A particle moves such that its acceleration is given by `a =- beta (x-2)` Here `beta` is positive constant and `x` is the position form origin. Time period of oscillation is

Options

A. `2pisqrt(beta)`
B. `2pi(1)/(sqrt(beta))`
C. `2pi sqrt(beta +2)`
D. `2pi sqrt((1)/(beta+2))`

Explanation

To find the time period of oscillation for a particle whose acceleration is given by the equation \( a = -\beta (x - 2) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Equation**:
The given equation for acceleration can be rewritten as:
\[ a = -\beta (x - 2) \]
This indicates that the particle experiences a restoring force proportional to its displacement from the position \( x = 2 \).

2. **Relating Acceleration to Displacement**:
We know that acceleration \( a \) can also be expressed in terms of displacement \( x \) as:
\[ a = \frac{d^2x}{dt^2} \]
Thus, we can set the two expressions for acceleration equal to each other:
\[ \frac{d^2x}{dt^2} = -\beta (x - 2) \]

3. **Shifting the Equilibrium Position**:
To simplify the equation, we can introduce a new variable \( y \) such that:
\[ y = x - 2 \]
Therefore, \( x = y + 2 \) and the equation becomes:
\[ \frac{d^2y}{dt^2} = -\beta y \]

4. **Identifying the Form of Simple Harmonic Motion**:
The equation \( \frac{d^2y}{dt^2} = -\beta y \) is a standard form of simple harmonic motion (SHM), where:
\[ \omega^2 = \beta \]
Here, \( \omega \) is the angular frequency.

5. **Finding the Time Period**:
The time period \( T \) of oscillation is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = \sqrt{\beta} \):
\[ T = \frac{2\pi}{\sqrt{\beta}} \]

### Final Answer:
The time period of oscillation is:
\[ T = \frac{2\pi}{\sqrt{\beta}} \]