Rationalizing the Denominator

podi

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Hello I try to understand how my professor did this just like that. I also added my attempt.
Thanks for your help!

Professor:

[MATH]\omega^{\prime}=\omega \sqrt{\frac{m_{e f f}}{m_{eff}+m}} \Rightarrow \frac{\Delta w}{w}=-\frac{m}{2 m_{eff}}[/MATH]
My attempt:

[MATH]=\frac{\omega \sqrt{\frac{m_{eff}}{m_{eff}+m}}-\omega}{\omega}=\frac{\not \omega\left[\sqrt{\frac{m_{eff}}{m_{eff}+m}}-1\right]}{\not\omega }[/MATH]
[MATH]=\frac{\sqrt{m_{eff}}}{\sqrt{m_{eff} +m}}-1=\frac{\sqrt{m_{e f f}}-\sqrt{m_{eff}+m}}{\sqrt{m_{eff}+m}}[/MATH]
[MATH]=\frac{\sqrt{m_{eff}}-\sqrt{m_{eff}+m}}{\sqrt{m_{\text {eff}}+m}} \frac{\sqrt{m_{eff}}+\sqrt{m_{eff}+m}}{\sqrt{m_{eff}}+\sqrt{m_{eff}+m}}[/MATH]
[MATH]=-\frac{m}{\sqrt{m_{eff}+m}\left(\sqrt{m_{eff}}+\sqrt{m_{eff}+m}\right)}[/MATH]
 
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Sorry, 4th line should be corrected like that.
[MATH]=\left(\frac{\sqrt{m_{eff}}-\sqrt{m_{eff}+m}}{\sqrt{m_{\text {eff}}+m}}\right)\left( \frac{\sqrt{m_{eff}}+\sqrt{m_{eff}+m}}{\sqrt{m_{eff}}+\sqrt{m_{eff}+m}}\right)[/MATH]
 
Your work looks fine.
My question is where did you get your first line from?
What happened to w'? Where did Δw come from?
 
Maybe it is better to rephrase my question. By explaining where what is [MATH]\Delta \omega[/MATH].

[MATH] = \frac{\Delta \omega}{\omega} = \frac{\omega' -\omega}{\omega} [/MATH]
[MATH] =\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} [/MATH]
[MATH] =\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}} [/MATH]
[MATH] =\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) [/MATH]
[MATH] =-\frac{m}{\sqrt{m_{e f f}+m}\left(\sqrt{m_{e f f}}+\sqrt{m_{e f f}+m}\right)} [/MATH]
I come up above result but I didn't understand how my professor found out below one?
[MATH]\frac{\Delta \omega}{\omega} \stackrel{?}{=} \frac{m}{2m_{eff}}[/MATH]
 
Hello I couldn't understand how my professor reached that result. Can anyone help me?

Given

[MATH]\omega'=\omega\sqrt{\frac{m_{e f f}}{m_{e f f}+m}}[/MATH]
Question


[MATH]\frac{\Delta \omega}{\omega} = \ ?[/MATH]
Answer of my professor


[MATH]\frac{\Delta \omega}{\omega} {=} \frac{m}{2m_{eff}}[/MATH]
My attempt but couldn't reach to his result.


[MATH]\frac{\Delta \omega}{\omega} = \frac{\omega' -\omega}{\omega}[/MATH]
[MATH]=\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} [/MATH]
[MATH] =\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}} [/MATH]
[MATH] =\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)[/MATH]
[MATH] =-\frac{m}{\sqrt{m_{e f f}+m}\left(\sqrt{m_{e f f}}+\sqrt{m_{e f f}+m}\right)} [/MATH]
How my professor found this answer?


[MATH]\frac{\Delta \omega}{\omega} \stackrel{?}{=} \frac{m}{2m_{eff}}[/MATH]
 
Rather than just restate the problem from the other thread, I think you need to put it in context. What was said before this? What is meant by \(\Delta\omega\) in context? (It has to refer to the change caused by something, or between two cases.)

Your work suggests that m (in the denominator) is being taken to 0, but then it shouldn't be present in the answer.

In general, we can't tell you how someone else did something! In particular, it could be that he is wrong! But at the least, we need to know what he knows about the problem.
 
Context is everything. I wonder if your professor said that \( m_{eff} \) is very much greater than \( m \) ?
If so then this could be used to approximate the value of \( \sqrt{m_{eff}+m} \)
 
Rather than just restate the problem from the other thread, I think you need to put it in context. What was said before this? What is meant by \(\Delta\omega\) in context? (It has to refer to the change caused by something, or between two cases.)

Your work suggests that m (in the denominator) is being taken to 0, but then it shouldn't be present in the answer.

In general, we can't tell you how someone else did something! In particular, it could be that he is wrong! But at the least, we need to know what he knows about the problem.
Sorry for restating problem but I realized that I couldn't express the problem clearly and unfortunately I had no more right to edit or delete the thread. Instead I wanted to report previous one and express my question in a more clear manner in another thread. However, lesson is taken I'll be more careful next time.
 
Context is everything. I wonder if your professor said that \( m_{eff} \) is very much greater than \( m \) ?
If so then this could be used to approximate the value of \( \sqrt{m_{eff}+m} \)
I think it is because of that even if I didn't see such kind of expression in my notes from the context I can see that [MATH]m<<m_eff[/MATH]. So probably as you said it is an approximation rather than exact solution of the problem. Your answer made me realize something I completely miss out, thank you.
 
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