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

### Step 1: Convert \( y_1 \) to cosine form
We can express \( y_1 \) in terms of cosine:
\[ y_1 = A_0 \sin(\omega t - kx) = A_0 \cos\left(\frac{\pi}{2} - (\omega t - kx)\right) = A_0 \cos\left(\omega t - kx + \frac{\pi}{2}\right) \]
This gives us:
\[ y_1 = A_0 \cos\left(\omega t - kx + \frac{\pi}{2}\right) \]

### Step 2: Identify the phases
Now we can identify the phases of the two waves:
- For \( y_1 \), the phase \( \phi_1 = \omega t - kx + \frac{\pi}{2} \)
- For \( y_2 \), the phase \( \phi_2 = \omega t - kx + \frac{\pi}{6} \)

### Step 3: Calculate the phase difference
Next, we calculate the phase difference \( \Delta \phi = \phi_2 - \phi_1 \):
\[ \Delta \phi = \left(\omega t - kx + \frac{\pi}{6}\right) - \left(\omega t - kx + \frac{\pi}{2}\right) = \frac{\pi}{6} - \frac{\pi}{2} \]
Converting \( \frac{\pi}{2} \) to sixths:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]
Thus,
\[ \Delta \phi = \frac{\pi}{6} - \frac{3\pi}{6} = -\frac{2\pi}{6} = -\frac{\pi}{3} \]

### Step 4: Calculate the resultant amplitude
Using the formula for the resultant amplitude of two waves with equal amplitudes \( A_0 \) and a phase difference \( \Delta \phi \):
\[ A_{resultant} = \sqrt{A_0^2 + A_0^2 + 2A_0A_0 \cos(\Delta \phi)} \]
Substituting \( \Delta \phi = -\frac{\pi}{3} \):
\[ A_{resultant} = \sqrt{A_0^2 + A_0^2 + 2A_0^2 \cos\left(-\frac{\pi}{3}\right)} \]
Since \( \cos(-\theta) = \cos(\theta) \):
\[ \cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Thus,
\[ A_{resultant} = \sqrt{A_0^2 + A_0^2 + 2A_0^2 \cdot \frac{1}{2}} = \sqrt{A_0^2 + A_0^2 + A_0^2} = \sqrt{3A_0^2} = A_0 \sqrt{3} \]

### Final Result
The resultant amplitude of oscillations is:
\[ A_{resultant} = A_0 \sqrt{3} \]