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

Options

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

Answer

To solve the problem step-by-step, we need to analyze the motion of a particle performing simple harmonic motion (SHM) and relate its velocity and acceleration at two different instances.

### Step 1: Understand the equations of SHM
In SHM, the position \( x \) of the particle can be expressed as:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( t \) is the time.

### Step 2: Find expressions for velocity and acceleration
The velocity \( v \) and acceleration \( a \) of the particle can be derived from the position function:
- The velocity \( v \) is given by the derivative of position with respect to time:
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t) \]
- The acceleration \( a \) is given by the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \]

### Step 3: Set up equations for two instances
Let’s denote the two instances as \( t_1 \) and \( t_2 \):
- At time \( t_1 \):
- Velocity \( v_1 = u \)
- Acceleration \( a_1 = \alpha \)

From the equations:
\[ u = A \omega \cos(\omega t_1) \quad \text{(1)} \]
\[ \alpha = -A \omega^2 \sin(\omega t_1) \quad \text{(2)} \]

- At time \( t_2 \):
- Velocity \( v_2 = v \)
- Acceleration \( a_2 = \beta \)

From the equations:
\[ v = A \omega \cos(\omega t_2) \quad \text{(3)} \]
\[ \beta = -A \omega^2 \sin(\omega t_2) \quad \text{(4)} \]

### Step 4: Relate the two instances
From equations (1) and (3), we can express \( \cos(\omega t_1) \) and \( \cos(\omega t_2) \):
\[ \cos(\omega t_1) = \frac{u}{A \omega} \quad \text{(5)} \]
\[ \cos(\omega t_2) = \frac{v}{A \omega} \quad \text{(6)} \]

From equations (2) and (4), we can express \( \sin(\omega t_1) \) and \( \sin(\omega t_2) \):
\[ \sin(\omega t_1) = -\frac{\alpha}{A \omega^2} \quad \text{(7)} \]
\[ \sin(\omega t_2) = -\frac{\beta}{A \omega^2} \quad \text{(8)} \]

### Step 5: Find the distance between the two positions
The distance \( x_1 - x_2 \) can be expressed as:
\[ x_1 - x_2 = A \sin(\omega t_1) - A \sin(\omega t_2 \]
Factoring out \( A \):
\[ x_1 - x_2 = A \left( \sin(\omega t_1) - \sin(\omega t_2) \right) \]

Using the sine difference identity:
\[ \sin(\omega t_1) - \sin(\omega t_2) = 2 \cos\left(\frac{\omega t_1 + \omega t_2}{2}\right) \sin\left(\frac{\omega t_1 - \omega t_2}{2}\right) \]

### Step 6: Substitute and simplify
Substituting equations (7) and (8) into the expression for distance:
\[ x_1 - x_2 = A \left(-\frac{\alpha}{A \omega^2} - \left(-\frac{\beta}{A \omega^2}\right)\right) \]
This simplifies to:
\[ x_1 - x_2 = \frac{A}{\omega^2} (\beta - \alpha) \]

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