A body is executing simple harmonic motion with frequency 'n' the frequency of its potential energy is

Options

A. 3n
B. 4n
C. n
D. 2n

Answer

To solve the problem of finding the frequency of the potential energy of a body executing simple harmonic motion (SHM) with frequency 'n', we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Simple Harmonic Motion (SHM)**:
- A body in SHM oscillates about an equilibrium position. The motion can be described by a sinusoidal function, where the displacement varies with time.

2. **Identifying the Frequency of Motion**:
- Given that the frequency of the body executing SHM is 'n', we denote this frequency as \( f = n \).

3. **Potential Energy in SHM**:
- The potential energy (PE) in SHM is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

4. **Behavior of Potential Energy**:
- The potential energy is zero at the equilibrium position (where \( x = 0 \)) and reaches its maximum value at the extreme positions (maximum displacement).

5. **Time Period of SHM**:
- The time period \( T \) of SHM is related to the frequency by:
\[ T = \frac{1}{f} = \frac{1}{n} \]

6. **Frequency of Potential Energy**:
- The potential energy reaches its maximum value when the displacement is maximum and is zero at the equilibrium position.
- The potential energy oscillates twice during one complete cycle of the motion (once while moving to the maximum displacement and once while returning).
- Therefore, the frequency of the potential energy is double that of the frequency of the motion:
\[ f_{PE} = 2f = 2n \]

### Conclusion:
Thus, the frequency of the potential energy of the body executing simple harmonic motion with frequency 'n' is \( 2n \).