Work, Energy and Power NCERT Solutions Class 11 Physics - Solved Exercise Question 6.4

Question 6.4:

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.

Solution:
Total energy of the particle, E = 1 J

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