Two particles `P_(1)` and `P_(2)` are performing `SHM` along the same line about the same mean position , initial they are at their extreme position. If the time period of each particle is `12` sec and the difference of their amplitude is `12 cm` then find the minimum time after which the separation between the particle becomes `6 cm`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
We have two particles \( P_1 \) and \( P_2 \) performing Simple Harmonic Motion (SHM) along the same line about the same mean position. Both particles start at their extreme positions, and the difference in their amplitudes is given as \( 12 \, \text{cm} \).

### Step 2: Define the parameters
Let:
- \( A_1 \) = Amplitude of particle \( P_1 \)
- \( A_2 \) = Amplitude of particle \( P_2 \)
- Given that \( A_2 - A_1 = 12 \, \text{cm} \)
- Time period \( T = 12 \, \text{s} \)

The angular frequency \( \omega \) can be calculated as:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6} \, \text{rad/s} \]

### Step 3: Write the equations of motion
The equations of motion for the two particles can be written as:
- For particle \( P_1 \):
\[ x_1 = A_1 \cos(\omega t) \]
- For particle \( P_2 \):
\[ x_2 = A_2 \cos(\omega t) \]

### Step 4: Calculate the separation
The separation \( S \) between the two particles is given by:
\[ S = |x_2 - x_1| = |A_2 \cos(\omega t) - A_1 \cos(\omega t)| = |(A_2 - A_1) \cos(\omega t)| \]
Substituting \( A_2 - A_1 = 12 \, \text{cm} \):
\[ S = |12 \cos(\omega t)| \]

### Step 5: Set the separation equal to 6 cm
We need to find the time when the separation is \( 6 \, \text{cm} \):
\[ |12 \cos(\omega t)| = 6 \]
This simplifies to:
\[ \cos(\omega t) = \frac{6}{12} = \frac{1}{2} \]

### Step 6: Solve for \( \omega t \)
The cosine function gives us:
\[ \cos(\omega t) = \frac{1}{2} \implies \omega t = \frac{\pi}{3} \text{ or } \omega t = \frac{5\pi}{3} \text{ (but we need the first occurrence)} \]

### Step 7: Substitute \( \omega \) and solve for \( t \)
Substituting \( \omega = \frac{\pi}{6} \):
\[ \frac{\pi}{6} t = \frac{\pi}{3} \]
Solving for \( t \):
\[ t = \frac{\pi/3}{\pi/6} = \frac{1}{3} \times 6 = 2 \, \text{s} \]

### Conclusion
The minimum time after which the separation between the particles becomes \( 6 \, \text{cm} \) is \( 2 \, \text{s} \).

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