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 find 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: Express Velocity in Terms of SHM Parameters
In SHM, the velocity \( v \) of the particle can be expressed as:
\[ v = A \omega \cos(\omega t) \]
where:
- \( A \) is the amplitude of the motion,
- \( \omega \) is the angular frequency, and
- \( t \) is the time.

### Step 3: Substitute Velocity into the Kinetic Energy Formula
Substituting the expression for velocity into the kinetic energy formula gives:
\[ KE = \frac{1}{2} m (A \omega \cos(\omega t))^2 \]
This simplifies to:
\[ KE = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]

### Step 4: Use the Trigonometric Identity
We can use the trigonometric identity for \( \cos^2(\theta) \):
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
Applying this identity to our kinetic energy expression:
\[ KE = \frac{1}{2} m A^2 \omega^2 \left(\frac{1 + \cos(2\omega t)}{2}\right) \]
This simplifies to:
\[ KE = \frac{1}{4} m A^2 \omega^2 + \frac{1}{4} m A^2 \omega^2 \cos(2\omega t) \]

### Step 5: Relate Angular Frequency to Frequency
Since \( \omega = 2\pi v \), we can express the kinetic energy in terms of frequency \( v \):
\[ KE = \frac{1}{4} m A^2 (2\pi v)^2 + \frac{1}{4} m A^2 (2\pi v)^2 \cos(2(2\pi v)t) \]
This simplifies to:
\[ KE = \frac{1}{4} m A^2 (4\pi^2 v^2) + \frac{1}{4} m A^2 (4\pi^2 v^2) \cos(4\pi vt) \]

### Conclusion
The kinetic energy of the particle executing SHM varies periodically with time and has a frequency that is double the original frequency \( v \). Therefore, the kinetic energy changes periodically with a frequency of \( 2v \).

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