# Thread: Show that eqn1 equals eqn2

1. ## 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

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

James

2. 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.

3. Hello James
Originally Posted by bobred
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

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

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

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

4. Hi

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

James