Rationalizing the Denominator

Question

Hello I couldn't get how my professor solved this question. Thanks for your help in advance!

**Given**

$$

\omega'=\omega\sqrt{\frac{m_{e f f}}{m_{e f f}+m}}

$$

**Question**

$$

\frac{\Delta \omega}{\omega} = \ ?

$$

\

**Answer of my professor**

$$

\frac{\Delta \omega}{\omega} {=} \frac{m}{2m_{eff}}

$$

\

**My attempt but couldn't reach to his result.**

$$

\frac{\Delta \omega}{\omega} = \frac{\omega' -\omega}{\omega}

$$

$$

=\frac{\omega \sqrt{\frac{m_{e f f}}{m_{e f f}+m}}-\omega}{\omega}=\frac{\not \omega\left[\sqrt{\frac{m_{e f f}}{m_{e f f}+m}}-1\right]}{\not\omega}

$$

$$

=\frac{\sqrt{m_{e f f}}}{\sqrt{m_{e f f}+m}}-1=\frac{\sqrt{m_{e f f}}-\sqrt{m_{e f f}+m}}{\sqrt{m_{e f f}+m}}

$$

$$

=\left(\frac{\sqrt{m_{e f f}}-\sqrt{m_{e f f}+m}}{\sqrt{m_{e f f}+m}}\right) \left( \frac{\sqrt{m_{e f f}}+\sqrt{m_{e f f}+m}}{\sqrt{m_{e f f}}+\sqrt{m_{e f f}+m}} \right)

$$

$$

=-\frac{m}{\sqrt{m_{e f f}+m}\left(\sqrt{m_{e f f}}+\sqrt{m_{e f f}+m}\right)}

$$

\

**How my professor found? **

$$

\frac{\Delta \omega}{\omega} \stackrel{?}{=} \frac{m}{2m_{eff}}

$$

Answer 1

There's code.

Re-post in text or post an image with the question written.