A particle is executing SHM according to the equation `x = A cos omega t`. Average speed of the particle during the interval `0 le t le (pi)/(6omega)` is

Options

A. `(sqrt(3)Aomega)/(2)`
B. `(sqrt(3) A omega)/(4)`
C. `(3Aomega)/(pi)`
D. `(3 A omega)/(pi)(2 - sqrt(3))`

Explanation

To find the average speed of a particle executing Simple Harmonic Motion (SHM) described by the equation \( x = A \cos(\omega t) \) over the interval \( 0 \leq t \leq \frac{\pi}{6\omega} \), we can follow these steps:

### Step 1: Determine the displacement at \( t = 0 \)
At \( t = 0 \):
\[ x(0) = A \cos(0) = A \]

### Step 2: Determine the displacement at \( t = \frac{\pi}{6\omega} \)
At \( t = \frac{\pi}{6\omega} \):
\[ x\left(\frac{\pi}{6\omega}\right) = A \cos\left(\omega \cdot \frac{\pi}{6\omega}\right) = A \cos\left(\frac{\pi}{6}\right) = A \cdot \frac{\sqrt{3}}{2} \]

### Step 3: Calculate the total displacement
The total displacement \( \Delta x \) during the time interval from \( t = 0 \) to \( t = \frac{\pi}{6\omega} \) is:
\[ \Delta x = x\left(\frac{\pi}{6\omega}\right) - x(0) = \left(A \cdot \frac{\sqrt{3}}{2}\right) - A = A \left(\frac{\sqrt{3}}{2} - 1\right) \]

### Step 4: Calculate the time interval
The time interval \( \Delta t \) is:
\[ \Delta t = \frac{\pi}{6\omega} - 0 = \frac{\pi}{6\omega} \]

### Step 5: Calculate the average speed
The average speed \( v_{avg} \) is defined as the total displacement divided by the total time:
\[ v_{avg} = \frac{\Delta x}{\Delta t} = \frac{A\left(\frac{\sqrt{3}}{2} - 1\right)}{\frac{\pi}{6\omega}} = \frac{6A\left(\frac{\sqrt{3}}{2} - 1\right)\omega}{\pi} \]

### Step 6: Simplify the expression
To express this in a more simplified form:
\[ v_{avg} = \frac{6A(\sqrt{3} - 2)}{2\pi} \cdot \omega = \frac{3A(\sqrt{3} - 2)\omega}{\pi} \]

Thus, the average speed of the particle during the interval \( 0 \leq t \leq \frac{\pi}{6\omega} \) is:
\[ v_{avg} = \frac{3A(\sqrt{3} - 2)\omega}{\pi} \]