The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx /2, where k is the force constant of the oscillator. For k = 0.5 N m', the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1J moving under this potential must 'turn back' when it reaches x =+ 2 m.
The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx /2, where k is the force constant of the oscillator. For k = 0.5 N m', the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1J moving under this potential must 'turn back' when it reaches x =+ 2 m.
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter8: Potential Energy And Conservation Of Energy
Section: Chapter Questions
Problem 73AP: A mysterious force acts on all particles along a particular line and always points towards a...
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