Two particles `P` and `Q` describe simple harmonic motions of same period , same amplitude along the same line about the same equilibrium position `O`. When `P` and `Q` are on opposite sides of `O` at the same distance from `O` they have the same speed of `1.2m//s` in the same direction, when their displacements are the same they have the same speed of `1.6m//s` in opposite directions .The maximum velocity in `m//s` of either particle is

Options

A. `2.8`
B. `2.5`
C. `2.4`
D. `2`

Explanation

To solve the problem step by step, we will use the information given about the two particles \( P \) and \( Q \) that are undergoing simple harmonic motion (SHM).

### Step 1: Understand the Motion of Particles
Both particles \( P \) and \( Q \) have the same amplitude \( A \) and angular frequency \( \omega \), and they oscillate about the same equilibrium position \( O \).

### Step 2: Analyze the First Condition
When \( P \) and \( Q \) are on opposite sides of \( O \) at the same distance from \( O \), they have the same speed of \( 1.2 \, \text{m/s} \) in the same direction.

Using the formula for velocity in SHM:
\[ v = A \omega \cos(\theta) \]
For this case, we denote the angle as \( \theta \):
\[ A \omega \cos(\theta) = 1.2 \quad (1) \]

### Step 3: Analyze the Second Condition
When their displacements are the same, they have the same speed of \( 1.6 \, \text{m/s} \) but in opposite directions.

Using the same formula for velocity:
\[ v = A \omega \sin(\theta) \]
For this case, we have:
\[ A \omega \sin(\theta) = 1.6 \quad (2) \]

### Step 4: Divide the Equations
Now, we can divide equation (2) by equation (1):
\[ \frac{A \omega \sin(\theta)}{A \omega \cos(\theta)} = \frac{1.6}{1.2} \]
This simplifies to:
\[ \tan(\theta) = \frac{1.6}{1.2} = \frac{4}{3} \]

### Step 5: Find \( \theta \)
From \( \tan(\theta) = \frac{4}{3} \), we can find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 6: Use the Trigonometric Identity
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we can find \( \sin(\theta) \) and \( \cos(\theta) \):
Let \( \sin(\theta) = 4k \) and \( \cos(\theta) = 3k \) for some \( k \):
\[ (4k)^2 + (3k)^2 = 1 \implies 16k^2 + 9k^2 = 1 \implies 25k^2 = 1 \implies k = \frac{1}{5} \]
Thus,
\[ \sin(\theta) = \frac{4}{5}, \quad \cos(\theta) = \frac{3}{5} \]

### Step 7: Substitute Back to Find \( A \omega \)
Now we can substitute back into either equation (1) or (2) to find \( A \omega \):
Using equation (1):
\[ A \omega \cos(\theta) = 1.2 \implies A \omega \cdot \frac{3}{5} = 1.2 \]
Solving for \( A \omega \):
\[ A \omega = 1.2 \cdot \frac{5}{3} = 2.0 \, \text{m/s} \]

### Final Answer
The maximum velocity \( A \omega \) of either particle is:
\[ \boxed{2.0 \, \text{m/s}} \]