# Show that eqn1 equals eqn2

• Jun 23rd 2010, 01:53 AM
bobred
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
• Jun 23rd 2010, 02:50 AM
Ackbeet
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.
• Jun 23rd 2010, 02:54 AM
Hello James
Quote:

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

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

• Jun 23rd 2010, 03:39 AM
bobred
Hi

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

James