A particle performs linear S.H.M. At a particular instant, velocity of the particle is `u` and acceleration is `alpha` while at another instant, velocity is `v` and acceleration `beta` `(0ltalphaltbeta)`. The distance between the two position is

Options

A. `(u^(2)-v^(2))/(alpha+beta)`
B. `(u^(2)+v^(2))/(alpha+beta)`
C. `(u^(2)-v^(2))/(alpha-beta)`
D. `(u^(2)+v^(2))/(alpha-beta)`

Answer

To solve the problem, we need to determine the distance between two positions of a particle performing linear simple harmonic motion (SHM) given its velocities and accelerations at those positions.

### Step-by-Step Solution:

1. **Understanding SHM Relations**:
- In SHM, the acceleration \( a \) at a position \( x \) is given by:
\[ a = -\omega^2 x \]
- The velocity \( v \) at a position \( x \) is given by:
\[ v^2 = \omega^2 A^2 - x^2 \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Setting Up the Equations**:
- For the first position \( x_1 \):
- Velocity \( u \) and acceleration \( \alpha \):
\[ \alpha = -\omega^2 x_1 \quad \text{(1)} \]
\[ u^2 = \omega^2 A^2 - x_1^2 \quad \text{(2)} \]
- For the second position \( x_2 \):
- Velocity \( v \) and acceleration \( \beta \):
\[ \beta = -\omega^2 x_2 \quad \text{(3)} \]
\[ v^2 = \omega^2 A^2 - x_2^2 \quad \text{(4)} \]

3. **Relating Accelerations**:
- From equations (1) and (3):
\[ \alpha + \beta = -\omega^2 (x_1 + x_2) \]
- Rearranging gives:
\[ x_1 + x_2 = -\frac{\alpha + \beta}{\omega^2} \quad \text{(5)} \]

4. **Relating Velocities**:
- Subtract equation (4) from equation (2):
\[ u^2 - v^2 = (\omega^2 A^2 - x_1^2) - (\omega^2 A^2 - x_2^2) \]
- Simplifying gives:
\[ u^2 - v^2 = x_2^2 - x_1^2 \]
- This can be factored as:
\[ u^2 - v^2 = (x_2 - x_1)(x_2 + x_1) \quad \text{(6)} \]

5. **Substituting for \( x_1 + x_2 \)**:
- From equation (5), substitute \( x_1 + x_2 \) into equation (6):
\[ u^2 - v^2 = (x_2 - x_1)\left(-\frac{\alpha + \beta}{\omega^2}\right) \]
- Rearranging gives:
\[ x_2 - x_1 = \frac{u^2 - v^2}{-\frac{\alpha + \beta}{\omega^2}} = \frac{(u^2 - v^2) \omega^2}{\alpha + \beta} \]

6. **Final Result**:
- The distance between the two positions \( x_2 - x_1 \) is:
\[ x_2 - x_1 = \frac{u^2 - v^2}{\alpha + \beta} \]

### Conclusion:
The distance between the two positions is given by:
\[ \boxed{\frac{u^2 - v^2}{\alpha + \beta}} \]