A particle executes SHM with frequency 4 Hz. Frequency with which its PE oscillates is

Options

A. 4 Hz
B. 2 Hz
C. 6 Hz
D. 8 Hz

Answer

To solve the problem of finding the frequency with which the potential energy (PE) of a particle executing simple harmonic motion (SHM) oscillates, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given frequency**: The particle executes SHM with a frequency \( f = 4 \, \text{Hz} \).

2. **Relate frequency to angular frequency**: The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 4 = 8\pi \, \text{rad/s} \]

3. **Potential Energy in SHM**: The potential energy \( PE \) in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement. In SHM, the displacement \( x \) can be expressed as:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude.

4. **Substituting for displacement**: Substitute \( x \) into the potential energy equation:
\[ PE = \frac{1}{2} k (A \sin(\omega t))^2 = \frac{1}{2} k A^2 \sin^2(\omega t) \]

5. **Using the identity for sine squared**: We can use the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
\[ PE = \frac{1}{2} k A^2 \cdot \frac{1 - \cos(2\omega t)}{2} = \frac{1}{4} k A^2 (1 - \cos(2\omega t)) \]

6. **Identifying the frequency of potential energy oscillation**: The term \( \cos(2\omega t) \) indicates that the potential energy oscillates with a frequency that is double the frequency of the displacement. Thus, the frequency of the potential energy \( f_{PE} \) is:
\[ f_{PE} = \frac{2\omega}{2\pi} = 2f \]

7. **Calculating the frequency of potential energy**: Since \( f = 4 \, \text{Hz} \):
\[ f_{PE} = 2 \times 4 = 8 \, \text{Hz} \]

### Final Answer:
The frequency with which the potential energy oscillates is \( 8 \, \text{Hz} \).

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