When a particle oscillates in simple harmonic motion, both in potential energy and kinetic energy vary sinusoidally with time. If `v` be the frequency of the motion of the particle, the frequency associated with the kinetic energy is `:`

Options

A. `4v`
B. `2v`
C. `v` is highest when the spring comes back to natural position
D. `(v)/(2)`

Explanation

To solve the problem, we need to analyze the relationship between the motion of a particle in simple harmonic motion (SHM) and the variations in its kinetic energy (KE) and potential energy (PE).

### Step-by-Step Solution:

1. **Understanding Simple Harmonic Motion (SHM)**:
- In SHM, a particle oscillates about a mean position. The motion can be described by a sinusoidal function.
- The displacement of the particle from the mean position can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Kinetic Energy (KE) in SHM**:
- The kinetic energy of the particle is given by:
\[ KE = \frac{1}{2} m v^2 \]
- The velocity \( v \) can be derived from the displacement:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
- Substituting \( v(t) \) into the kinetic energy formula:
\[ KE = \frac{1}{2} m (A \omega \cos(\omega t + \phi))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi) \]

3. **Potential Energy (PE) in SHM**:
- The potential energy is given by:
\[ PE = \frac{1}{2} k x^2 \]
- Using Hooke's law, where \( k \) is the spring constant, we can express it in terms of displacement:
\[ PE = \frac{1}{2} k (A \sin(\omega t + \phi))^2 = \frac{1}{2} k A^2 \sin^2(\omega t + \phi) \]

4. **Frequency of Kinetic Energy**:
- The kinetic energy varies as \( \cos^2(\omega t + \phi) \). The function \( \cos^2 \) can be expressed using the double angle identity:
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
- Therefore, the kinetic energy can be rewritten as:
\[ KE = \frac{1}{4} m A^2 \omega^2 + \frac{1}{2} m A^2 \omega^2 \cos(2(\omega t + \phi)) \]
- This shows that the kinetic energy oscillates with a frequency of \( 2\omega \).

5. **Relating Angular Frequency to Frequency**:
- The frequency \( v \) is related to the angular frequency \( \omega \) by:
\[ \omega = 2\pi v \]
- Thus, the frequency associated with kinetic energy becomes:
\[ \text{Frequency of KE} = 2v \]

### Final Answer:
The frequency associated with the kinetic energy is \( 2v \).