    $\displaystyle M=\frac {m \Omega^2}{\sqrt{4 \alpha^2km \Omega^2+k^2-m^2 \Omega^4}}$

    but always have unwanted variables leftover... any help please.
    Hi

    You can see that $\displaystyle \Omega$ is present only in $\displaystyle \beta$ expression therefore you can replace $\displaystyle \Omega$ by $\displaystyle \beta \omega$

    and that $\displaystyle r$ is present only in $\displaystyle \alpha$ expression therefore you can replace $\displaystyle r$ by $\displaystyle 2\alpha \sqrt{mk}$

    $\displaystyle M=\frac {m\Omega^2}{\sqrt{r^2 \Omega^2+(k-m \Omega^2)^2}}$

    $\displaystyle M=\frac {m\beta^2 \omega^2}{\sqrt{4 \alpha^2 m k \beta^2 \omega^2+(k-m \beta^2 \omega^2)^2}}$

    Dividing numerator and debominator by $\displaystyle m \omega^2$

    $\displaystyle M=\frac {\beta^2}{\sqrt{4 \alpha^2 \beta^2 \frac{k}{m \omega^2}+(\frac{k}{m \omega^2}- \beta^2)^2 }}$

    Using $\displaystyle \frac{k}{m \omega^2} = 1$

    $\displaystyle M=\frac {\beta^2}{\sqrt{(1-\beta^2)^2+4\alpha^2\beta^2}}$
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  3. #3
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    thanks, great help!
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  4. #4
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    Lexington, MA (USA)
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    Hello, robocombot!

    Show that:
    . . $\displaystyle M \;=\;\dfrac {m\Omega^2}{\sqrt{r^2 \Omega^2+(k-m \Omega^2)^2}} \;=\;\dfrac {\beta^2}{\sqrt{(1-\beta^2)^2+4\alpha^2\beta^2}}$

    using the following; .$\displaystyle \begin{Bmatrix}
    \alpha &=&\frac{r}{2\sqrt{mk}} & [1] \\ \\[-3mm]
    \omega &=& \sqrt{\frac{k}{m} & [2] \\ \\[-3mm]
    \beta&=&\frac{\Omega}{\omega}& [3] \end{Bmatrix}$

    $\displaystyle \text{From [2] and [3]: }\;\beta \:=\:\dfrac{\Omega}{\sqrt{\frac{k}{m}}} \:=\:\Omega\sqrt{\dfrac{m}{k}} \quad\Rightarrow\quad \beta^2 \:=\:\dfrac{m\Omega^2}{k} $


    I started with the right side . . .


    $\displaystyle M \;=\;\dfrac{\beta^2}{\sqrt{(1-\beta^2)^2 + 4\alpha^2\beta^2}} $


    . . .$\displaystyle \;=\; \dfrac{ \dfrac{m\Omega^2}{k}} {\sqrt{\left(1-\dfrac{m\Omega^2}{k}\right)^2 + 4\left(\dfrac{r^2}{4mk}\right)\left(\dfrac{m\Omega ^2}{k}\right)}}
    $


    . . .$\displaystyle =\; \frac{ \dfrac{m\Omega^2}{k}} {\sqrt{\left(\dfrac{k-m\Omega^2}{k}\right)^2 + \dfrac{r^2\Omega^2}{k^2}}} $


    . . .$\displaystyle =\; \frac{\dfrac{m\Omega^2}{k}} {\dfrac{\sqrt{(k-m\Omega^2)^2 + r^2\Omega^2}}{k}} $


    . . .$\displaystyle =\;\dfrac{m\Omega^2}{\sqrt{r^2\Omega^2 + (k-m\Omega^2)^2}} $

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