The displacement of two particles executing SHM are represented by equations, `y_(1)=2 sin (10 t + theta), y_(2)=3 cos 10 t`. The phase difference between the velocity of these particles is

Options

A. `theta`
B. `-theta`
C. `theta + pi//2`
D. `theta - pi//2 `

Explanation

To find the phase difference between the velocities of the two particles executing simple harmonic motion (SHM), we will follow these steps:

### Step 1: Write down the displacement equations
The displacement equations for the two particles are given as:
1. \( y_1 = 2 \sin(10t + \theta) \)
2. \( y_2 = 3 \cos(10t) \)

### Step 2: Find the velocity equations
The velocity of a particle in SHM can be found by differentiating the displacement with respect to time \( t \).

For particle 1:
\[ v_1 = \frac{dy_1}{dt} = \frac{d}{dt}(2 \sin(10t + \theta)) = 2 \cdot 10 \cos(10t + \theta) = 20 \cos(10t + \theta) \]

For particle 2:
\[ v_2 = \frac{dy_2}{dt} = \frac{d}{dt}(3 \cos(10t)) = 3 \cdot (-10 \sin(10t)) = -30 \sin(10t) \]

### Step 3: Express the velocities in terms of sine
We can express \( v_2 \) in terms of cosine to find the phase difference more easily. Recall that:
\[ \sin(10t) = \cos\left(10t - \frac{\pi}{2}\right) \]
Thus,
\[ v_2 = -30 \sin(10t) = -30 \cos\left(10t - \frac{\pi}{2}\right) \]

### Step 4: Identify the phase angles
Now we have:
- For \( v_1 \): The phase angle is \( 10t + \theta \)
- For \( v_2 \): The phase angle is \( 10t - \frac{\pi}{2} \)

### Step 5: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two velocities is given by:
\[ \Delta \phi = \text{Phase of } v_1 - \text{Phase of } v_2 \]
Substituting the phase angles:
\[ \Delta \phi = (10t + \theta) - \left(10t - \frac{\pi}{2}\right) = \theta + \frac{\pi}{2} \]

### Final Result
Thus, the phase difference between the velocities of the two particles is:
\[ \Delta \phi = \theta + \frac{\pi}{2} \]

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