The two waves are represented by
`y_(1)= 10^(-6) sin(100t + (x)/(50)+ 0.5)m`
`Y_(2) =10^(-2) cos(100t + (x)/(50))m`
where x is ihn metres and t in seconds. The phase difference between the waves is approximately:

Options

A. 1.07 rad
B. 2.07 rad
C. 0.5 rad
D. 1.5 rad

Answer

To find the phase difference between the two waves given by the equations:

1. **Identify the wave equations**:
- Wave 1: \( y_1 = 10^{-6} \sin(100t + \frac{x}{50} + 0.5) \)
- Wave 2: \( y_2 = 10^{-2} \cos(100t + \frac{x}{50}) \)

2. **Convert the second wave to sine form**:
- We know that \( \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \). Therefore, we can rewrite \( y_2 \) as:
\[ y_2 = 10^{-2} \sin\left(100t + \frac{x}{50} + \frac{\pi}{2}\right) \]

3. **Identify the phase of each wave**:
- For wave 1, the phase \( \phi_1 \) is:
\[ \phi_1 = 100t + \frac{x}{50} + 0.5 \]
- For wave 2, the phase \( \phi_2 \) is:
\[ \phi_2 = 100t + \frac{x}{50} + \frac{\pi}{2} \]

4. **Calculate the phase difference**:
- The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_1 - \phi_2 \]
- Substituting the expressions for \( \phi_1 \) and \( \phi_2 \):
\[ \Delta \phi = \left(100t + \frac{x}{50} + 0.5\right) - \left(100t + \frac{x}{50} + \frac{\pi}{2}\right) \]
- Simplifying this gives:
\[ \Delta \phi = 0.5 - \frac{\pi}{2} \]

5. **Calculate the numerical value**:
- We know \( \frac{\pi}{2} \approx 1.57 \):
\[ \Delta \phi \approx 0.5 - 1.57 = -1.07 \]
- Taking the magnitude of the phase difference:
\[ |\Delta \phi| \approx 1.07 \]

6. **Final answer**:
- The phase difference between the two waves is approximately \( 1.07 \) radians.