Two simple harmonic motion are represented by equations
`y_(1) = 4 sin (10 t + phi) rArr y_(2) = 5 cos 10t`
What is the phase difference between their velocities ?

Options

A. `phi`
B. `-phi`
C. `(phi+(pi)/(2))`
D. `(phi-(pi)/(2))`

Explanation

To find the phase difference between the velocities of the two simple harmonic motions represented by the equations \( y_1 = 4 \sin(10t + \phi) \) and \( y_2 = 5 \cos(10t) \), we can follow these steps:

### Step 1: Differentiate the Displacement Equations to Find Velocities

1. **For \( y_1 \)**:
\[ y_1 = 4 \sin(10t + \phi) \]
To find the velocity \( v_1 \), we differentiate \( y_1 \) with respect to time \( t \):
\[ v_1 = \frac{dy_1}{dt} = 4 \cdot 10 \cos(10t + \phi) = 40 \cos(10t + \phi) \]

2. **For \( y_2 \)**:
\[ y_2 = 5 \cos(10t) \]
To find the velocity \( v_2 \), we differentiate \( y_2 \) with respect to time \( t \):
\[ v_2 = \frac{dy_2}{dt} = -5 \cdot 10 \sin(10t) = -50 \sin(10t) \]

### Step 2: Rewrite the Velocity Equations in Terms of Cosine

The velocity \( v_2 \) can be rewritten using the sine to cosine conversion:
\[ v_2 = -50 \sin(10t) = -50 \cos\left(10t - \frac{\pi}{2}\right) \]

### Step 3: Identify the Phase of Each Velocity

From the equations obtained:
- The phase of \( v_1 \) is \( 10t + \phi \).
- The phase of \( v_2 \) is \( 10t - \frac{\pi}{2} \).

### Step 4: Calculate the Phase Difference

The phase difference \( \Delta \phi \) between the two velocities can be calculated as:
\[ \Delta \phi = \text{Phase of } v_1 - \text{Phase of } v_2 = (10t + \phi) - \left(10t - \frac{\pi}{2}\right) \]
This simplifies to:
\[ \Delta \phi = \phi + \frac{\pi}{2} \]

### Conclusion

The phase difference between the velocities \( v_1 \) and \( v_2 \) is \( \phi + \frac{\pi}{2} \).