A particle moves such that its acceleration a is given by a = -bx , where x is the displacement from equilibrium positionand is a constant. The period of oscillation is

Options

A. `2 pi sqrtb`
B. `(2 pi)/ sqrtb`
C. `2pi/b`
D. `2 sqrt(pi/ b)`

Answer

To find the period of oscillation for a particle moving under the influence of the acceleration given by \( a = -bx \), we can follow these steps:

### Step 1: Understand the equation of motion
The equation \( a = -bx \) indicates that the acceleration \( a \) is directly proportional to the displacement \( x \) from the equilibrium position, with a negative sign. This is characteristic of simple harmonic motion (SHM).

**Hint:** Recognize that the negative sign indicates that the acceleration is directed towards the equilibrium position, which is a key feature of SHM.

### Step 2: Relate acceleration to SHM
In SHM, the acceleration can be expressed as:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency. By comparing this with the given equation \( a = -bx \), we can identify that:
\[ \omega^2 = b \]

**Hint:** Remember that in SHM, the angular frequency \( \omega \) relates to the acceleration and displacement.

### Step 3: Find the angular frequency
From the relation \( \omega^2 = b \), we can express the angular frequency \( \omega \) as:
\[ \omega = \sqrt{b} \]

**Hint:** The angular frequency is the square root of the proportionality constant in the acceleration equation.

### Step 4: Calculate the period of oscillation
The period \( T \) of oscillation in SHM is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting our expression for \( \omega \):
\[ T = \frac{2\pi}{\sqrt{b}} \]

**Hint:** The period is inversely proportional to the angular frequency.

### Step 5: Final expression for the period
Thus, the period of oscillation for the particle is:
\[ T = 2\pi \sqrt{\frac{1}{b}} \]

**Hint:** Ensure that you understand how the period relates to the constant \( b \) in the acceleration equation.

### Conclusion
The final answer for the period of oscillation is:
\[ T = 2\pi \sqrt{\frac{1}{b}} \]