A particle executes S.H.M., according to the displacement equation x = 6 sin `(3pit+pi//6)`m. Then the magnitude of its acceleration at t = 2 s is

Options

A. `3pi^(2)m//s^(2)`
B. `9pi^(2)m//s^(2)`
C. `18pi^(2)m//s^(2)`
D. `27pi^(2)m//s^(2)`

Explanation

To solve the problem, we need to find the magnitude of the acceleration of a particle executing simple harmonic motion (SHM) at a specific time, given its displacement equation.

**Step-by-Step Solution:**

1. **Identify the given displacement equation:**
The displacement of the particle is given by:
\[ x(t) = 6 \sin(3\pi t + \frac{\pi}{6}) \text{ m} \]

2. **Determine the time at which we need to find the acceleration:**
We need to find the acceleration at \( t = 2 \) seconds.

3. **Calculate the displacement at \( t = 2 \) seconds:**
Substitute \( t = 2 \) into the displacement equation:
\[ x(2) = 6 \sin(3\pi \cdot 2 + \frac{\pi}{6}) = 6 \sin(6\pi + \frac{\pi}{6}) \]
Using the property of sine, \( \sin(6\pi + \theta) = \sin(\theta) \):
\[ x(2) = 6 \sin(\frac{\pi}{6}) = 6 \cdot \frac{1}{2} = 3 \text{ m} \]

4. **Find the velocity by differentiating the displacement equation:**
The velocity \( v(t) \) is the first derivative of the displacement:
\[ v(t) = \frac{dx}{dt} = 6 \cdot 3\pi \cos(3\pi t + \frac{\pi}{6}) = 18\pi \cos(3\pi t + \frac{\pi}{6}) \]

5. **Find the acceleration by differentiating the velocity equation:**
The acceleration \( a(t) \) is the derivative of the velocity:
\[ a(t) = \frac{dv}{dt} = -18\pi \cdot 3\pi \sin(3\pi t + \frac{\pi}{6}) = -54\pi^2 \sin(3\pi t + \frac{\pi}{6}) \]

6. **Substitute \( t = 2 \) into the acceleration equation:**
We already know \( \sin(3\pi \cdot 2 + \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2} \):
\[ a(2) = -54\pi^2 \cdot \frac{1}{2} = -27\pi^2 \text{ m/s}^2 \]

7. **Determine the magnitude of the acceleration:**
The magnitude of the acceleration is:
\[ |a(2)| = 27\pi^2 \text{ m/s}^2 \]

**Final Answer:**
The magnitude of the acceleration at \( t = 2 \) seconds is:
\[ 27\pi^2 \text{ m/s}^2 \]

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