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 (T) is 12 seconds and the amplitude is \( a \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Time period \( T = 12 \) seconds
- Time \( t = 2 \) seconds
- Amplitude \( a \)

2. **Calculate Angular Frequency (\( \omega \)):**
The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \, \text{rad/s} \]

3. **Determine the Displacement at \( t = 2 \) seconds:**
The displacement \( x \) of the particle at any time \( t \) in S.H.M is given by:
\[ x = a \sin(\omega t) \]
Substituting \( \omega \) and \( t \):
\[ x = a \sin\left(\frac{\pi}{6} \cdot 2\right) = a \sin\left(\frac{\pi}{3}\right) \]

4. **Calculate \( \sin\left(\frac{\pi}{3}\right) \):**
We know that:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Therefore, substituting this value back:
\[ x = a \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}a}{2} \]

5. **Determine the Shortest Distance from the Mean Position:**
Since the particle moves from the mean position to the maximum displacement and back, the shortest distance traveled from the mean position to the point of displacement \( x \) is simply \( |x| \):
\[ \text{Shortest distance} = \frac{\sqrt{3}a}{2} \]

### Final Answer:
The shortest distance traveled by the particle from the mean position in 2 seconds is:
\[ \frac{\sqrt{3}a}{2} \]