A particle executes simple harmonic motion of period `16 s`. Two seconds later after it passes through the center of oscillation its velocity is found to be `2 m//s`. Find the amplitude.

Explanation

To solve the problem, we need to find the amplitude of a particle executing simple harmonic motion (SHM) given its period and velocity at a specific time after passing through the center.

### Step-by-Step Solution:

1. **Identify the Period (T)**:
The period of the motion is given as \( T = 16 \, \text{s} \).

2. **Calculate Angular Frequency (\( \omega \))**:
The angular frequency is calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{16} = \frac{\pi}{8} \, \text{rad/s} \]

3. **Determine the Time After Passing Through the Mean Position**:
The particle passes through the center at \( t = 0 \). We need to find the velocity at \( t = 2 \, \text{s} \).

4. **Use the Velocity Formula in SHM**:
The velocity \( V \) in simple harmonic motion is given by:
\[ V = \omega A \cos(\omega t) \]
Where \( A \) is the amplitude and \( t \) is the time.

5. **Substitute Known Values**:
At \( t = 2 \, \text{s} \), the velocity is given as \( V = 2 \, \text{m/s} \). Substituting the known values into the velocity equation:
\[ 2 = \left(\frac{\pi}{8}\right) A \cos\left(\frac{\pi}{8} \cdot 2\right) \]

6. **Calculate \( \cos(\omega t) \)**:
First, calculate \( \omega t \):
\[ \omega t = \frac{\pi}{8} \cdot 2 = \frac{\pi}{4} \]
Now, find \( \cos\left(\frac{\pi}{4}\right) \):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

7. **Substitute \( \cos(\omega t) \) into the Velocity Equation**:
Now substitute \( \cos\left(\frac{\pi}{4}\right) \) back into the velocity equation:
\[ 2 = \left(\frac{\pi}{8}\right) A \left(\frac{1}{\sqrt{2}}\right) \]

8. **Solve for Amplitude \( A \)**:
Rearranging the equation to solve for \( A \):
\[ A = \frac{2 \cdot 8 \sqrt{2}}{\pi} = \frac{16\sqrt{2}}{\pi} \]
Now, calculate the numerical value:
\[ A \approx \frac{16 \cdot 1.414}{3.14} \approx \frac{22.624}{3.14} \approx 7.2 \, \text{m} \]

### Final Answer:
The amplitude \( A \) is approximately \( 7.2 \, \text{m} \).