Solving Energy of a Spring: Find Velocity at 2/3 E

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SUMMARY

The discussion focuses on calculating the velocity of an object attached to a spring when its potential energy is two-thirds of the total mechanical energy (E). The key equation used is E = KE + PE_s, where the potential energy (PE) at this state is 2/3 E, leading to the conclusion that kinetic energy (KE) is 1/3 E. The correct formula derived for velocity is v = sqrt(k/m) A, where k is the spring constant, m is the mass, and A is the maximum displacement. The initial attempt at the solution was incorrect due to a misunderstanding of energy distribution at different points in the oscillation.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with potential energy (PE) and kinetic energy (KE) equations
  • Knowledge of spring constants and mass in oscillatory systems
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Review the principles of simple harmonic motion (SHM) and energy conservation
  • Study the derivation of velocity in oscillatory systems using energy methods
  • Explore the relationship between spring constant (k), mass (m), and amplitude (A) in SHM
  • Practice solving similar problems involving energy distribution in oscillating systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to energy in spring systems.

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Homework Statement



An object of mass m attached to a spring of force constant oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.

What is the object's velocity when its potential energy is 2/3 E ?


Homework Equations



E=KE+PE_s



The Attempt at a Solution



1/2kA^2 = 2/3(1/2kA^2) + 1/3(1/2mv^2)

solving for v...

1/6mv^2 = 1/2kA^2 - 1/3kA^2

1/6mv^2 = 1/6 kA^2

mv^2 = kA^2

v= sqrt(k/m) A

But it's wrong...any help? Thanks!
 
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velocity when its potential energy is 2/3 E
when PE = 2/3*E then KE = 1/3*E
At maximum displacement, v = 0, KE = 0, so E = PE =1/2*k*A²

1/2kA^2 = 2/3(1/2kA^2) + 1/3(1/2mv^2)
This says E = 2/3*(maxPE) + 1/3*(maxKE) and is only true when v is the maximum velocity. Not the velocity you are looking for in the question.
 

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