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 metres and t is in seconds. The phase difference between them is

Options

A. 1.0 radian
B. 1.25 radian
C. 1.57 radian
D. 0.57 radian

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: Identify the phase of each wave
The phase of the first wave \( y_1 \) can be expressed as:
\[ \phi_1 = \omega t + kx + 0.57 \]

The phase of the second wave \( y_2 \) can be expressed as:
\[ \phi_2 = \omega t + kx \]

### Step 2: Convert the cosine function to sine
To compare the phases directly, we can convert the cosine function in \( y_2 \) to a sine function. We know that:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_2 \) as:
\[ y_2 = a \sin\left(\omega t + kx + \frac{\pi}{2}\right) \]
This means the phase for \( y_2 \) is:
\[ \phi_2 = \omega t + kx + \frac{\pi}{2} \]

### Step 3: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two waves is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the expressions for \( \phi_1 \) and \( \phi_2 \):
\[ \Delta \phi = \left(\omega t + kx + \frac{\pi}{2}\right) - \left(\omega t + kx + 0.57\right) \]

### Step 4: Simplify the expression
When we simplify this, we find:
\[ \Delta \phi = \frac{\pi}{2} - 0.57 \]

### Step 5: Calculate the numerical value
Using the approximation \( \pi \approx 3.14 \):
\[ \frac{\pi}{2} \approx 1.57 \]
Thus:
\[ \Delta \phi \approx 1.57 - 0.57 = 1.00 \text{ radians} \]

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

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