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Math Help - Show that eqn1 equals eqn2

  1. #1
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    Show that eqn1 equals eqn2

    Hi

    I am asked to show that

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

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

    \alpha=r/(2\sqrt{mk}), \beta=\Omega/\sqrt{k/m}

    I get the following, but I am not sure where to go or if I have made a mistake

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

    Any ideas? Thanks in advance.
    James
    Last edited by bobred; June 23rd 2010 at 02:39 AM.
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  2. #2
    A Plied Mathematician
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    As an overall strategy, I would plug in \alpha and \beta into your second equation, and leave the first equation alone. Simplify the second equation, and hopefully you'll see that it equals the first.
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  3. #3
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    Hello James
    Quote Originally Posted by bobred View Post
    Hi

    I am asked to show that

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

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

    \alpha=r/(2\sqrt{mk}), \beta=\Omega/\sqrt{k/m}

    I get the following, but I am not sure where to go or if I have made a mistake

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

    Any ideas? Thanks in advance.
    James
    You are almost right, but you have lost a power of 2 on the first k inside the square root sign.

    When you take out the k from (k-\beta^2k)^2, you get k^2(1-\beta^2)^2.

    So you should have
    M=\dfrac{\beta^{2}k}{\sqrt{k^2(1-\beta^{2})^{2}+4\alpha^{2}k^{2}\beta^{2}}}
    =\dfrac{\beta^{2}k}{\sqrt{k^2\Big((1-\beta^{2})^{2}+4\alpha^{2}\beta^{2}\Big)}}

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

    Grandad
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  4. #4
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    Hi

    Thanks to you both I have got there both ways now.

    James
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