The period of a particle in SHM is `8 s`. At `t = 0` it is in its equilibrium position.
(a) Compare the distance travelled in the first `4s` and the second `4s`.
(b) Compare the distance travelled in the first `2s` and the second `2s`.

Explanation

To solve the problem, we will analyze the motion of a particle in Simple Harmonic Motion (SHM) based on the given period and the distances traveled in specified time intervals.

### Given:
- Period \( T = 8 \, \text{s} \)
- At \( t = 0 \), the particle is at its equilibrium position.

### Part (a): Compare the distance traveled in the first 4 seconds and the second 4 seconds.

1. **Understanding the Motion:**
- The period \( T = 8 \, \text{s} \) means the particle completes one full cycle in 8 seconds.
- Half of the period \( \frac{T}{2} = 4 \, \text{s} \) is the time taken to move from the equilibrium position to one extreme and back to the equilibrium position.

2. **Distance Traveled in the First 4 Seconds:**
- From \( t = 0 \) to \( t = 4 \, \text{s} \):
- The particle moves from the equilibrium position (O) to one extreme position (P) and back to the equilibrium position (O).
- Distance traveled \( d_1 = OP + PO = A + A = 2A \).

3. **Distance Traveled in the Second 4 Seconds:**
- From \( t = 4 \, \text{s} \) to \( t = 8 \, \text{s} \):
- The particle moves from the equilibrium position (O) to the opposite extreme position (Q) and back to the equilibrium position (O).
- Distance traveled \( d_2 = OQ + QO = A + A = 2A \).

4. **Comparison:**
- Since \( d_1 = 2A \) and \( d_2 = 2A \), we find that:
\[ d_1 = d_2 \]
- Therefore, the distance traveled in the first 4 seconds is equal to the distance traveled in the second 4 seconds.

### Part (b): Compare the distance traveled in the first 2 seconds and the second 2 seconds.

1. **Understanding the Motion:**
- The time interval of 2 seconds corresponds to \( \frac{T}{4} = 2 \, \text{s} \), which is one-fourth of the total period.

2. **Distance Traveled in the First 2 Seconds:**
- From \( t = 0 \) to \( t = 2 \, \text{s} \):
- The particle moves from the equilibrium position (O) to the maximum amplitude position (P).
- Distance traveled \( d_1 = OP = A \).

3. **Distance Traveled in the Second 2 Seconds:**
- From \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \):
- The particle moves from the maximum amplitude position (P) back to the equilibrium position (O).
- Distance traveled \( d_2 = PO = A \).

4. **Comparison:**
- Since \( d_1 = A \) and \( d_2 = A \), we find that:
\[ d_1 = d_2 \]
- Therefore, the distance traveled in the first 2 seconds is equal to the distance traveled in the second 2 seconds.

### Summary of Results:
- (a) The distance traveled in the first 4 seconds is equal to the distance traveled in the second 4 seconds.
- (b) The distance traveled in the first 2 seconds is equal to the distance traveled in the second 2 seconds.