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. The ratio `(AX)/(BX)` is

Options

A. `(2sqrt(2)-1)/(2sqrt(2)+1)`
B. `(sqrt(3)-1)/(sqrt(3)+1)`
C. `(3sqrt(2)-2)/(3sqrt(2)+2)`
D. `(1)/(3)`

Explanation

To solve the problem, we need to find the ratio \( \frac{AX}{BX} \) where A and B are points on the path of a particle executing Simple Harmonic Motion (SHM) and the velocity is zero at these points. Given that the particle crosses point X at intervals of 2 seconds and 4 seconds with a speed of 5 m/s, we can proceed as follows:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Points A and B are the extreme positions of the SHM where the velocity is zero.
- The particle crosses point X at intervals of 2 seconds and 4 seconds, indicating its motion back and forth.

2. **Determine the Time Period**:
- The total time taken to go from X back to X after reaching point A and then B is \( 2 + 4 = 6 \) seconds.
- Therefore, the time period \( T \) of the SHM is 6 seconds.

3. **Calculate Half the Time Period**:
- The time taken to move from the mean position (equilibrium) to point A or B is half of the time period:
\[ \frac{T}{2} = \frac{6}{2} = 3 \text{ seconds} \]

4. **Determine Time to Reach Point X**:
- Let \( t_0 \) be the time taken to go from the mean position to point X. The total time to go from the mean position to A and back to X is:
\[ t_0 + t_0 + 3 = 4 \text{ seconds} \]
- Simplifying gives:
\[ 2t_0 + 3 = 4 \implies 2t_0 = 1 \implies t_0 = \frac{1}{2} \text{ seconds} \]

5. **Calculate the Angular Frequency**:
- The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} \text{ rad/s} \]

6. **Calculate the Displacement from Mean Position**:
- The displacement \( x \) from the mean position after \( t_0 \) seconds is:
\[ x = A \sin(\omega t_0) = A \sin\left(\frac{\pi}{3} \cdot \frac{1}{2}\right) = A \sin\left(\frac{\pi}{6}\right) = A \cdot \frac{1}{2} \]

7. **Express AX and BX**:
- The distances from the mean position to points A and B are:
\[ AX = A - x \quad \text{and} \quad BX = A + x \]
- Substituting for \( x \):
\[ AX = A - \frac{A}{2} = \frac{A}{2} \quad \text{and} \quad BX = A + \frac{A}{2} = \frac{3A}{2} \]

8. **Calculate the Ratio**:
- Now, we can find the ratio \( \frac{AX}{BX} \):
\[ \frac{AX}{BX} = \frac{\frac{A}{2}}{\frac{3A}{2}} = \frac{1}{3} \]

### Final Answer:
The ratio \( \frac{AX}{BX} \) is \( \frac{1}{3} \).