The potential energy of a particle executing `SHM` change from maximum to minimum in `5 s`. Then the time period of `SHM` is:

Options

A. `5s`
B. `10s`
C. `15s`
D. `20s`

Explanation

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

### Step 1: Understand the motion of the particle in SHM
In Simple Harmonic Motion (SHM), the potential energy of the particle is maximum when it is at the extreme positions (amplitude) and minimum when it is at the mean position (equilibrium).

### Step 2: Identify the time taken for the change in potential energy
The problem states that the potential energy changes from maximum to minimum in 5 seconds. This means that the particle moves from one extreme position to the mean position.

### Step 3: Determine the complete cycle of SHM
In one complete cycle of SHM, the particle moves from:
1. Maximum potential energy (extreme position)
2. To minimum potential energy (mean position)
3. Back to maximum potential energy (the other extreme position)
4. And finally returns to minimum potential energy (mean position)

Thus, the particle completes a full cycle in the following sequence:
- From maximum to minimum (5 seconds)
- From minimum to maximum (5 seconds)
- Total time for a full cycle = 5 seconds + 5 seconds = 10 seconds.

### Step 4: Calculate the time period of SHM
The time period (T) of SHM is the time taken to complete one full cycle. From our calculation:
- The time taken for one complete cycle is 10 seconds.

### Conclusion
Therefore, the time period of the SHM is **10 seconds**.

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