Two particles are executing identical simple harmonic motions described by the equations `x_1=acos(omegat+(pi)/(6))` and `x_2=acos(omegat+(pi)/(3))`. The minimum interval of time between the particles crossing the respective mean positions is `(pi)/(2omega)`

Options

A. `(pi)/(2omega)`
B. `(pi)/(3omega)`
C. `(pi)/(4omega)`
D. `(pi)/(6omega)`

Explanation

To solve the problem, we need to find the minimum interval of time between the two particles crossing their respective mean positions. The equations of motion for the two particles are given as:

1. \( x_1 = a \cos(\omega t + \frac{\pi}{6}) \)
2. \( x_2 = a \cos(\omega t + \frac{\pi}{3}) \)

### Step 1: Determine the time when \( x_1 \) crosses the mean position
To find when \( x_1 \) crosses the mean position (where \( x_1 = 0 \)), we set the equation to zero:

\[ 0 = a \cos(\omega t_1 + \frac{\pi}{6}) \]

This implies:

\[ \cos(\omega t_1 + \frac{\pi}{6}) = 0 \]

The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):

\[ \omega t_1 + \frac{\pi}{6} = \frac{(2n + 1)\pi}{2} \]

For the minimum time, we take \( n = 0 \):

\[ \omega t_1 + \frac{\pi}{6} = \frac{\pi}{2} \]

Solving for \( t_1 \):

\[ \omega t_1 = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

Thus,

\[ t_1 = \frac{\pi}{3\omega} \]

### Step 2: Determine the time when \( x_2 \) crosses the mean position
Next, we find when \( x_2 \) crosses the mean position (where \( x_2 = 0 \)):

\[ 0 = a \cos(\omega t_2 + \frac{\pi}{3}) \]

This implies:

\[ \cos(\omega t_2 + \frac{\pi}{3}) = 0 \]

Using the same logic as before:

\[ \omega t_2 + \frac{\pi}{3} = \frac{(2m + 1)\pi}{2} \]

For the minimum time, we take \( m = 0 \):

\[ \omega t_2 + \frac{\pi}{3} = \frac{\pi}{2} \]

Solving for \( t_2 \):

\[ \omega t_2 = \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi}{6} - \frac{2\pi}{6} = \frac{\pi}{6} \]

Thus,

\[ t_2 = \frac{\pi}{6\omega} \]

### Step 3: Calculate the minimum interval of time between the two crossings
Now, we find the interval of time between the two particles crossing their mean positions:

\[ \Delta t = t_1 - t_2 = \frac{\pi}{3\omega} - \frac{\pi}{6\omega} \]

Finding a common denominator:

\[ \Delta t = \frac{2\pi}{6\omega} - \frac{\pi}{6\omega} = \frac{2\pi - \pi}{6\omega} = \frac{\pi}{6\omega} \]

### Final Result
The minimum interval of time between the particles crossing their respective mean positions is:

\[ \Delta t = \frac{\pi}{6\omega} \]