The time taken by a particle performing SHM to pass from point A and B where it is velocities are same is 2s. After another 2 s it returns to B. The time period oscillation is (in seconds)

Options

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

Answer

To solve the problem, let's break it down step by step:

### Step 1: Understanding the Motion
The particle is performing Simple Harmonic Motion (SHM). In SHM, the particle oscillates about a mean position, and its velocity is the same at two points A and B. This implies that points A and B are equidistant from the mean position.

### Step 2: Time Taken from A to B
It is given that the time taken to travel from point A to point B is 2 seconds.

### Step 3: Time Taken to Return to B
After reaching point B, the particle takes another 2 seconds to return to point B. This means that the total time taken to go from A to B and then back to B is 4 seconds (2 seconds to go from A to B and 2 seconds to return to B).

### Step 4: Analyzing the Motion
Since the particle takes the same amount of time to go from B to the maximum displacement (let's call it C) and then back to B, we can deduce that:
- Time taken from A to the mean position (M) = 1 second
- Time taken from M to B = 1 second
- Time taken from B to C (maximum displacement) = 2 seconds
- Time taken from C back to M = 2 seconds
- Time taken from M back to B = 1 second

### Step 5: Total Time for Half Oscillation
The total time for half an oscillation (from A to C and back to B) is:
- A to M (1 second) + M to B (1 second) + B to C (2 seconds) + C to M (2 seconds) = 6 seconds.

### Step 6: Total Time Period Calculation
Since the time taken for half an oscillation is 4 seconds (from A to B and back to B), the time period (T) of one complete oscillation is:
\[ T = 2 \times \text{(time for half oscillation)} = 2 \times 4 \text{ seconds} = 8 \text{ seconds} \]

### Final Answer
Thus, the time period of oscillation is **8 seconds**.

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