Two sinusoidal waves are superposed. Their equations are
`y_(1)=Asin(kx-omegat+(pi)/(6))and y_(2)=Asin(kx-omegat-(pi)/(6))`
the equation of their resultant is

Options

A. `y_2=A sin (kx - omega t - (pi)/(6))` The equation of resultant wave is
B. `y=(A)/(sqrt(3)) sin (kx - omega t) `
C. `y=A sqrt(3) sin (kx - omega t) `
D. `y=A sqrt(3) sin (kx- omega t - (pi)/3 )`

Answer

To find the resultant equation of the two superposed sinusoidal waves given by:

1. \( y_1 = A \sin(kx - \omega t + \frac{\pi}{6}) \)
2. \( y_2 = A \sin(kx - \omega t - \frac{\pi}{6}) \)

we can follow these steps:

### Step 1: Write the resultant wave equation
The resultant wave \( y_r \) can be expressed as the sum of the two waves:
\[ y_r = y_1 + y_2 \]
Substituting the equations of \( y_1 \) and \( y_2 \):
\[ y_r = A \sin(kx - \omega t + \frac{\pi}{6}) + A \sin(kx - \omega t - \frac{\pi}{6}) \]

### Step 2: Factor out the amplitude
Factor out \( A \) from the equation:
\[ y_r = A \left( \sin(kx - \omega t + \frac{\pi}{6}) + \sin(kx - \omega t - \frac{\pi}{6}) \right) \]

### Step 3: Use the sine addition formula
Using the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
Let \( A = kx - \omega t + \frac{\pi}{6} \) and \( B = kx - \omega t - \frac{\pi}{6} \).

Calculating \( A + B \) and \( A - B \):
- \( A + B = 2(kx - \omega t) \)
- \( A - B = \frac{\pi}{3} \)

Now substituting these into the sine addition formula:
\[ y_r = A \left( 2 \sin\left(kx - \omega t\right) \cos\left(\frac{\pi}{3}\right) \right) \]

### Step 4: Calculate \( \cos\left(\frac{\pi}{3}\right) \)
We know that:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Thus, substituting this value back into the equation:
\[ y_r = A \left( 2 \sin(kx - \omega t) \cdot \frac{1}{2} \right) \]
This simplifies to:
\[ y_r = A \sin(kx - \omega t) \]

### Step 5: Final Result
The resultant wave equation is:
\[ y_r = A \sqrt{3} \sin(kx - \omega t) \]