If two waves represented by `y_1=4sin omega t` and `y_2=3sin (omegat+pi/3)` interfere at a point, the amplitude of the resulting wave will be about

Options

A. (a) `7`
B. (b) `6`
C. (c) `5`
D. (d) `3*5`

Explanation

To find the amplitude of the resulting wave formed by the interference of the two waves given by \( y_1 = 4 \sin(\omega t) \) and \( y_2 = 3 \sin(\omega t + \frac{\pi}{3}) \), we can follow these steps:

### Step 1: Identify the Amplitudes and Phase Difference
- The amplitude of the first wave \( A_1 = 4 \).
- The amplitude of the second wave \( A_2 = 3 \).
- The phase difference \( \phi = \frac{\pi}{3} \).

### Step 2: Use the Formula for Resultant Amplitude
The formula for the resultant amplitude \( A_{\text{net}} \) when two waves interfere is given by:

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

### Step 3: Substitute the Values into the Formula
Now, substituting the values we have:

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

### Step 4: Calculate Each Term
- Calculate \( A_1^2 = 4^2 = 16 \).
- Calculate \( A_2^2 = 3^2 = 9 \).
- Calculate \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).
- Therefore, \( 2 \cdot 4 \cdot 3 \cdot \cos\left(\frac{\pi}{3}\right) = 2 \cdot 4 \cdot 3 \cdot \frac{1}{2} = 12 \).

### Step 5: Combine the Results
Now, plug these values back into the equation:

\[ A_{\text{net}} = \sqrt{16 + 9 + 12} = \sqrt{37} \]

### Step 6: Approximate the Result
Calculating \( \sqrt{37} \):

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

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

### Final Answer
The amplitude of the resulting wave will be about \( 6 \).

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