Displacement x of simple harmonic oscillation

Click For Summary
SUMMARY

The discussion focuses on calculating the displacement x in simple harmonic oscillation when the kinetic energy K is one-sixth of the potential energy U. Given an amplitude A of 10 cm, the equations used include KE = (1/2) k A² and U = (1/2) k x². The correct relationship derived is A² = (7/6) x², leading to a displacement x of approximately 0.0926 m. The clarification on the terminology of kinetic energy (KE) versus energy in the context of springs is also addressed.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with kinetic and potential energy equations
  • Knowledge of the spring constant (k) in oscillatory systems
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of energy conservation in simple harmonic motion
  • Learn about the role of the spring constant (k) in oscillatory systems
  • Explore the implications of massless springs in energy calculations
  • Investigate the relationship between amplitude and displacement in harmonic oscillators
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for examples of energy relationships in simple harmonic oscillation.

deezy
Messages
18
Reaction score
0

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
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
9
Views
2K
Replies
13
Views
2K
Replies
16
Views
2K
Replies
31
Views
2K
Replies
2
Views
959
Replies
5
Views
1K
Replies
1
Views
1K
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K