A and B are two points on the path of a particle executing SHM in a straight line, at which its velocity is zero. Starting from a certain point X `(AXltBX)` the particle crosses this pint again at successive intervals of 2 seconds and 4 seconds with a speed of `5m//s`
Q. Amplitude of oscillation is

Options

A. `(10sqrt(3))/(pi)m`
B. `(4sqrt(2))/(pi)m`
C. `(9sqrt(2))/(pi)m`
D. `(17sqrt(2))/(pi)m`

Explanation

To find the amplitude of the oscillation of a particle executing Simple Harmonic Motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
- Points A and B are where the velocity of the particle is zero, indicating these are the extreme positions of the SHM.
- The particle crosses point X at intervals of 2 seconds and 4 seconds with a speed of 5 m/s.

2. **Identifying the Time Period:**
- The time taken to go from point X to point A (or B) is 2 seconds.
- The total time taken to return to point X after reaching point B is 4 seconds.
- Therefore, the total time period (T) for one complete cycle is the sum of these intervals:
\[ T = 2 \text{ seconds} + 2 \text{ seconds} = 4 \text{ seconds} \]

3. **Calculating Angular Frequency:**
- The angular frequency (ω) is given by:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s} \]

4. **Using the Velocity Formula:**
- The velocity (v) of a particle in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
- Here, \(v\) is given as 5 m/s, and \(x\) is the displacement from the mean position to point X.

5. **Finding the Displacement (x):**
- The time taken to move from the mean position to point X is \(t_0 = 1\) second (half of 2 seconds).
- The displacement \(x\) can be calculated using:
\[ x = A \sin(\omega t_0) = A \sin\left(\frac{\pi}{2} \cdot 1\right) = A \sin\left(\frac{\pi}{2}\right) = A \]
- Thus, \(x = A\).

6. **Substituting Values into Velocity Equation:**
- Substitute \(x = A\) into the velocity equation:
\[ 5 = \frac{\pi}{2} \sqrt{A^2 - A^2} \]
- This simplifies to:
\[ 5 = \frac{\pi}{2} \sqrt{A^2 - A^2} = 0 \quad \text{(not useful)} \]
- Instead, we need to find \(x\) at a different point, say \(x = \frac{A}{2}\) (since the particle crosses X at a point between A and B).

7. **Revising the Velocity Equation:**
- Let’s consider \(x = \frac{A}{2}\):
\[ 5 = \frac{\pi}{2} \sqrt{A^2 - \left(\frac{A}{2}\right)^2} \]
- This simplifies to:
\[ 5 = \frac{\pi}{2} \sqrt{A^2 - \frac{A^2}{4}} = \frac{\pi}{2} \sqrt{\frac{3A^2}{4}} = \frac{\pi A \sqrt{3}}{4} \]

8. **Solving for Amplitude (A):**
- Rearranging gives:
\[ 5 = \frac{\pi A \sqrt{3}}{4} \quad \Rightarrow \quad A = \frac{20}{\pi \sqrt{3}} \]

9. **Final Calculation:**
- To find the numerical value of A:
\[ A \approx \frac{20}{3.14 \cdot 1.732} \approx \frac{20}{5.441} \approx 3.68 \text{ meters} \]

### Final Answer:
The amplitude of oscillation is approximately \(3.68\) meters.