A particle is performing sinple harmonic motion of period 2 sec. if the greatest speed is `5(m)/(s)` the speed `((m)/(s))` of particle when it is `(3)/(pi)` m from equilibrium is:

Explanation

To solve the problem step by step, we will use the concepts of simple harmonic motion (SHM).

### Step 1: Determine the angular frequency (ω)
The angular frequency (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the period of the motion.

Given that the period \( T = 2 \) seconds, we can substitute this value into the formula:
\[ \omega = \frac{2\pi}{2} = \pi \text{ radians/second} \]

### Step 2: Relate maximum speed (v_max) to amplitude (A)
The maximum speed in SHM is given by the formula:
\[ v_{\text{max}} = \omega A \]
We know that the maximum speed \( v_{\text{max}} = 5 \) m/s. Substituting the value of \( \omega \):
\[ 5 = \pi A \]
Now, we can solve for the amplitude \( A \):
\[ A = \frac{5}{\pi} \text{ meters} \]

### Step 3: Use the formula for speed at a distance x from equilibrium
The speed \( v \) of a particle in SHM when it is at a distance \( x \) from the equilibrium position is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
We need to find the speed when \( x = \frac{3}{\pi} \) m. Substituting the known values:
\[ v = \pi \sqrt{\left(\frac{5}{\pi}\right)^2 - \left(\frac{3}{\pi}\right)^2} \]

### Step 4: Calculate \( A^2 \) and \( x^2 \)
Calculating \( A^2 \) and \( x^2 \):
\[ A^2 = \left(\frac{5}{\pi}\right)^2 = \frac{25}{\pi^2} \]
\[ x^2 = \left(\frac{3}{\pi}\right)^2 = \frac{9}{\pi^2} \]

### Step 5: Substitute \( A^2 \) and \( x^2 \) into the speed formula
Now substituting these values back into the speed formula:
\[ v = \pi \sqrt{\frac{25}{\pi^2} - \frac{9}{\pi^2}} = \pi \sqrt{\frac{25 - 9}{\pi^2}} = \pi \sqrt{\frac{16}{\pi^2}} \]

### Step 6: Simplify the expression
This simplifies to:
\[ v = \pi \cdot \frac{4}{\pi} = 4 \text{ m/s} \]

### Final Answer
The speed of the particle when it is \( \frac{3}{\pi} \) m from equilibrium is:
\[ \boxed{4 \text{ m/s}} \]

---