The acceleration displacement graph of a particle executing simple harmonic motion is shown in figure. The time period of simple harmonic motion is

Options

A. `(4 pi)/(sqrt(3))s`
B. `(2 pi)/(sqrt(3))s`
C. The given graph doesn't represent simple harmonic motion
D. information is insufficient

Explanation

To find the time period of a particle executing simple harmonic motion (SHM) from the acceleration-displacement graph, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
In SHM, the acceleration \( a \) is related to the displacement \( x \) by the equation:
\[ a = -\omega^2 x \]
This indicates that the acceleration is directly proportional to the displacement but in the opposite direction.

2. **Identify the Graph Type**:
The graph of acceleration versus displacement is a straight line with a negative slope. This can be expressed in the form \( y = mx + c \), where \( y \) is acceleration \( a \) and \( x \) is displacement \( x \).

3. **Determine the Slope**:
The slope of the line \( m \) can be defined as:
\[ m = \frac{dy}{dx} \]
Since the graph shows a negative slope, we can denote it as:
\[ m = -\omega^2 \]

4. **Calculate the Slope from the Graph**:
If the angle \( \theta \) made with the horizontal is given (for example, \( 37^\circ \)), we can find the slope using:
\[ m = -\tan(\theta) \]
For \( \theta = 37^\circ \):
\[ \tan(37^\circ) = \frac{3}{4} \]
Therefore, the slope \( m \) becomes:
\[ m = -\frac{3}{4} \]

5. **Relate the Slope to Angular Frequency**:
From the relationship \( m = -\omega^2 \), we have:
\[ -\omega^2 = -\frac{3}{4} \]
Thus, we can write:
\[ \omega^2 = \frac{3}{4} \]

6. **Find Angular Frequency**:
Taking the square root gives:
\[ \omega = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

7. **Relate Angular Frequency to Time Period**:
The angular frequency \( \omega \) is related to the time period \( T \) by the equation:
\[ \omega = \frac{2\pi}{T} \]
Therefore, we can rearrange this to find \( T \):
\[ T = \frac{2\pi}{\omega} \]

8. **Substitute the Value of \( \omega \)**:
Substituting \( \omega = \frac{\sqrt{3}}{2} \):
\[ T = \frac{2\pi}{\frac{\sqrt{3}}{2}} = \frac{4\pi}{\sqrt{3}} \]

9. **Final Answer**:
The time period of the simple harmonic motion is:
\[ T = \frac{4\pi}{\sqrt{3}} \]