I fail to see how the integral over dE should be equal to pi. I've tried multiple subsitution but I can't seem to find it.
Anyone out there that can shed some light on this?
>>7642110
it is trivial as the book says
>>7642115
Actually is says elementary..
Can you show me the steps or are you just here to be funny?
>>7642110
Does anyone has any idea?
Book and page?
>>7642163
Mechanics (Volume 1) by Landau en Lifshitz, Third edition.
Page 28.
The chapter, or paragraph, is called "Determination of the potential energy from the period of oscillation"
>>7642110
Do a substitution to reduce it to the integral
[math]\int_0^1 \frac{\mathrm{d}u}{\sqrt{(1-u)u}}[/math]
>>7642168
>lifshitz.
Huehuehue.
>>7642174
Can you make a picture?
>>7642174
What am I doing wrong?
>>7642224
Am I taking the wrong steps?
No one here that can help me?
I don't know about the minus sign you get though senpai
>>7642515
First of, Thanks for helping!
Doesn't the minus sign come from the change in variables?
And what inverse function comes out of the integral in your case? Can't seem to read it properly.
>>7642224
You have immaculate and stylish handwriting.
>>7642604
Whenever you're faced with an integral in physics which is in terms of physical quantities, you're first step should almost always be some (linear) change of variables to a dimensionless parameter. Then the resulting integral which just equal some number multiplied by the physical constants, and whether you can solve the integral or not is usually of little importance to the physics. In this case, you have an integral over energy which depends on the two dimensionful energy values alpha (which I'll call a because I'm too lazy to latex) and U.
Now the easiest dimensionless change of variables is a linear one, z, such that z(E=U)=0 and z(E=a)=1, which you can figure out is just given by z = (E-U)/(a-U). Then you end up with a an integral with no parameters which you can look up or solve with a sin(t^2) subsitution.
>>7642635
Thanks for the advice! I'll try that.
>>7642635
Works out like a charm!
Thanks again
