A particle executing simple harmonic motion with time period T. the time period with which its kinetic energy oscillates is

Options

A. T
B. 2T
C. 4T
D. `T/2`

Explanation

To solve the problem, we need to determine the time period of the kinetic energy of a particle executing simple harmonic motion (SHM) with a given time period \( T \).

### Step-by-Step Solution:

1. **Understanding Simple Harmonic Motion (SHM)**:
- A particle in SHM oscillates between two extreme positions (maximum displacement) and passes through a mean position (equilibrium position).
- The time period \( T \) is the time taken to complete one full cycle of motion.

2. **Velocity in SHM**:
- The velocity of the particle is maximum at the mean position and zero at the extreme positions.
- The velocity \( v(t) \) can be expressed as a function of time, typically in the form \( v(t) = A \omega \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

3. **Kinetic Energy in SHM**:
- The kinetic energy \( KE \) of the particle is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
- Substituting the expression for velocity, we get:
\[ KE(t) = \frac{1}{2} m (A \omega \cos(\omega t + \phi))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi) \]

4. **Analyzing the Kinetic Energy Oscillation**:
- The kinetic energy oscillates between 0 (when the velocity is 0 at the extreme positions) and a maximum value (when the velocity is maximum at the mean position).
- The kinetic energy reaches its maximum value when \( \cos^2(\omega t + \phi) \) is at its peak (which occurs every half cycle of the motion).

5. **Determining the Time Period of Kinetic Energy**:
- Since the kinetic energy reaches its maximum value twice during one complete cycle (once going towards the mean position and once coming back), the time period of the kinetic energy oscillation is half of the time period of the motion.
- Therefore, the time period of the kinetic energy oscillation is:
\[ T_{KE} = \frac{T}{2} \]

### Final Answer:
The time period with which the kinetic energy oscillates is \( \frac{T}{2} \).

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