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

Given

• Potential energy: \( V(x) = \dfrac{1}{2} k x^{2} \)
• Force constant: \( k = 0.5 \,\text{N m}^{-1} \)
• Total mechanical energy of the particle: \( E = 1 \,\text{J} \)

Step 1: Write V(x) explicitly

Using \( k = 0.5 \,\text{N m}^{-1} \),

\( V(x) = \dfrac{1}{2} \cdot 0.5 \cdot x^{2} = 0.25 x^{2} \,\text{J} \)

Step 2: Condition at turning points

At a turning point, the particle’s velocity is zero, so its kinetic energy is zero and the total energy is purely potential:

At turning point: \( E = V(x) \)

Given \( E = 1 \,\text{J} \), we set

\( 1 = 0.25 x^{2} \)

Step 3: Solve for x

\( 0.25 x^{2} = 1 \Rightarrow x^{2} = \dfrac{1}{0.25} = 4 \Rightarrow x = \pm 2 \,\text{m} \)

Thus, the particle must turn back at \( x = +2 \,\text{m} \) and \( x = -2 \,\text{m} \).

Physical explanation

• The total energy line \( E = 1 \,\text{J} \) intersects the potential energy curve \( V(x) = 0.25 x^{2} \) at \( x = \pm 2 \,\text{m} \); beyond these points, \( V(x) > E \), which is not allowed for real motion.
• Hence, as the particle approaches \( \pm 2 \,\text{m} \), its kinetic energy falls to zero, it momentarily stops, and then reverses direction, giving the turning points of the simple harmonic motion.