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 two waves given by the equations \( y_1 = A_1 \sin(\omega t - kx) \) and \( y_2 = A_2 \cos(\omega t - kx) \), we can follow these steps:

### Step 1: Rewrite the cosine wave
We start with the second wave \( y_2 = A_2 \cos(\omega t - kx) \). We can express the cosine function in terms of sine:
\[ y_2 = A_2 \cos(\omega t - kx) = A_2 \sin\left(\omega t - kx + \frac{\pi}{2}\right) \]
This transformation allows us to express both waves in terms of sine.

### Step 2: Write the superposition of the two waves
Now we can write the superposition of the two waves:
\[ y = y_1 + y_2 = A_1 \sin(\omega t - kx) + A_2 \sin\left(\omega t - kx + \frac{\pi}{2}\right) \]

### Step 3: Use the sine addition formula
Using the sine addition formula, we can combine these two sine functions. The resulting wave can be expressed in the form:
\[ y = A_r \sin\left(\omega t - kx + \phi\right) \]
where \( A_r \) is the resultant amplitude and \( \phi \) is the phase constant.

### Step 4: Calculate the resultant amplitude
To find \( A_r \), we can use the Pythagorean theorem since the two amplitudes can be considered as perpendicular components:
\[ A_r = \sqrt{A_1^2 + A_2^2} \]

### Step 5: Calculate the phase constant
The phase constant \( \phi \) can be found using the tangent function:
\[ \tan(\phi) = \frac{A_2}{A_1} \]
Thus, we can express \( \phi \) as:
\[ \phi = \tan^{-1}\left(\frac{A_2}{A_1}\right) \]

### Step 6: Write the final equation of the resulting wave
Now, we can write the final equation for the resulting wave:
\[ y = \sqrt{A_1^2 + A_2^2} \sin\left(\omega t - kx + \tan^{-1}\left(\frac{A_2}{A_1}\right)\right) \]

### Final Result
The equation of the resulting wave on superposition is:
\[ y = \sqrt{A_1^2 + A_2^2} \sin\left(\omega t - kx + \tan^{-1}\left(\frac{A_2}{A_1}\right)\right) \]

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