If a waveform has the equation `y_1=A_1 sin (omegat-kx)` & `y_2=A_2 cos (omegat-kx)` , find the equation of the resulting wave on superposition.

Answer

To find the equation of the resulting wave from the superposition of the two given waves \( y_1 = A_1 \sin(\omega t - kx) \) and \( y_2 = A_2 \cos(\omega t - kx) \), we can follow these steps:

### Step 1: Rewrite the cosine function in terms of sine
The first step is to express \( y_2 \) in terms of sine. We know that:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_2 \) as:
\[ y_2 = A_2 \cos(\omega t - kx) = A_2 \sin\left(\omega t - kx + \frac{\pi}{2}\right) \]

### Step 2: Set up the superposition of the two waves
Now we have:
\[ y_1 = A_1 \sin(\omega t - kx) \]
\[ y_2 = A_2 \sin\left(\omega t - kx + \frac{\pi}{2}\right) \]
The resultant wave \( y \) from the superposition of \( y_1 \) and \( y_2 \) is given by:
\[ y = y_1 + y_2 = A_1 \sin(\omega t - kx) + A_2 \sin\left(\omega t - kx + \frac{\pi}{2}\right) \]

### Step 3: Use the sine addition formula
Using the sine addition formula, we can express the resultant wave:
\[ y = A_1 \sin(\omega t - kx) + A_2 \left(\sin(\omega t - kx) \cos\left(\frac{\pi}{2}\right) + \cos(\omega t - kx) \sin\left(\frac{\pi}{2}\right)\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \), we have:
\[ y = A_1 \sin(\omega t - kx) + A_2 \cos(\omega t - kx) \]

### Step 4: Find the resultant amplitude and phase
The two waves can be considered as perpendicular components in a right triangle where:
- One side is \( A_1 \) (along the sine direction)
- The other side is \( A_2 \) (along the cosine direction)

The resultant amplitude \( A \) can be calculated using the Pythagorean theorem:
\[ A = \sqrt{A_1^2 + A_2^2} \]

### Step 5: Determine the phase of the resultant wave
The phase \( \phi \) of the resultant wave can be determined using:
\[ \tan(\phi) = \frac{A_2}{A_1} \]

### Final Result
Thus, the equation of the resultant wave can be expressed as:
\[ y = A \sin(\omega t - kx + \phi) \]
where \( A = \sqrt{A_1^2 + A_2^2} \) and \( \phi = \tan^{-1}\left(\frac{A_2}{A_1}\right) \).