For a simple harmonic vibrator frequency n, the frequency of kinetic energy changing completely to potential energy is

Options

A. n/2
B. n
C. 2n
D. 4n

Explanation

To solve the problem, we need to understand the relationship between kinetic energy (KE) and potential energy (PE) in a simple harmonic oscillator (SHO). The question asks for the frequency at which kinetic energy completely changes to potential energy.

### Step-by-Step Solution:

1. **Understanding Simple Harmonic Motion (SHM)**:
- A simple harmonic oscillator moves back and forth around an equilibrium position. The total mechanical energy in SHM is the sum of kinetic energy and potential energy.

2. **Energy in SHM**:
- At the equilibrium position, the kinetic energy is maximum, and potential energy is zero.
- At the maximum displacement (amplitude), the potential energy is maximum, and kinetic energy is zero.

3. **Energy Transformation**:
- As the oscillator moves from the equilibrium position to the maximum displacement, kinetic energy is converted into potential energy.
- Conversely, as it moves back towards the equilibrium position, potential energy is converted back into kinetic energy.

4. **Frequency of Energy Transformation**:
- In one complete cycle of SHM, the oscillator moves from the equilibrium position to the maximum displacement (where KE = 0 and PE = max), then back to equilibrium (where KE = max and PE = 0), and then to the opposite maximum displacement (where KE = 0 and PE = max again).
- This means that in one complete cycle, the kinetic energy changes to potential energy twice (once while moving to the maximum displacement and once while returning).

5. **Calculating the Frequency**:
- If the frequency of the oscillator is \( n \), it means that it completes \( n \) cycles in one second.
- Since the transformation of energy occurs twice in each cycle, the frequency at which kinetic energy completely changes to potential energy is \( 2n \).

6. **Final Answer**:
- Therefore, the frequency of kinetic energy changing completely to potential energy is \( 2n \).

### Conclusion:
The answer to the question is \( 2n \).

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