If the two waves represented dy `y_(1)=4cos omegat` and `y_(2)=3 cos(omegat+pi//3)` interfere at a point, then the amplitude of the resulting wave will be about

Options

A. 7
B. 5
C. 6
D. `3.5`

Explanation

To find the amplitude of the resulting wave when two waves interfere, we can use the formula for the resultant amplitude \( R_a \) of two waves with amplitudes \( A_1 \) and \( A_2 \) and a phase difference \( \phi \):

\[ R_a = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]

### Step 1: Identify the amplitudes and phase difference
From the given equations:
- \( y_1 = 4 \cos(\omega t) \) implies \( A_1 = 4 \)
- \( y_2 = 3 \cos(\omega t + \frac{\pi}{3}) \) implies \( A_2 = 3 \)
- The phase difference \( \phi = \frac{\pi}{3} \)

### Step 2: Substitute the values into the formula
Now we substitute \( A_1 \), \( A_2 \), and \( \phi \) into the formula:

\[ R_a = \sqrt{4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cdot \cos\left(\frac{\pi}{3}\right)} \]

### Step 3: Calculate the squares of the amplitudes
Calculate \( 4^2 \) and \( 3^2 \):

\[ 4^2 = 16 \quad \text{and} \quad 3^2 = 9 \]

### Step 4: Calculate \( \cos\left(\frac{\pi}{3}\right) \)
We know that:

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 5: Substitute and simplify
Now substitute back into the equation:

\[ R_a = \sqrt{16 + 9 + 2 \cdot 4 \cdot 3 \cdot \frac{1}{2}} \]

Calculating the product:

\[ 2 \cdot 4 \cdot 3 \cdot \frac{1}{2} = 12 \]

So we have:

\[ R_a = \sqrt{16 + 9 + 12} \]

### Step 6: Final calculation
Now sum the values:

\[ R_a = \sqrt{37} \]

### Step 7: Approximate the square root
Calculating \( \sqrt{37} \):

\[ \sqrt{37} \approx 6.08 \]

### Conclusion
Thus, the amplitude of the resulting wave is approximately \( 6 \).