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

Answer

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

1. \( y_1 = 10^{-6} \sin \left(100t + \frac{x}{50} + 0.5\right) \)
2. \( y_2 = 10^{-6} \cos \left(100t + \frac{x}{50}\right) \)

we will follow these steps:

### Step 1: Identify the phase of each wave
The phase of the first wave \( y_1 \) can be expressed as:
\[ \phi_1 = 100t + \frac{x}{50} + 0.5 \]
The phase of the second wave \( y_2 \) can be expressed as:
\[ \phi_2 = 100t + \frac{x}{50} \]

### Step 2: Convert the cosine function to sine
To compare the two phases directly, we can convert the cosine function in \( y_2 \) to sine. We know that:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_2 \) as:
\[ y_2 = 10^{-6} \sin\left(100t + \frac{x}{50} + \frac{\pi}{2}\right) \]
This means the phase of \( y_2 \) can be expressed as:
\[ \phi_2 = 100t + \frac{x}{50} + \frac{\pi}{2} \]

### Step 3: Calculate the phase difference
Now we can find the phase difference \( \Delta \phi \) between the two waves:
\[ \Delta \phi = \phi_1 - \phi_2 \]
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 4: Calculate the numerical value
Now we need to calculate \( \Delta \phi \):
\[ \Delta \phi = 0.5 - \frac{3.14}{2} = 0.5 - 1.57 = -1.07 \]
Since we are interested in the magnitude of the phase difference, we take the absolute value:
\[ |\Delta \phi| = 1.07 \text{ radians} \]

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

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