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Question #42805
The Question
A particle of mass and charge is attached to the lower end of a massless spring of spring constant . The upper end of the spring is fixed to a rigid support. The system is initially at rest in its vertical equilibrium position under gravity. At , a time-varying vertical electric field is switched on, pointing downwards. Given that the angular frequency of the electric field is , what is the total mechanical energy of the oscillator (sum of kinetic, elastic potential, and gravitational potential energies) at time , relative to its initial equilibrium state at ?
Options
- A.
- B.
- C.
- D.
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Intuition_Breaker
Train your brain to resist "obvious" answers that are designed to deceive your intuition.
Concepts Behind This Question
Work against Conservative Force|Vertical Oscillations|Vector Calculus
Why This Question Is Tricky
Common Mistakes
This question tests the misapplication of energy conservation in time-varying fields.
Students often incorrectly assume that because the electrostatic force is conservative at any given instant, they can define a time-dependent potential energy and conserve the total energy .
Since and , this flawed logic leads to the conclusion that .
However, the Hamiltonian of the system explicitly depends on time, meaning total mechanical energy is not conserved; the electric field does net work on the oscillator over the half-cycle, which must be calculated via the Work-Energy Theorem or by explicitly solving the non-autonomous differential equation.
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Electrostatic Potential and Capacitance - Class 12 |
| Concept | Work against Conservative Force |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |