Two simple harmonic are represented by the equation `y_(1)=0.1 sin (100pi+(pi)/3) and y_(2)=0.1 cos pit`.
The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is.

Options

A. `-pi/6`
B. `pi/3`
C. `-pi/3`
D. `pi/6`

Explanation

To find the phase difference of the velocity of particle 1 with respect to the velocity of particle 2, we will follow these steps:

### Step 1: Write down the equations of motion
The equations of motion for the two particles are given as:
- \( y_1 = 0.1 \sin(100\pi t + \frac{\pi}{3}) \)
- \( y_2 = 0.1 \cos(\pi t) \)

### Step 2: Differentiate to find the velocities
The velocity \( v \) is given by the time derivative of the displacement \( y \):
- For particle 1:
\[ v_1 = \frac{dy_1}{dt} = 0.1 \cdot 100\pi \cos(100\pi t + \frac{\pi}{3}) = 10\pi \cos(100\pi t + \frac{\pi}{3}) \]

- For particle 2:
\[ v_2 = \frac{dy_2}{dt} = 0.1 \cdot (-\sin(\pi t)) = -0.1\sin(\pi t) \]

### Step 3: Convert \( v_2 \) to a cosine function
To find the phase difference, it is easier to express \( v_2 \) in terms of cosine. We can use the identity:
\[ \sin(\theta) = \cos\left(\theta - \frac{\pi}{2}\right) \]
Thus, we can write:
\[ v_2 = -0.1 \sin(\pi t) = 0.1 \cos\left(\pi t + \frac{\pi}{2}\right) \]

### Step 4: Identify the phase angles
Now we can identify the phase angles:
- For \( v_1 \): The phase angle is \( \phi_1 = 100\pi t + \frac{\pi}{3} \)
- For \( v_2 \): The phase angle is \( \phi_2 = \pi t + \frac{\pi}{2} \)

### Step 5: Find the phase difference
The phase difference of the velocities \( v_1 \) with respect to \( v_2 \) is given by:
\[ \Delta \phi = \phi_1 - \phi_2 \]

### Step 6: Substitute the phase angles
Substituting the phase angles into the equation:
\[ \Delta \phi = (100\pi t + \frac{\pi}{3}) - (\pi t + \frac{\pi}{2}) \]
\[ = (100\pi t - \pi t) + \left(\frac{\pi}{3} - \frac{\pi}{2}\right) \]
\[ = 99\pi t + \left(\frac{2\pi}{6} - \frac{3\pi}{6}\right) \]
\[ = 99\pi t - \frac{\pi}{6} \]

### Step 7: Determine the phase difference
Since we are interested in the phase difference independent of time, we can focus on the constant term:
\[ \Delta \phi = -\frac{\pi}{6} \]

### Final Answer
Thus, the phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is:
\[ \Delta \phi = -\frac{\pi}{6} \]