When a particle executes SHM oscillates with a frequency v, then the kinetic energy of the particle

Options

A. change periodically with a frequency of v
B. changes periodically with a frequency of 2v
C. changes periodically with a frequency of v/2
D. remain constant

Explanation

To solve the problem regarding the kinetic energy of a particle executing simple harmonic motion (SHM) with a frequency \( v \), we can follow these steps:

### Step 1: Understand the Kinetic Energy Formula
The kinetic energy (KE) of a particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

### Step 2: Define the Motion of the Particle
For a particle in SHM, the displacement \( x \) can be expressed as:
\[ x = a \sin(\omega t) \]
where \( a \) is the amplitude of the motion and \( \omega \) is the angular frequency.

### Step 3: Find the Velocity
The velocity \( v \) of the particle is the time derivative of displacement:
\[ v = \frac{dx}{dt} = \frac{d}{dt}(a \sin(\omega t)) = a \omega \cos(\omega t) \]

### Step 4: Substitute Velocity into the Kinetic Energy Formula
Now, substituting the expression for velocity into the kinetic energy formula:
\[ KE = \frac{1}{2} m (a \omega \cos(\omega t))^2 \]
\[ KE = \frac{1}{2} m a^2 \omega^2 \cos^2(\omega t) \]

### Step 5: Analyze the Periodicity of the Kinetic Energy
The term \( \cos^2(\omega t) \) has a period of \( \frac{T}{2} \), where \( T \) is the period of the motion. The period \( T \) is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
Thus, the period of \( \cos^2(\omega t) \) is:
\[ T_{\cos^2} = \frac{T}{2} = \frac{\pi}{\omega} \]

### Step 6: Relate Period to Frequency
Since frequency \( f \) is the reciprocal of the period, we have:
\[ f = \frac{1}{T} \]
If the period of the kinetic energy is \( \frac{T}{2} \), then the frequency of the kinetic energy will be:
\[ f_{KE} = \frac{1}{T/2} = 2f \]
Given that the original frequency is \( v \), the frequency of the kinetic energy will be \( 2v \).

### Conclusion
Therefore, the kinetic energy of the particle changes periodically with the frequency of \( 2v \).

### Final Answer
The correct option is that the kinetic energy changes periodically with the frequency of \( 2v \).

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