Two waves are represented by the equations `y_1=asin(omegat=kx+0.57)m` and `y_2=acos(omegat+kx)`m where x is in metre and t in second. The phase difference between them is

Options

A. `1.25 rad `
B. `1.57 rad`
C. `0.57 rad`
D. `1.0 rad`

Explanation

To find the phase difference between the two waves represented by the equations \( y_1 = a \sin(\omega t - kx + 0.57) \) and \( y_2 = a \cos(\omega t + kx) \), we can follow these steps:

### Step 1: Rewrite the second wave in terms of sine
The cosine function can be rewritten in terms of sine:
\[ y_2 = a \cos(\omega t + kx) = a \sin\left(\omega t + kx + \frac{\pi}{2}\right) \]
This is because \( \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \).

### Step 2: Identify the phase of each wave
Now we can identify the phases of both waves:
- For \( y_1 \):
\[ \text{Phase of } y_1 = \omega t - kx + 0.57 \]
- For \( y_2 \):
\[ \text{Phase of } y_2 = \omega t + kx + \frac{\pi}{2} \]

### Step 3: Extract the phase constants
From the equations:
- The phase constant for \( y_1 \) is \( \phi_1 = 0.57 \).
- The phase constant for \( y_2 \) is \( \phi_2 = \frac{\pi}{2} \).

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two waves is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the values we found:
\[ \Delta \phi = \frac{\pi}{2} - 0.57 \]

### Step 5: Convert \( \frac{\pi}{2} \) to a decimal
Using the approximation \( \pi \approx 3.14 \):
\[ \frac{\pi}{2} \approx \frac{3.14}{2} \approx 1.57 \]
Thus:
\[ \Delta \phi \approx 1.57 - 0.57 = 1.00 \]

### Final Answer
The phase difference between the two waves is approximately:
\[ \Delta \phi \approx 1.00 \text{ radians} \]