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.