`X_(1)` and `X_(2)` are two points on the path of a particle executing `SHM` in a straight line, at which its velocity is zero. Starting from a certain point `X(X_(1)XltX_(2)X)` then particle crosses this point again at successive intervals of `2s` and `4s` with a speed of `5m//s`.
The time period of `SHM` is

Options

A. `6s`
B. `3s`
C. `4s`
D. `2s`

Explanation

To find the time period of the simple harmonic motion (SHM) for the given problem, we can follow these steps:

### Step 1: Understand the Problem
We have two points \(X_1\) and \(X_2\) where the particle's velocity is zero. The particle crosses a certain point \(X\) (where \(X_1 < X < X_2\)) at intervals of 2 seconds and 4 seconds with a speed of 5 m/s.

### Step 2: Identify the Time Intervals
The particle takes:
- 2 seconds to move from \(X\) to \(X_1\) (and back to \(X\)),
- 4 seconds to move from \(X\) to \(X_2\) (and back to \(X\)).

### Step 3: Determine the Total Time for One Complete Cycle
The total time taken for the particle to go from \(X_1\) to \(X_2\) and back to \(X_1\) is the sum of the time intervals:
- Time from \(X\) to \(X_1\) and back to \(X\): 2 seconds (up and down)
- Time from \(X\) to \(X_2\) and back to \(X\): 4 seconds (up and down)

Thus, the total time for one complete cycle (from \(X_1\) to \(X_2\) and back to \(X_1\)) is:
\[ T = 2 + 4 = 6 \text{ seconds} \]

### Step 4: Conclusion
The time period \(T\) of the simple harmonic motion is 6 seconds.

### Final Answer
The time period of SHM is \(6 \text{ seconds}\).

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