Displacement x of simple harmonic oscillation

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Homework Help Overview

The problem involves simple harmonic oscillation, specifically determining the displacement \( x \) when the kinetic energy \( K \) is one-sixth of the potential energy \( U \). The amplitude of the oscillation is given as \( A = 10 \) cm.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic and potential energy in the context of simple harmonic motion, attempting to derive the displacement using energy equations. Questions arise regarding the correctness of the kinetic energy formula and the need for additional equations related to energy.

Discussion Status

Some participants have provided alternative approaches to the problem, with one suggesting a revised equation relating kinetic and potential energy. There is an ongoing exploration of the definitions and terminology used, particularly concerning kinetic energy in the context of a spring system.

Contextual Notes

Participants note the assumption that the spring is massless, which influences the interpretation of kinetic energy in the system. There is also a discussion about the potential confusion surrounding the abbreviation for kinetic energy.

deezy
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Homework Statement


The amplitude of simple harmonic oscillation is A = 10 cm. Find a displacement x when K = 1/6 U. Here K is a kinetic energy and U is a potential energy.


Homework Equations


KE = \frac{1}{2} k A^2
U = \frac{1}{2} k x^2

The Attempt at a Solution



I'm not sure the correct method of doing this problem but here is what I have attempted based on an example from my notes:

KE = \frac{1}{6} U
\frac{1}{2} k A^2 = \frac{1}{6} \frac{1}{2} k x^2
A^2 = \frac{1}{6} x^2
(0.1)^2 = \frac{1}{6} x^2
x \approx 0.24 m
 
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Your formula for kinetic energy isn't correct.
 
You also need another equation (or law) regarding energy.
 
This may be it, I think:

K = \frac{1}{6} U
U = \frac{1}{2} kx^2
KE = U + K = U + \frac{1}{6} U = \frac{7}{6} U = \frac {7}{6} \frac {1}{2} kx^2
KE = \frac {1}{2} kA^2
\frac{1}{2} kA^2 = \frac {7}{6} \frac {1}{2} kx^2
A^2 = \frac{7}{6}x^2
(0.1)^2 = \frac {7}{6}x^2
x \approx 0.0926 m
 
Looks good.

By the way, what does KE stand for? In my first post, I mistakenly thought you were referring to the kinetic energy as KE is a common abbreviation for it.
 
I was using KE as the kinetic energy of the spring... wasn't sure if energy in a spring should be referred to as kinetic energy or just energy.

Should it just be E for energy?
 
You're using the approximation that the spring is massless, so it has no kinetic energy. It only has potential energy. Your KE is the total energy of the system.
 

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