In the previous problem, the displacement of the particle from the mean position corresponding to the instant mentioned is

Options

A. `(5)/(pi)m`
B. `(5sqrt3)/(pi)m`
C. `(10sqrt3)/(pi)m`
D. `(5sqrt3)/(2pi)m`

Explanation

To solve the problem of finding the displacement of a particle from the mean position in simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the displacement of a particle in SHM at a specific time, given that the maximum velocity (B_max) is 5 m/s.

2. **Identify the Formula**: The displacement \( x \) in SHM can be expressed as:
\[ x = A \cos(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time,
- \( \phi \) is the phase constant.

3. **Relate Maximum Velocity to Amplitude and Angular Frequency**: The maximum velocity \( B_{max} \) is given by the formula:
\[ B_{max} = A \omega \]
Since \( B_{max} = 5 \, \text{m/s} \), we can express this as:
\[ A \omega = 5 \]

4. **Determine the Angular Frequency**: The angular frequency \( \omega \) can be related to the period \( T \) of the motion:
\[ \omega = \frac{2\pi}{T} \]
If we assume a time period \( T \) of 4 seconds, then:
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{rad/s} \]

5. **Calculate Amplitude**: Substitute \( \omega \) back into the maximum velocity equation:
\[ A \cdot \frac{\pi}{2} = 5 \implies A = \frac{5 \cdot 2}{\pi} = \frac{10}{\pi} \, \text{m} \]

6. **Determine the Phase Constant**: If we assume a phase constant \( \phi \) of \( \frac{\pi}{6} \) (as mentioned in the transcript), we can now substitute \( t = 4 \) seconds into the displacement equation:
\[ x = A \cos(\omega t + \phi) = \frac{10}{\pi} \cos\left(\frac{\pi}{2} \cdot 4 + \frac{\pi}{6}\right) \]

7. **Calculate the Argument of the Cosine**:
\[ \frac{\pi}{2} \cdot 4 = 2\pi \quad \text{and} \quad 2\pi + \frac{\pi}{6} = 2\pi + \frac{\pi}{6} = 2\pi + 0.5236 \approx 2\pi + 0.524 \]
Since cosine is periodic with a period of \( 2\pi \):
\[ \cos(2\pi + \frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \]

8. **Final Calculation of Displacement**:
\[ x = \frac{10}{\pi} \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{\pi} \, \text{m} \]

### Final Answer:
The displacement of the particle from the mean position at \( t = 4 \) seconds is:
\[ x = \frac{5\sqrt{3}}{\pi} \, \text{m} \]