The time period of a particle executing S.H.M.is 12 s. The shortest distance travelledby it from mean position in 2 second is ( amplitude is a )

Options

A. `(a)/(2)`
B. `(A)/(sqrt(2))`
C. `(sqrt(3)a)/(2)`
D. a

Answer

To solve the problem, we need to find the shortest distance traveled by a particle executing Simple Harmonic Motion (S.H.M.) from the mean position in 2 seconds, given that the time period is 12 seconds and the amplitude is \( a \).

### Step-by-Step Solution:

1. **Identify the Time Period and Calculate Angular Frequency**:
The time period \( T \) is given as 12 seconds. The relationship between the time period and angular frequency \( \omega \) is given by:
\[ T = \frac{2\pi}{\omega} \]
Rearranging this gives:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6} \, \text{rad/s} \]

**Hint**: Remember that the angular frequency is related to the time period by the formula \( \omega = \frac{2\pi}{T} \).

2. **Write the Equation of Motion**:
The equation for the position \( x \) of a particle in S.H.M. starting from the mean position is:
\[ x(t) = a \sin(\omega t) \]
Substituting the value of \( \omega \):
\[ x(t) = a \sin\left(\frac{\pi}{6} t\right) \]

**Hint**: The sine function describes the oscillation of the particle in S.H.M.

3. **Calculate the Position at \( t = 2 \) seconds**:
Substitute \( t = 2 \) seconds into the equation:
\[ x(2) = a \sin\left(\frac{\pi}{6} \cdot 2\right) = a \sin\left(\frac{\pi}{3}\right) \]
We know that:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Therefore:
\[ x(2) = a \cdot \frac{\sqrt{3}}{2} \]

**Hint**: Use the known values of sine for common angles to simplify calculations.

4. **Determine the Shortest Distance from the Mean Position**:
The distance from the mean position to the position at \( t = 2 \) seconds is simply the absolute value of \( x(2) \):
\[ \text{Distance} = |x(2)| = a \cdot \frac{\sqrt{3}}{2} \]

**Hint**: The shortest distance from the mean position is the absolute value of the position calculated.

5. **Final Answer**:
Thus, the shortest distance traveled by the particle from the mean position in 2 seconds is:
\[ \frac{\sqrt{3}}{2} a \]

The correct option is \( \frac{\sqrt{3}}{2} a \).