On the superposition of the two waves given as `y_1=A_0 sin( omegat-kx)` and `y_2=A_0 cos ( omega t -kx+(pi)/6)` , the resultant amplitude of oscillations will be

Options

A. `sqrt(3)A_0`
B. `(A_0)/(2) `
C. `A_0`
D. `(3)/(2)A_0`

Answer

To find the resultant amplitude of the two waves given by the equations \( y_1 = A_0 \sin(\omega t - kx) \) and \( y_2 = A_0 \cos(\omega t - kx + \frac{\pi}{6}) \), we can follow these steps:

### Step 1: Identify the amplitudes
The amplitude of the first wave \( y_1 \) is:
\[ A_1 = A_0 \]
The amplitude of the second wave \( y_2 \) is:
\[ A_2 = A_0 \]

### Step 2: Determine the phase difference
The second wave can be rewritten in terms of sine. We know that:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can express \( y_2 \) as:
\[ y_2 = A_0 \sin\left(\omega t - kx + \frac{\pi}{6} + \frac{\pi}{2}\right) = A_0 \sin\left(\omega t - kx + \frac{2\pi}{3}\right) \]
This means the phase difference \( \phi \) between \( y_1 \) and \( y_2 \) is:
\[ \phi = \frac{2\pi}{3} \]

### Step 3: Use the formula for resultant amplitude
The formula for the resultant amplitude \( R \) when two waves superpose is given by:
\[ R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]
Substituting the values:
\[ R = \sqrt{A_0^2 + A_0^2 + 2 A_0 A_0 \cos\left(\frac{2\pi}{3}\right)} \]

### Step 4: Calculate \( \cos\left(\frac{2\pi}{3}\right) \)
We know that:
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]
Now substituting this value into the equation for \( R \):
\[ R = \sqrt{A_0^2 + A_0^2 + 2 A_0^2 \left(-\frac{1}{2}\right)} \]
This simplifies to:
\[ R = \sqrt{A_0^2 + A_0^2 - A_0^2} = \sqrt{A_0^2} = A_0 \]

### Conclusion
The resultant amplitude of the oscillations is:
\[ R = A_0 \]