The phase difference between two waves represented by `y_(1)=10 ^(-6) sin [100 t+(x//50)+0.5]m, y_(2)=10^(-6) cos[100t+(x//50)]m`where x is expressed in metres and t is expressed in seconds, is approximately

Options

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

Answer

To find the phase difference between the two waves represented by the equations \( y_1 = 10^{-6} \sin\left(100t + \frac{x}{50} + 0.5\right) \) and \( y_2 = 10^{-6} \cos\left(100t + \frac{x}{50}\right) \), we will follow these steps:

### Step 1: Identify the phase of each wave
The phase of the first wave \( y_1 \) is given by:
\[ \phi_1 = 100t + \frac{x}{50} + 0.5 \]
The phase of the second wave \( y_2 \) can be rewritten in terms of sine using the identity \( \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \):
\[ y_2 = 10^{-6} \cos\left(100t + \frac{x}{50}\right) = 10^{-6} \sin\left(100t + \frac{x}{50} + \frac{\pi}{2}\right) \]
Thus, the phase of the second wave is:
\[ \phi_2 = 100t + \frac{x}{50} + \frac{\pi}{2} \]

### Step 2: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two waves is given by:
\[ \Delta \phi = \phi_1 - \phi_2 \]
Substituting the expressions for \( \phi_1 \) and \( \phi_2 \):
\[ \Delta \phi = \left(100t + \frac{x}{50} + 0.5\right) - \left(100t + \frac{x}{50} + \frac{\pi}{2}\right) \]
This simplifies to:
\[ \Delta \phi = 0.5 - \frac{\pi}{2} \]

### Step 3: Convert \( \frac{\pi}{2} \) to decimal
We know that \( \frac{\pi}{2} \approx 1.57 \). Therefore, we can compute:
\[ \Delta \phi \approx 0.5 - 1.57 = -1.07 \]
Taking the absolute value gives:
\[ |\Delta \phi| \approx 1.07 \text{ radians} \]

### Final Answer
The phase difference between the two waves is approximately \( 1.07 \) radians.

---