When a particle oscillates simple harmonically, its potential energy varies periodically. If the frequency of oscillation of the particle is n, the frequency of potential energy variation is

Options

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

Answer

To solve the problem step by step, we need to analyze the relationship between the oscillation of a particle in simple harmonic motion (SHM) and the variation of its potential energy.

### Step 1: Understand the Position of the Particle
In simple harmonic motion, the position \( x \) of a particle can be expressed as:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude, and \( \omega \) is the angular frequency. The frequency \( n \) is related to the angular frequency by:
\[ \omega = 2\pi n \]

### Step 2: Write the Expression for Potential Energy
The potential energy \( PE \) in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant. Substituting the expression for \( x(t) \):
\[ PE = \frac{1}{2} k (A \sin(\omega t))^2 \]

### Step 3: Simplify the Potential Energy Expression
Substituting \( \omega = 2\pi n \):
\[ PE = \frac{1}{2} k A^2 \sin^2(2\pi n t) \]
Using the trigonometric identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
\[ PE = \frac{1}{2} k A^2 \cdot \frac{1 - \cos(4\pi n t)}{2} \]
This simplifies to:
\[ PE = \frac{1}{4} k A^2 (1 - \cos(4\pi n t)) \]

### Step 4: Identify the Frequency of Potential Energy Variation
From the expression \( PE = \frac{1}{4} k A^2 (1 - \cos(4\pi n t)) \), we can see that the term \( \cos(4\pi n t) \) oscillates with a frequency of:
\[ \text{Frequency of } PE = \frac{4\pi n}{2\pi} = 2n \]

### Conclusion
Thus, the frequency of potential energy variation is \( 2n \).

### Final Answer
The frequency of potential energy variation is \( 2n \).

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