Question:
A particle in SHM has amplitude \( 2 \text{ m} \) and total energy \( 8 \text{ J} \). Its potential energy at displacement \( 1 \text{ m} \) is:
A particle in SHM has amplitude \( 2 \text{ m} \) and total energy \( 8 \text{ J} \). Its potential energy at displacement \( 1 \text{ m} \) is:
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For energy problems in SHM, remember that Potential Energy scales quadratically with displacement (\(U \propto x^2\)). If the displacement becomes half of the maximum amplitude (\(x = A/2\)), the potential energy will always drop to exactly one-fourth (\(1/4\)) of the total energy!
Updated On: Jun 3, 2026
- \( 1 \text{ J} \)
- \( 2 \text{ J} \)
- \( 4 \text{ J} \)
- \( 8 \text{ J} \)
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The Correct Option is B
Solution and Explanation
Concept:
In Simple Harmonic Motion (SHM), the restoring force acting on the particle is directly proportional to its displacement from the mean position. The mechanical energy of the system oscillates between kinetic and potential forms.
The key equations governing the energy configurations are:
• Total Mechanical Energy (\(E\)): The total energy remains constant throughout the motion and depends on the amplitude \(A\): \[ E = \frac{1}{2}m\omega^2 A^2 \]
• Potential Energy (\(U\)): The potential energy at any displacement \(x\) from the mean equilibrium position is given by: \[ U = \frac{1}{2}m\omega^2 x^2 \] By taking the ratio of these two quantities, we can eliminate the system constants (\(m\) and \(\omega\)).
Step 1: Identifying the parameters given in the question.
We are given:
• Amplitude of oscillation, \( A = 2 \text{ m} \)
• Total mechanical energy, \( E = 8 \text{ J} \)
• Instantaneous displacement, \( x = 1 \text{ m} \)
Step 2: Establishing the ratio between potential energy and total energy.
Dividing the potential energy expression by the total energy expression: \[ \frac{U}{E} = \frac{\frac{1}{2}m\omega^2 x^2}{\frac{1}{2}m\omega^2 A^2} = \frac{x^2}{A^2} \] Rearranging the formula to solve for the instantaneous potential energy \(U\): \[ U = E \cdot \left(\frac{x}{A}\right)^2 \]
Step 3: Substituting the numerical values into the ratio relationship.
Substitute \(E = 8\), \(x = 1\), and \(A = 2\) into our equation: \[ U = 8 \cdot \left(\frac{1}{2}\right)^2 \] \[ U = 8 \cdot \frac{1}{4} = 2 \text{ J} \] This perfectly matches option (B).
• Total Mechanical Energy (\(E\)): The total energy remains constant throughout the motion and depends on the amplitude \(A\): \[ E = \frac{1}{2}m\omega^2 A^2 \]
• Potential Energy (\(U\)): The potential energy at any displacement \(x\) from the mean equilibrium position is given by: \[ U = \frac{1}{2}m\omega^2 x^2 \] By taking the ratio of these two quantities, we can eliminate the system constants (\(m\) and \(\omega\)).
Step 1: Identifying the parameters given in the question.
We are given:
• Amplitude of oscillation, \( A = 2 \text{ m} \)
• Total mechanical energy, \( E = 8 \text{ J} \)
• Instantaneous displacement, \( x = 1 \text{ m} \)
Step 2: Establishing the ratio between potential energy and total energy.
Dividing the potential energy expression by the total energy expression: \[ \frac{U}{E} = \frac{\frac{1}{2}m\omega^2 x^2}{\frac{1}{2}m\omega^2 A^2} = \frac{x^2}{A^2} \] Rearranging the formula to solve for the instantaneous potential energy \(U\): \[ U = E \cdot \left(\frac{x}{A}\right)^2 \]
Step 3: Substituting the numerical values into the ratio relationship.
Substitute \(E = 8\), \(x = 1\), and \(A = 2\) into our equation: \[ U = 8 \cdot \left(\frac{1}{2}\right)^2 \] \[ U = 8 \cdot \frac{1}{4} = 2 \text{ J} \] This perfectly matches option (B).
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