The potential energy function for a particle executing linear simple h

1.	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 Nm-1, the graph of V(x) versus x is shown in fig. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = ± 2 m.
Answer» At any instant, the total energy of the oscillator is the sum its K.E. and P.E. i.e., E = K.E. + P.E. = 1/2 mv² + 1/2 kx² the particle turns at the instant, when its velocity becomes zero, i.e., v = 0 E = 0 + 1/2 kx² As E = 1 J and k = 1/2 N/m or, 1 = 1/2 x 1/2 x² or, x² = 4 ⇒ x = ± 2m

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 Nm-1, the graph of V(x) versus x is shown in fig. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = ± 2 m.

Answer»

At any instant, the total energy of the oscillator is the sum its K.E. and P.E. i.e.,

E = K.E. + P.E. = 1/2 mv² + 1/2 kx²

the particle turns at the instant, when its velocity becomes zero, i.e., v = 0

E = 0 + 1/2 kx²

As E = 1 J and k = 1/2 N/m

or, 1 = 1/2 x 1/2 x² or, x² = 4 ⇒ x = ± 2m