Question

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.

Answer 1

Explanation:

Particle energy E = 1 J

K = 0.5 N m-1

K.E =$\frac{1}{2}$mv2

Based on law of conservation of energy:

E = V + K

1 = $\frac{1}{2}$ kx^2 + $\frac{1}{2}$ mv^2

Velocity becomes zero when it turns back

1 = $\frac{1}{2}$kx2

$\frac{1}{2}$x 0.5x^2 = 1

X^2 = 4

X = $\pm 2$

Thus, on reaching x = ±2 m, the particle turns back.