Thepotential energy function for a particle executing linear simplehar

1.	Thepotential energy function for a particle executing linear simpleharmonic motion is given by V(x)=kx2/2,where kis the force constant of theoscillator. For k= 0.5 N m–1,the graph of V(x)versus xis shown in Fig. 6.12. Show thata particle of total energy 1 J moving under this potential must ‘turnback’ when it reaches x= ± 2 m.
Answer» The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx²/2, where k is the force constant of the oscillator. For k = 0.5 N m^–1, 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.

Thepotential energy function for a particle executing linear simpleharmonic motion is given by V(x)=kx2/2,where kis the force constant of theoscillator. For k= 0.5 N m–1,the graph of V(x)versus xis shown in Fig. 6.12. Show thata particle of total energy 1 J moving under this potential must ‘turnback’ when it reaches x= ± 2 m.

Answer»

The
potential energy function for a particle executing linear simple
harmonic motion is given by V(x)
=kx²/2,
where k
is the force constant of the
oscillator. For k
= 0.5 N m^–1,
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.

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