A particle is executing S.H.M.if `v_1` and `v_2` are the velocities of the particle at distances `x_1` and `x_2` from mean position respectively , then time period of oscillation is

Options

A. `2πsqrt(((x_2^2+x_1^2))/((V_1^2+V_2^2)))`
B. `2πsqrt(((x_2^2-x_1^2))/((V_1^2-V_2^2)))`
C. `2πsqrt(((x_2^2-x_1^2))/((V_2^2-V_1^2)))`
D. `2πsqrt(((x_1^2-x_2^2))/((V_1^2+V_2^2)))`

Answer

To solve the problem, we need to derive the time period of a particle executing Simple Harmonic Motion (S.H.M) based on its velocities at two different positions from the mean position.

### Step-by-Step Solution:

1. **Understanding the Velocity in S.H.M**:
The velocity \( v \) of a particle in S.H.M 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.

2. **Applying the Formula for Two Positions**:
For two different positions \( x_1 \) and \( x_2 \), we can write:
\[ v_1 = \omega \sqrt{A^2 - x_1^2} \quad \text{(1)} \]
\[ v_2 = \omega \sqrt{A^2 - x_2^2} \quad \text{(2)} \]

3. **Squaring Both Equations**:
Squaring both equations (1) and (2) gives:
\[ v_1^2 = \omega^2 (A^2 - x_1^2) \quad \text{(3)} \]
\[ v_2^2 = \omega^2 (A^2 - x_2^2) \quad \text{(4)} \]

4. **Rearranging the Equations for Amplitude**:
From equations (3) and (4), we can express \( A^2 \):
\[ A^2 = \frac{v_1^2}{\omega^2} + x_1^2 \quad \text{(5)} \]
\[ A^2 = \frac{v_2^2}{\omega^2} + x_2^2 \quad \text{(6)} \]

5. **Setting Equations (5) and (6) Equal**:
Since both expressions equal \( A^2 \), we can set them equal to each other:
\[ \frac{v_1^2}{\omega^2} + x_1^2 = \frac{v_2^2}{\omega^2} + x_2^2 \]

6. **Rearranging to Isolate Terms**:
Rearranging gives:
\[ \frac{v_1^2 - v_2^2}{\omega^2} = x_2^2 - x_1^2 \]

7. **Solving for \(\omega\)**:
Rearranging further, we find:
\[ \omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \]

8. **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 \( \omega \):
\[ T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} \]

### Final Answer:
The time period of oscillation is given by:
\[ T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} \]