Period of Oscillation Homework Statement

AI Thread Summary
The discussion revolves around deriving the period of oscillation for a symmetric potential system, expressed as T=√(8m)∫(0 to a) (dx/√(V(a)-V(x))). The participant attempts to set up the integral using Hamilton's equation and the conservation of energy, noting that at t=0, the particle is released from rest at position x=a. They express confusion about integrating to find T and how the factor √(8m) is obtained. The response emphasizes the importance of separating variables and integrating both sides correctly, guiding the participant towards the algebraic steps needed to arrive at the solution. The thread concludes with a reminder about forum etiquette regarding homework help.
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Homework Statement



Assume that the potential is symmetric with respect to zero and the system has amplitude ##a##, show that the period is given by : ##T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}.##

Homework Equations



##E = \frac12 m(\frac{dx}{dt})^2+V(x)##

The Attempt at a Solution



For a particle, I know that at ##t=0## if we release it from rest at position ##x=a## we then have ##\frac{dx}{dt}=0## at ##t=0## and thus ##E=V(a)##. So when the particle reaches the origin for the first time it has gone through one quarter of a period of the oscillator. Thus, I have to integrate with respect to t from ##0## to ##\frac{T}{4}## and rearrange the equation ##E## for ##\frac{dx}{dt}##. But from here I am not sure how to set it up properly to get ##T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}.##
 
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Hi. Indeed you already have everything to set up your integral:
If you substitute your value of E in Hamilton's equation, can you separate the variables (functions of x on one side, functions of t on the other)?
 
Goddar said:
Hi. Indeed you already have everything to set up your integral:
If you substitute your value of E in Hamilton's equation, can you separate the variables (functions of x on one side, functions of t on the other)?

I have ##V(a) -V(x) = \frac12m(\frac{dx}{dt}2) \implies 2\sqrt{V(a)-V(x)} = m\frac{dx}{dt} \implies \implies \sqrt{\frac{2}{m}}dt = \frac{dx}{\sqrt{V(a)-V(x)}}## but I am confused on how they to get ##T##.
 
What about integrating both sides? You know the limits of the integrals (for t and x) from your first post, so what do you get?
 
When I am integrate I am not sure how they got the ##\sqrt{8m}##.
 
What do you get when you integrate dt from 0 to T/4?
 
Goddar said:
What do you get when you integrate dt from 0 to T/4?

We will get ##T/4## if we integrate dt from 0 to T/4.
 
Ok, then what is (T/4)⋅(2/m)1/2?
This is just algebra...
 
Goddar said:
Ok, then what is (T/4)⋅(2/m)1/2?
This is just algebra...
Oh wow.. how dumb am I. Thank you very much!
 
  • #10
Can you show step by step procedure? I am confused
 
  • #11
juliocezario30 said:
Can you show step by step procedure? I am confused
The thread is a year old. Quite possibly neither participant still uses PF. If you have been given the same homework problem, please follow forum rules by posting your own attempt. A new thread might be best.
 
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