algebraic manipulation

**robocombot** · July 10th 2010, 05:17 AM

Hello,

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

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

to:

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

Using the following;

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

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

$\beta=\Omega/\omega$

I have got as far as;

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

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

**running-gag** · July 10th 2010, 05:43 AM

Originally Posted by robocombot

Hello,

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

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

to:

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

Using the following;

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

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

$\beta=\Omega/\omega$

I have got as far as;

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

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

Hi

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

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

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

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

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

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

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

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

**robocombot** · July 10th 2010, 05:53 AM

thanks, great help!

**Soroban** · July 10th 2010, 06:29 AM

Hello, robocombot!

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

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

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

I started with the right side . . .

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

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

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

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

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

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