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. 5.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Answer 1

Total energy of the particle, E = 1 J
Force constant, k = 0.5 N m–1
Kinetic energy of the particle, K =$\frac{1 }{ 2}$ mv2
According to the conservation law: E = V + K
1 =$\frac{ 1 }{ 2}$ kx2 + $\frac{ 1 }{ 2}$ mv2
At the moment of ‘turn back’, velocity (and hence K) becomes zero.
∴ 1 = $\frac{1 }{2 }$ kx2
$\frac{1 }{2 }$ × 0.5 x2 = 1
x2 = 4
x = ± 2
Hence, the particle turns back when it reaches x = ± 2m.