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.

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Chapter8: Potential Energy And Conservation Of Energy
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6.4 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',
V(x)
the graph of V(x) versus x is shown
in Fig. 6.12. Show that a particle of
total energy 1 J moving under this
potential must 'turn back' when it
reaches x = + 2 m.
Fig. 6.12
Transcribed Image Text:6.4 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', V(x) the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = + 2 m. Fig. 6.12
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