Time period of a harmonic oscillator

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Homework Help Overview

The discussion revolves around the time period of a harmonic oscillator, specifically examining an integral that involves amplitude and energy parameters. Participants are analyzing the mathematical representation of the problem and questioning the nature of the system described.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the expression for the integral and its components, particularly the role of the amplitude A. There is also a discussion about the nature of the system, with some questioning whether it qualifies as a harmonic oscillator based on the value of n.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some have provided insights into the relationship between the restoring force and the definition of a harmonic oscillator, while others are focused on the mathematical aspects of the integral.

Contextual Notes

There is a suggestion that the problem may involve a 'trick' question regarding the definition of a harmonic oscillator, particularly when n is not equal to 2. Participants are also considering the implications of using Gamma functions in the integral.

VVS2000
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Homework Statement
Given is the potential energy of the harmonic oscillator: U=a|x|^n, amplititude is A
Find the time period of this harmonic oscillator
Relevant Equations
E=(1/2)m(dx/dt)^2 + a|x|^n
20200918_164850.jpg
 
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Your result is written as
4A\sqrt{\frac{m}{2E}}\int_0^1 \frac{dx}{\sqrt{1-x^n}}
where amplitude A is
A=(\frac{E}{a})^{\frac{1}{n}}
 
anuttarasammyak said:
Your result is written as
4A\sqrt{\frac{m}{2E}}\int_0^1 \frac{dx}{\sqrt{1-x^n}}
where amplitude A is
A=(\frac{E}{a})^{\frac{1}{n}}
No, I have'nt written 4A. It's 4. A is inside the root in the denominator.
But how do you solve that integral?
 
anuttarasammyak said:
Your result is written as
4A\sqrt{\frac{m}{2E}}\int_0^1 \frac{dx}{\sqrt{1-x^n}}
where amplitude A is
A=(\frac{E}{a})^{\frac{1}{n}}
No, I have'nt written 4A. It's 4. A is inside the root in the denominator.
But how do you solve that integral?
 
I made a replacement of x/A ##\rightarrow## x.

The definite integral is a function of ##\Gamma## functions of ##\frac{1}{n}##.
 
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Harmonic oscillator in classical physics are not systems subject to an force/ente proportional to its "displacement"? So n shouldn't be two?
Or it is not a harmonic oscillator?
 
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As noted by LCSphysicist (#6), this is NOT a harmonic oscillator (except when n=2). The word 'harmonic' is specifically used twice. Is there a possibility this is a 'trick' question?

A harmonic oscillator is one which has a restoring given by:
##\vec F = -k\vec x ##
If I were answering I would demonstrate that only n=2 gives a harmonic oscillator. (Hint, what is the relationship between ##\vec F ## and U?)

Then express in the usual SHM equation format:
##\frac {d^2 x }{dt^2} + \omega^2 x = 0##
You then get the period from ##\omega##.

Just a thought.
 

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