A particle is performing simple harmonic motion along the x axis about the origin. The amplitude of oscillation is a. A large number of photographs of the particle are shot at regular intervals of time with a high speed camera. It was found that photographs having the particle at `x_(1) + Deltax` were maximum in number and photographs having the particle at `x_(2) + Deltax` were least in number. What are values of `x_(1) " and " x_(2)`?

Options

A. `x_(1) = +- a, x_(2) = 0`
B. `x_(1) = 0, x_(2) = +- a`
C. `x_(1) = +- (a)/(2), x_(2) = 0`
D. None of these

Answer

To solve the problem, we need to analyze the behavior of a particle undergoing simple harmonic motion (SHM) along the x-axis.

### Step-by-Step Solution:

1. **Understanding Simple Harmonic Motion (SHM)**:
- In SHM, the particle oscillates about an equilibrium position (in this case, the origin, x = 0).
- The maximum displacement from the equilibrium position is called the amplitude (A). Thus, the particle oscillates between -A and +A.

2. **Identifying Maximum and Minimum Positions**:
- The problem states that photographs of the particle at `x1 + Δx` were maximum in number. This indicates that the particle spends more time near this position.
- Conversely, photographs at `x2 + Δx` were least in number, indicating that the particle spends less time near this position.

3. **Analyzing Particle's Motion**:
- In one complete cycle of SHM (time period T), the particle moves from -A to +A and back to -A.
- The particle moves fastest at the equilibrium position (x = 0) and slows down as it approaches the maximum displacement (±A).

4. **Determining x1 and x2**:
- Since the particle spends the most time near the extremes of its motion (±A), we conclude that:
- **x1** (where photographs are maximum) is at the extreme positions: **x1 = ±A**.
- The particle spends the least time at the equilibrium position (x = 0), hence:
- **x2** (where photographs are minimum) is at the equilibrium position: **x2 = 0**.

### Final Values:
- **x1 = ±A**
- **x2 = 0**