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

Options

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

Answer

To solve the problem of finding the time taken by a particle executing simple harmonic motion (SHM) to go from \( x = -\frac{A}{2} \) to \( x = +\frac{A}{2} \), we will follow these steps:

### Step 1: Set up the displacement equation
The displacement-time equation of the particle is given as:
\[ x = A \sin(\omega t + \frac{\pi}{6}) \]

### Step 2: Find the time \( t_1 \) when \( x = -\frac{A}{2} \)
We set the displacement equal to \(-\frac{A}{2}\):
\[ -\frac{A}{2} = A \sin(\omega t_1 + \frac{\pi}{6}) \]
Dividing both sides by \( A \):
\[ -\frac{1}{2} = \sin(\omega t_1 + \frac{\pi}{6}) \]
The sine function equals \(-\frac{1}{2}\) at angles:
\[ \omega t_1 + \frac{\pi}{6} = -\frac{\pi}{6} + 2n\pi \quad \text{(for any integer } n\text{)} \]
Solving for \( t_1 \):
\[ \omega t_1 = -\frac{\pi}{6} - \frac{\pi}{6} + 2n\pi \]
\[ \omega t_1 = -\frac{\pi}{3} + 2n\pi \]
Taking \( n = 0 \):
\[ t_1 = -\frac{\pi}{3\omega} \]

### Step 3: Find the time \( t_2 \) when \( x = +\frac{A}{2} \)
Now we set the displacement equal to \(+\frac{A}{2}\):
\[ \frac{A}{2} = A \sin(\omega t_2 + \frac{\pi}{6}) \]
Dividing both sides by \( A \):
\[ \frac{1}{2} = \sin(\omega t_2 + \frac{\pi}{6}) \]
The sine function equals \(\frac{1}{2}\) at angles:
\[ \omega t_2 + \frac{\pi}{6} = \frac{\pi}{6} + 2m\pi \quad \text{(for any integer } m\text{)} \]
Solving for \( t_2 \):
\[ \omega t_2 = \frac{\pi}{6} - \frac{\pi}{6} + 2m\pi \]
\[ \omega t_2 = 2m\pi \]
Taking \( m = 0 \):
\[ t_2 = 0 \]

### Step 4: Calculate the time taken to go from \( x = -\frac{A}{2} \) to \( x = +\frac{A}{2} \)
The time taken \( \Delta t \) is given by:
\[ \Delta t = t_2 - t_1 \]
Substituting the values we found:
\[ \Delta t = 0 - \left(-\frac{\pi}{3\omega}\right) \]
\[ \Delta t = \frac{\pi}{3\omega} \]

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

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