Two waves represented by `y=asin(omegat-kx)` and `y=acos(omegat-kx)` are superposed. The resultant wave will have an amplitude.

Options

A. a
B. `sqrt(2)a`
C. `2a`
D. 0

Answer

To find the amplitude of the resultant wave formed by the superposition of the two given waves \( y = A \sin(\omega t - kx) \) and \( y = A \cos(\omega t - kx) \), we can follow these steps:

### Step 1: Identify the amplitudes of the waves
The first wave is given by:
\[ y_1 = A \sin(\omega t - kx) \]
The amplitude \( A_1 \) of this wave is \( A \).

The second wave is given by:
\[ y_2 = A \cos(\omega t - kx) \]
The amplitude \( A_2 \) of this wave is also \( A \).

### Step 2: Convert the cosine wave to sine form
To compare the two waves, we can express the cosine function in terms of sine. The cosine function can be written as:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite the second wave as:
\[ y_2 = A \cos(\omega t - kx) = A \sin\left(\omega t - kx + \frac{\pi}{2}\right) \]

### Step 3: Determine the phase difference
Now we have:
- \( y_1 = A \sin(\omega t - kx) \)
- \( y_2 = A \sin\left(\omega t - kx + \frac{\pi}{2}\right) \)

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

### Step 4: Use the formula for resultant amplitude
The formula for the resultant amplitude \( A_R \) when two waves are superimposed is given by:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]
Substituting the values we have:
- \( A_1 = A \)
- \( A_2 = A \)
- \( \phi = \frac{\pi}{2} \)

Thus, we get:
\[ A_R = \sqrt{A^2 + A^2 + 2 A \cdot A \cdot \cos\left(\frac{\pi}{2}\right)} \]

### Step 5: Simplify the expression
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), the equation simplifies to:
\[ A_R = \sqrt{A^2 + A^2 + 0} \]
\[ A_R = \sqrt{2A^2} \]
\[ A_R = \sqrt{2} A \]

### Final Answer
The amplitude of the resultant wave is:
\[ A_R = \sqrt{2} A \]