Amplitude of oscillation of a particle that executes `SHM` is `2cm`. Its displacement from its mean position in a time equal to `1//6^(th)` of its time period is

Options

A. `sqrt(2)cm`
B. `sqrt(3)cm`
C. `(1)/(sqrt(2))cm`
D. `(1)/(sqrt(3))cm`

Explanation

To solve the problem, we need to find the displacement of a particle executing Simple Harmonic Motion (SHM) after a time equal to \( \frac{1}{6} \) of its time period, given that the amplitude of oscillation is \( 2 \, \text{cm} \).

### Step-by-Step Solution:

1. **Understand the SHM Equation**:
The displacement \( x \) of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time,
- \( \phi \) is the phase constant.

2. **Given Values**:
- Amplitude \( A = 2 \, \text{cm} \).
- Time \( t = \frac{T}{6} \), where \( T \) is the time period of the oscillation.

3. **Calculate Angular Frequency**:
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]

4. **Substitute Time into the Displacement Equation**:
Substitute \( t = \frac{T}{6} \) into the displacement equation:
\[ x\left(\frac{T}{6}\right) = A \sin\left(\omega \cdot \frac{T}{6}\right) \]
Substituting for \( \omega \):
\[ x\left(\frac{T}{6}\right) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{6}\right) \]
Simplifying this gives:
\[ x\left(\frac{T}{6}\right) = A \sin\left(\frac{2\pi}{6}\right) = A \sin\left(\frac{\pi}{3}\right) \]

5. **Calculate the Sine Value**:
The sine of \( \frac{\pi}{3} \) is:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

6. **Final Calculation of Displacement**:
Now, substitute \( A = 2 \, \text{cm} \):
\[ x\left(\frac{T}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{cm} \]

### Conclusion:
The displacement of the particle from its mean position after a time equal to \( \frac{1}{6} \) of its time period is \( \sqrt{3} \, \text{cm} \).