Average energy mass on a spring

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SUMMARY

The mean potential energy (PE) of a mass on a spring with spring constant k and maximum displacement x0 is calculated as half of the maximum potential energy. The maximum potential energy is defined as 1/2 * k * x0^2. The average potential energy over time is derived from the instantaneous potential energy formula, PE = 1/2 * k * x^2 = 1/2 * k * x0^2 * cos^2(ωt + φ), where the average of cos^2(ωt + φ) over a complete cycle is 1/2. Therefore, the mean potential energy is confirmed to be 1/4 * k * x0^2.

PREREQUISITES
  • Understanding of Hooke's Law and spring mechanics
  • Familiarity with potential energy equations
  • Knowledge of trigonometric functions and their averages
  • Basic concepts of harmonic motion
NEXT STEPS
  • Study the derivation of potential energy in harmonic oscillators
  • Learn about the energy conservation principles in mechanical systems
  • Explore the relationship between angular frequency and spring constant
  • Investigate the effects of damping on potential energy in springs
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for clear explanations of potential energy concepts in spring systems.

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Homework Statement



Hi, I hope this isn't a silly question. I am looking to find the mean potential energy of a mass on a spring with spring constant k and maximum displacement x0.

Homework Equations





The Attempt at a Solution


I know the maximum energy is 1/2*kx0^2 so would the mean potential energy be half that?
 
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Yes, the mean PE will be half of the maximum , but not just like that.
PE at any instant is 1/2 kx2 = 1/2kx02cos2(ωt+ø) ,
avg. of cos2(ωt+ø) over a time period is 1/2.
 
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ah okay thanks!
 

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