Two waves are represented by the equations
`y_(1)=a sin (omega t+kx+0.785)`
and `y_(2)=a cos (omega t+kx)`
where, x is in meter and t in second
The phase difference between them and resultant amplitude due to their superposition are

Options

A. 45∘ and 1.84 a
B. 30∘ and a
C. 30∘ and 2a
D. 45∘ and a

Answer

To solve the problem, we need to find the phase difference between the two waves and the resultant amplitude due to their superposition. Let's break down the solution step by step.

### Step 1: Identify the wave equations
The two wave equations given are:
1. \( y_1 = a \sin(\omega t + kx + 0.785) \)
2. \( y_2 = a \cos(\omega t + kx) \)

### Step 2: Convert the second wave equation to sine form
We know that the cosine function can be expressed in terms of the sine function:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_2 \) as:
\[ y_2 = a \sin\left(\omega t + kx + \frac{\pi}{2}\right) \]

### Step 3: Identify the phase of each wave
From the equations:
- The phase of \( y_1 \) is \( \phi_1 = \omega t + kx + 0.785 \)
- The phase of \( y_2 \) is \( \phi_2 = \omega t + kx + \frac{\pi}{2} \)

### 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 phases:
\[ \Delta \phi = \left(\omega t + kx + \frac{\pi}{2}\right) - \left(\omega t + kx + 0.785\right) \]
This simplifies to:
\[ \Delta \phi = \frac{\pi}{2} - 0.785 \]

### Step 5: Convert 0.785 to radians
We know that \( 0.785 \) radians is equivalent to \( \frac{\pi}{4} \). Thus:
\[ \Delta \phi = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \text{ radians} \]

### Step 6: Convert phase difference to degrees
To convert radians to degrees:
\[ \Delta \phi = \frac{\pi}{4} \times \frac{180}{\pi} = 45^\circ \]

### Step 7: Calculate the resultant amplitude
The amplitudes of both waves are equal:
- \( A_1 = a \)
- \( A_2 = a \)

The formula for the resultant amplitude \( A_R \) when two waves of equal amplitude interfere is:
\[ A_R = 2A \cos\left(\frac{\Delta \phi}{2}\right) \]
Substituting the values:
\[ A_R = 2a \cos\left(\frac{\pi/4}{2}\right) = 2a \cos\left(\frac{\pi}{8}\right) \]

### Step 8: Calculate \( \cos\left(\frac{\pi}{8}\right) \)
Using a calculator or trigonometric tables, we find:
\[ \cos\left(\frac{\pi}{8}\right) \approx 0.9239 \]
Thus:
\[ A_R \approx 2a \times 0.9239 \approx 1.8478a \approx 1.84a \]

### Final Answer
The phase difference between the waves is \( \frac{\pi}{4} \) radians (or \( 45^\circ \)), and the resultant amplitude due to their superposition is approximately \( 1.84a \).