A particle is executing SHM along a straight line. Its velocities at distances `x_(1)` and `x_(2)` from the mean position are `v_(1)` and `v_(2)`, respectively. Its time period is

Options

A. `2pi sqrt((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))`
B. `2pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`
C. `2pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`
D. `2pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`

Answer

To solve the problem, we need to find the time period of a particle executing Simple Harmonic Motion (SHM) given its velocities at two different positions from the mean position.

### Step-by-Step Solution:

1. **Understanding the relationship between velocity and displacement in SHM**:
The velocity \( v \) of a particle in SHM can be expressed using the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( \omega \) is the angular frequency, \( A \) is the amplitude, and \( x \) is the displacement from the mean position.

2. **Setting up the equations for the two positions**:
For the two positions \( x_1 \) and \( x_2 \) with corresponding velocities \( v_1 \) and \( v_2 \), we can write:
\[ v_1 = \omega \sqrt{A^2 - x_1^2} \]
\[ v_2 = \omega \sqrt{A^2 - x_2^2} \]

3. **Squaring both equations**:
Squaring both sides of the equations gives us:
\[ v_1^2 = \omega^2 (A^2 - x_1^2) \]
\[ v_2^2 = \omega^2 (A^2 - x_2^2) \]

4. **Rearranging the equations**:
Rearranging these equations, we have:
\[ \omega^2 A^2 - \omega^2 x_1^2 = v_1^2 \quad \text{(1)} \]
\[ \omega^2 A^2 - \omega^2 x_2^2 = v_2^2 \quad \text{(2)} \]

5. **Subtracting the two equations**:
Subtract equation (2) from equation (1):
\[ v_1^2 - v_2^2 = \omega^2 (x_2^2 - x_1^2) \]

6. **Solving for \(\omega^2\)**:
Rearranging gives:
\[ \omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \]

7. **Finding the time period \( T \)**:
The time period \( T \) is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \):
\[ T = \frac{2\pi}{\sqrt{\frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}}} \]

8. **Simplifying the expression for \( T \)**:
This can be rewritten as:
\[ T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} \]

### Final Answer:
Thus, the time period \( T \) of the particle executing SHM is given by:
\[ T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} \]