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. `2pisqrt((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2))`
B. `2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2))`
C. `2pisqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2))`
D. `2pisqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2))`

Explanation

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 distances from the mean position. Let's break down the solution step by step.

### Step 1: Understand the velocity equation in SHM
The velocity \( v \) of a particle in SHM at a distance \( x \) from the mean position is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude of the motion.

### Step 2: Write the equations for the given distances
For the distances \( x_1 \) and \( x_2 \), we can write the following equations based on the velocities \( v_1 \) and \( v_2 \):
1. At distance \( x_1 \):
\[ v_1 = \omega \sqrt{A^2 - x_1^2} \quad \text{(Equation 1)} \]
2. At distance \( x_2 \):
\[ v_2 = \omega \sqrt{A^2 - x_2^2} \quad \text{(Equation 2)} \]

### Step 3: Square both equations
To eliminate the square root, we square both equations:
1. From Equation 1:
\[ v_1^2 = \omega^2 (A^2 - x_1^2) \quad \text{(Equation 3)} \]
2. From Equation 2:
\[ v_2^2 = \omega^2 (A^2 - x_2^2) \quad \text{(Equation 4)} \]

### Step 4: Subtract the two equations
Now, we subtract Equation 4 from Equation 3:
\[ v_1^2 - v_2^2 = \omega^2 \left((A^2 - x_1^2) - (A^2 - x_2^2)\right) \]
This simplifies to:
\[ v_1^2 - v_2^2 = \omega^2 (x_2^2 - x_1^2) \]

### Step 5: Solve for \( \omega^2 \)
Rearranging the equation gives us:
\[ \omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \]

### Step 6: Find 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 \( \omega \):
\[ T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} \]

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