A particle moves such that acceleration is given by `a= -4x`. The period of oscillation is

Options

A. `pi`
B. `(2)/(pi)`
C. `(1)/(pi)`
D. `2pi`

Answer

To solve the problem, we need to find the period of oscillation for a particle whose acceleration is given by the equation \( a = -4x \).

### Step-by-Step Solution:

1. **Identify the Form of the Equation**:
The acceleration \( a \) of a particle in simple harmonic motion (SHM) is generally given by the equation:
\[ a = -\omega^2 x \]
Here, \( \omega \) is the angular frequency and \( x \) is the displacement from the mean position.

2. **Compare the Given Equation**:
We are given:
\[ a = -4x \]
By comparing this with the standard form of SHM, we can identify:
\[ -\omega^2 = -4 \]

3. **Solve for Angular Frequency \( \omega \)**:
From the comparison, we can deduce:
\[ \omega^2 = 4 \]
Taking the square root of both sides gives:
\[ \omega = 2 \, \text{rad/s} \]

4. **Calculate the Period of Oscillation**:
The period \( T \) of oscillation is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \]

5. **Final Answer**:
Therefore, the period of oscillation is:
\[ T = \pi \, \text{seconds} \]