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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-6 Work, Energy And Power




Q.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.



 

 

Answer:-

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) × 0.5x2 = 1
x2 = 4
x = ±2
Hence, the particle turns back when it reaches x = ± 2 m.