What Physical Situation is Governed by the Standard SHO Potential?

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Homework Help Overview

The discussion revolves around the standard simple harmonic oscillator (SHO) potential, expressed as \( V = \frac{1}{2} m \omega^2 x^2 \). Participants are exploring examples of physical situations that this potential describes, particularly in the context of quantum mechanics and classical mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the SHO potential and Hooke's Law, questioning the conditions under which the constants relate to each other. There is an exploration of the meaning of angular frequency and its connection to the potential energy expression.

Discussion Status

The discussion is active, with participants providing insights into the relationship between the SHO potential and classical mechanics principles. Some guidance has been offered regarding the interpretation of angular frequency and its derivation from Hooke's Law, but multiple interpretations and clarifications are still being explored.

Contextual Notes

Participants are navigating the definitions and relationships between variables in the context of quantum mechanics and classical oscillators, with some uncertainty about the application of these concepts in specific physical situations.

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Homework Statement


My QM book says that the standard SHO potential is [tex]m \omega^2 x^2/2[/tex]. Can someone give me an example of a physical situation that is governed by this potential. It seems rather out of nowhere to me...


Homework Equations





The Attempt at a Solution

 
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Any situation that is governed by Hookes Law: f=-kx, will have a potential V=kx^2/2
 
Yes but when is k = m \omega x? I guess \omega = g works, but
omega is usually an angle.
 
You are probably thinking of the oscillator potential energy using the formula:

[tex]V=1/2kx^2[/tex]

Remember that Omega is the angular frequency of the oscillation. Also, remember that for an oscillator obeying Hooke's Law the angular frequency is:

[tex]\omega = \sqrt{k/m}[/tex]

If you solve this for k, you should see where the m*omega^2 is coming from in the book's expression for the potential energy.
 

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