Que.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 oftotal energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
Question
Que.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 oftotal energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
Solution
The potential energy of a particle in simple harmonic motion is given by the equation V(x) = kx²/2. Here, k is the force constant of the oscillator and x is the displacement from the equilibrium position.
Given that k = 0.5 N m⁻¹ and the total energy of the particle is 1 J, we can set up the following equation:
1 J = (0.5 N m⁻¹) * x² / 2
Solving for x, we get:
x² = 2 * 1 J / (0.5 N m⁻¹) x² = 4 m²
Therefore, x = ± 2 m.
This means that when the particle reaches a displacement of x = ± 2 m, its potential energy equals its total energy. Since the particle cannot have more energy than it started with, it must 'turn back' at these points. This is consistent with the behavior of a particle in simple harmonic motion, which oscillates back and forth about the equilibrium position.
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