Three travelling waves are superimposed. The equations of the wave are
`y_(!)=A_(0) sin (kx-omegat), y_(2)=3 sqrt(2) A_(0) sin (kx-omegat+phi)` and `y_(3)=4 A_(0) cos (kx-omegat)`
find the value of `phi ("given "0 le phi le pi//2)` if the phase difference between the resultant wave and first wave is `pi//4`

Options

A. `pi/6`
B. `pi/3`
C. `pi/12`
D. None of these

Explanation

To solve the problem, we need to find the value of \(\phi\) given the equations of three waves and the phase difference between the resultant wave and the first wave. Let's break down the solution step by step.

### Step 1: Write the equations of the waves
The equations of the three waves are given as:
1. \(y_1 = A_0 \sin(kx - \omega t)\)
2. \(y_2 = 3\sqrt{2} A_0 \sin(kx - \omega t + \phi)\)
3. \(y_3 = 4 A_0 \cos(kx - \omega t)\)

### Step 2: Convert \(y_3\) to sine form
To combine the waves easily, we convert \(y_3\) into sine form using the identity \(\cos(\theta) = \sin(\theta + \frac{\pi}{2})\):
\[ y_3 = 4 A_0 \cos(kx - \omega t) = 4 A_0 \sin\left(kx - \omega t + \frac{\pi}{2}\right) \]

### Step 3: Draw the phasor diagram
We will represent each wave as a phasor:
- \(y_1\) has an amplitude of \(A_0\) along the x-axis.
- \(y_2\) has an amplitude of \(3\sqrt{2} A_0\) at an angle \(\phi\).
- \(y_3\) has an amplitude of \(4 A_0\) at an angle of \(\frac{\pi}{2}\) (90 degrees).

### Step 4: Resolve \(y_2\) into components
To find the resultant wave, we need to resolve \(y_2\) into its x and y components:
- The x-component of \(y_2\) is \(3\sqrt{2} A_0 \cos(\phi)\)
- The y-component of \(y_2\) is \(3\sqrt{2} A_0 \sin(\phi)\)

### Step 5: Calculate the total x and y components
Now we can calculate the total x and y components of the resultant wave:
- Total x-component \(R_x = A_0 + 3\sqrt{2} A_0 \cos(\phi)\)
- Total y-component \(R_y = 4 A_0 + 3\sqrt{2} A_0 \sin(\phi)\)

### Step 6: Find the phase difference
The phase difference between the resultant wave and the first wave is given as \(\frac{\pi}{4}\). This means:
\[ \tan\left(\frac{\pi}{4}\right) = \frac{R_y}{R_x} \]
Since \(\tan\left(\frac{\pi}{4}\right) = 1\), we have:
\[ R_y = R_x \]

### Step 7: Set up the equation
Substituting the expressions for \(R_x\) and \(R_y\):
\[ 4 A_0 + 3\sqrt{2} A_0 \sin(\phi) = A_0 + 3\sqrt{2} A_0 \cos(\phi) \]

### Step 8: Simplify the equation
Dividing through by \(A_0\) (assuming \(A_0 \neq 0\)):
\[ 4 + 3\sqrt{2} \sin(\phi) = 1 + 3\sqrt{2} \cos(\phi) \]
Rearranging gives:
\[ 3\sqrt{2} \sin(\phi) - 3\sqrt{2} \cos(\phi) = -3 \]
Factoring out \(3\sqrt{2}\):
\[ 3\sqrt{2} (\sin(\phi) - \cos(\phi)) = -3 \]
Dividing both sides by \(3\sqrt{2}\):
\[ \sin(\phi) - \cos(\phi) = -\frac{1}{\sqrt{2}} \]

### Step 9: Square both sides
Squaring both sides to eliminate the sine and cosine:
\[ (\sin(\phi) - \cos(\phi))^2 = \left(-\frac{1}{\sqrt{2}}\right)^2 \]
Expanding the left side:
\[ \sin^2(\phi) - 2\sin(\phi)\cos(\phi) + \cos^2(\phi) = \frac{1}{2} \]
Using \(\sin^2(\phi) + \cos^2(\phi) = 1\):
\[ 1 - 2\sin(\phi)\cos(\phi) = \frac{1}{2} \]
Thus:
\[ -2\sin(\phi)\cos(\phi) = -\frac{1}{2} \]
This simplifies to:
\[ \sin(2\phi) = \frac{1}{2} \]

### Step 10: Solve for \(\phi\)
The general solution for \(\sin(2\phi) = \frac{1}{2}\) gives:
\[ 2\phi = \frac{\pi}{6} \quad \text{or} \quad 2\phi = \frac{5\pi}{6} \]
Thus:
\[ \phi = \frac{\pi}{12} \quad \text{or} \quad \phi = \frac{5\pi}{12} \]
Since we are given that \(0 \leq \phi \leq \frac{\pi}{2}\), we take:
\[ \phi = \frac{\pi}{12} \]

### Final Answer
\(\phi = \frac{\pi}{12}\)

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