Two waves having equaitons
`x_1=asin(omegat+phi_1)`,`x=asin(omega+phi_2)`
If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is

Options

A. `pi/6`
B. `(2pi)/3`
C. `pi/4`
D. `pi/3`

Answer

To solve the problem step by step, we will analyze the given wave equations and apply the principles of wave superposition.

### Step 1: Write down the wave equations
The two waves are given by:
- \( x_1 = A \sin(\omega t + \phi_1) \)
- \( x_2 = A \sin(\omega t + \phi_2) \)

### Step 2: Write the resultant wave equation
When two waves superimpose, the resultant displacement \( x \) can be expressed as:
\[ x = x_1 + x_2 = A \sin(\omega t + \phi_1) + A \sin(\omega t + \phi_2) \]

### Step 3: Use the trigonometric identity for sine addition
We can use the trigonometric identity:
\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
Let \( A = \omega t + \phi_1 \) and \( B = \omega t + \phi_2 \). Thus, we can rewrite the resultant wave:
\[ x = 2A \sin\left(\frac{(\omega t + \phi_1) + (\omega t + \phi_2)}{2}\right) \cos\left(\frac{(\omega t + \phi_1) - (\omega t + \phi_2)}{2}\right) \]
This simplifies to:
\[ x = 2A \sin\left(\omega t + \frac{\phi_1 + \phi_2}{2}\right) \cos\left(\frac{\phi_1 - \phi_2}{2}\right) \]

### Step 4: Identify the amplitude of the resultant wave
From the resultant wave equation, the amplitude \( A' \) of the resultant wave is given by:
\[ A' = 2A \cos\left(\frac{\phi_1 - \phi_2}{2}\right) \]

### Step 5: Set the condition for amplitude
According to the problem, the amplitude of the resultant wave remains equal to that of the individual waves:
\[ A' = A \]
Thus, we have:
\[ 2A \cos\left(\frac{\phi_1 - \phi_2}{2}\right) = A \]

### Step 6: Simplify the equation
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ 2 \cos\left(\frac{\phi_1 - \phi_2}{2}\right) = 1 \]

### Step 7: Solve for the phase difference
Now, we can solve for \( \cos\left(\frac{\phi_1 - \phi_2}{2}\right) \):
\[ \cos\left(\frac{\phi_1 - \phi_2}{2}\right) = \frac{1}{2} \]
The angle whose cosine is \( \frac{1}{2} \) is:
\[ \frac{\phi_1 - \phi_2}{2} = \frac{\pi}{3} \]
Thus, multiplying by 2 gives:
\[ \phi_1 - \phi_2 = \frac{2\pi}{3} \]

### Conclusion
The phase difference \( \Delta \phi \) between the two waves is:
\[ \Delta \phi = \frac{2\pi}{3} \]