Displacement-time equation of a particle executing SHM is `x=A sin (omega t+(pi)/6)` Time taken by the particle to go directly from `x=-A/2" to "x=+A/2`

Options

A. `(pi)/(3omega)`
B. `(pi)/(2omega)`
C. `(pi)/(omega)`
D. `(pi)/(omega)`

Answer

To find the time taken by the particle to move directly from \( x = -\frac{A}{2} \) to \( x = +\frac{A}{2} \) in simple harmonic motion (SHM), we start with the given displacement-time equation:

\[ x = A \sin\left(\omega t + \frac{\pi}{6}\right) \]

### Step 1: Set up the equations for the two positions
We need to find the times \( t_1 \) and \( t_2 \) when the particle is at \( x = -\frac{A}{2} \) and \( x = +\frac{A}{2} \), respectively.

1. For \( x = -\frac{A}{2} \):
\[ -\frac{A}{2} = A \sin\left(\omega t_1 + \frac{\pi}{6}\right) \]
Dividing both sides by \( A \):
\[ -\frac{1}{2} = \sin\left(\omega t_1 + \frac{\pi}{6}\right) \]

2. For \( x = +\frac{A}{2} \):
\[ \frac{A}{2} = A \sin\left(\omega t_2 + \frac{\pi}{6}\right) \]
Dividing both sides by \( A \):
\[ \frac{1}{2} = \sin\left(\omega t_2 + \frac{\pi}{6}\right) \]

### Step 2: Solve for \( t_1 \) and \( t_2 \)
From the equations derived:

1. For \( t_1 \):
\[ \sin\left(\omega t_1 + \frac{\pi}{6}\right) = -\frac{1}{2} \]
The angle whose sine is \(-\frac{1}{2}\) is:
\[ \omega t_1 + \frac{\pi}{6} = -\frac{\pi}{6} + 2n\pi \quad \text{(for } n \in \mathbb{Z}\text{)} \]
Solving for \( t_1 \):
\[ \omega t_1 = -\frac{\pi}{6} - \frac{\pi}{6} + 2n\pi \]
\[ \omega t_1 = -\frac{\pi}{3} + 2n\pi \]
\[ t_1 = \frac{-\pi/3 + 2n\pi}{\omega} \]

2. For \( t_2 \):
\[ \sin\left(\omega t_2 + \frac{\pi}{6}\right) = \frac{1}{2} \]
The angle whose sine is \(\frac{1}{2}\) is:
\[ \omega t_2 + \frac{\pi}{6} = \frac{\pi}{6} + 2m\pi \quad \text{(for } m \in \mathbb{Z}\text{)} \]
Solving for \( t_2 \):
\[ \omega t_2 = \frac{\pi}{6} - \frac{\pi}{6} + 2m\pi \]
\[ \omega t_2 = 2m\pi \]
\[ t_2 = \frac{2m\pi}{\omega} \]

### Step 3: Calculate the time difference \( t_2 - t_1 \)
Now we need to find the time taken to go from \( t_1 \) to \( t_2 \):
\[ t_2 - t_1 = \frac{2m\pi}{\omega} - \frac{-\frac{\pi}{3} + 2n\pi}{\omega} \]
\[ = \frac{2m\pi + \frac{\pi}{3} - 2n\pi}{\omega} \]

Assuming \( n = 0 \) and \( m = 1 \) (the first positive cycle):
\[ t_2 - t_1 = \frac{2\pi + \frac{\pi}{3}}{\omega} \]
To simplify:
\[ = \frac{\frac{6\pi}{3} + \frac{\pi}{3}}{\omega} = \frac{\frac{7\pi}{3}}{\omega} \]

However, we need to find the time taken to go from \( -\frac{A}{2} \) to \( +\frac{A}{2} \):
Using the angles:
\[ t_2 - t_1 = \frac{\pi}{3\omega} \]

### Final Answer
The time taken by the particle to go directly from \( x = -\frac{A}{2} \) to \( x = +\frac{A}{2} \) is:
\[ \boxed{\frac{\pi}{3\omega}} \]