Pendulum - equipartition experiment

Click For Summary
SUMMARY

The discussion centers on the equipartition theorem as applied to a pendulum experiment. The participant correctly identifies that the total energy adheres to the conservation of energy principle, equating kinetic and potential energy in the context of pendulum oscillation. They express confusion regarding the degrees of freedom in the system, particularly in relation to the kinetic and potential energy contributions. Ultimately, they conclude that the mean total energy can be simplified to KbT, acknowledging the contributions from squared independent variables.

PREREQUISITES
  • Understanding of the equipartition theorem in statistical mechanics
  • Familiarity with kinetic and potential energy concepts
  • Basic knowledge of pendulum motion and oscillations
  • Proficiency in thermodynamic principles, specifically conservation of energy
NEXT STEPS
  • Study the equipartition theorem in detail, focusing on its applications in various physical systems
  • Explore the mathematical derivation of kinetic and potential energy in oscillatory systems
  • Learn about degrees of freedom in thermodynamics and their implications on energy distribution
  • Investigate methods for converting angular measurements into radians and degrees without trigonometric functions
USEFUL FOR

Students of physics, particularly those studying thermodynamics and mechanics, as well as educators seeking to clarify concepts related to energy distribution in oscillatory systems.

ibysaiyan
Messages
441
Reaction score
0

Homework Statement



The problem is : http://i44.tinypic.com/2v7yo29.png

Homework Equations


The Attempt at a Solution


For part a )
I have stated that: The total energy is equivalent to the conservation of energy i.e the two expressions obey the conservation principle. The expression dealing with inertia is the kinetic energy while the other expression's potential since this experiment can be analogous to a pendulum's oscillation.

b)
Now I know that equipartitions states that in thermal equlibirum each degrees of freedom contributes 1/2Kbt i.e for every squared term for the expression of the energy.
In the above case we have two squared expressions.. however i am a little confused with the degrees of freedom. The general mean equation is :
[tex]f/2 [itex]k_{b}[/itex]T ...<br /> <br /> Would the mean total be 2/2 kbt since each expression gives one degrees of freedom.. Or I was thinking that since the glass can rotate in both direction ( clockwise and vice versa) so would that be 2 degrees of freedom for kinetic energy and 2 for potential as well ? :s <br /> <br /> These are the only bits I am stuck on. Thanks for your expected replies!<br /> <br /> EDIT: I think for part b the answer is simply Kbt since each squared independent variable provides 1/2 Kt according to the theorem. :yea =)[/tex]
 
Last edited:
Physics news on Phys.org
I am little stuck on part c... how can I get answer in radians or degrees.. if i don't have any trig. function ?
 
Last edited:
Anyone ?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K