Problem 35 - Master Resource Book in JEE Main Physics

Question

The potential energy function for a particle executing linear SHM is given by \(V(x)=\frac{1}{2} k x^{2}\) where \(k\) is the force constant of the oscillator. For \(k=0.5 \mathrm{~N} / \mathrm{m}\), the graph of \(V(x)\) versus \(x\) is shown in the figure. A particle of total energy \(E\) turns back when it reaches \(x=\pm x_{m}\). If \(V\) and \(K\) indicate the \(\mathrm{PE}\) and \(\mathrm{KE}\) respectively of the particle at \(x=\pm x_{m}\) then which of the following is correct? \(\quad\) [NCERT Exemplar] (a) \(V=0, \bar{K}=\mathrm{E}\) (b) \(V=\mathrm{E}, \bar{K}=0\) (c) \(V

Step-by-Step Solution

Answer

Option (b) is correct: \( V = E, \bar{K} = 0 \).

Step 1: Understand the Problem Setup

The potential energy function for a particle executing simple harmonic motion (SHM) is given by \( V(x) = \frac{1}{2} k x^2 \). Here, \( k = 0.5 \, \mathrm{N/m} \). We need to analyze its relationship with the particle's total energy \( E \) at \( x = \pm x_m \), where the particle turns back. The task is to determine the correct association between potential energy (PE), kinetic energy (KE), and total energy (E) at these points.

Step 2: Analyze Energy at Turning Points

At the turning points \( x = \pm x_m \), the velocity of the particle is zero, so the kinetic energy (KE), \( K = 0 \). Since the total mechanical energy \( E \) is the sum of potential energy (PE) \( V \) and kinetic energy (KE) \( K \), and \( K = 0 \) at the turning points, we have \( V = E \). Thus, at the turning points, all the energy is potential energy.

Step 3: Evaluate the Options

Given that at \( x = \pm x_m \):- The potential energy \( V = E \)- The kinetic energy \( K = 0 \)The correct option aligns with these conditions. Option (b) states \( V = E \), \( \bar{K} = 0 \), which matches our analysis.

Key Concepts

Potential Energy Function

In simple harmonic motion (SHM), the potential energy of a particle is crucial. It helps to understand how energy is stored within the system. The potential energy function for SHM is given by \( V(x) = \frac{1}{2} k x^2 \), where \( k \) is the force constant of the oscillator and \( x \) is the displacement from the equilibrium position.
This formula tells us that potential energy depends on the displacement squared.

• The more the displacement, the higher the potential energy.
• The potential energy is zero when the particle is at the equilibrium position (\( x = 0 \)).
• It reaches a maximum at the extremes of the motion (\( x = \pm x_m \)).

For example, with \( k = 0.5 \text{ N/m} \), as given in the exercise, the potential energy graph forms a parabola. This shows how potential energy rises as the particle moves away from the equilibrium.

Total Mechanical Energy

Total mechanical energy in SHM is the sum of potential and kinetic energies. This total energy remains constant as long as no external forces are acting on the system.
Total Mechanical Energy \( E \) is expressed as:

• \( E = V + K \)

where \( V \) is potential energy and \( K \) is kinetic energy. At any point in the motion, these energies shift back and forth, but their sum remains unchanged.

• At maximum displacement (\( \pm x_m \)), kinetic energy is zero and potential energy is maximum.
• At equilibrium (\( x = 0 \)), potential energy is zero and kinetic energy is maximum.

This conservation means that if one form of energy decreases, the other increases, ensuring their total remains constant.

Turning Points in SHM

Turning points in SHM are where the motion changes direction. At these points, all energy is potential. This means:

• The particle stops momentarily before reversing direction.
• Kinetic energy becomes zero because the velocity is zero.
• Potential energy equals the total mechanical energy \( V = E \).

These characteristics help identify the correct conditions at turning points. From the given options, the one where \( V = E \) and \( K = 0 \) fits this description. Therefore, option (b) is correct, aligning with our understanding that at turning points, all energy is stored as potential energy.

Kinetic Energy Analysis

Kinetic energy in SHM represents the energy a particle has due to its motion. It fluctuates with the particle's position:

• \( K = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity.
• At equilibrium, kinetic energy is at its peak because velocity is maximum.
• At turning points, kinetic energy drops to zero since the velocity is zero.

The interconversion between kinetic and potential energy ensures constant total energy. Understanding this fluctuation helps explain motion dynamics within SHM. When analyzing problems like the exercise, confirming kinetic energy values at specific points illuminates the energy balance and validates our calculations.