If the particle repeats its motion after a fixed time interval of 8 s then after how much time its maximum value of PE will be attained after attaining its minimum value ?

Options

A. 2s
B. 4s
C. 8s
D. 1s

Answer

To solve the problem, we need to understand the relationship between the potential energy (PE) and the position of the particle in oscillatory motion. The potential energy in a simple harmonic oscillator is maximum when the particle is at the extreme positions (maximum displacement) and minimum when the particle is at the mean position (equilibrium position).

### Step-by-Step Solution:

1. **Identify the Period of Motion**:
The problem states that the particle repeats its motion after a fixed time interval of 8 seconds. This time interval is the period (T) of the oscillation.
\[ T = 8 \text{ seconds} \]

2. **Determine the Time for Maximum and Minimum PE**:
In one complete cycle of oscillation:
- The potential energy is minimum when the particle is at the mean position (x = 0).
- The potential energy is maximum when the particle is at the extreme positions (x = ±A, where A is the amplitude).

3. **Analyze the Motion**:
- At \( t = 0 \), assume the particle is at the mean position (minimum PE).
- The particle will take a quarter of the period (T/4) to reach the maximum displacement (maximum PE).

4. **Calculate Time to Maximum PE**:
Since the period T is 8 seconds, we calculate T/4:
\[ \frac{T}{4} = \frac{8 \text{ seconds}}{4} = 2 \text{ seconds} \]

5. **Conclusion**:
Therefore, after attaining its minimum value of potential energy, the maximum value of potential energy will be attained after 2 seconds.

### Final Answer:
The maximum value of potential energy will be attained after **2 seconds**.

---