The displacement of the interfaring light waves are ` y_1 =4 sin omega t and y_2=3sin (omegat +(pi)/( 2)) ` What is the amplitude of the resultant wave?

Options

A. 5
B. 7
C. 1
D. 0

Answer

To find the amplitude of the resultant wave from the given wave equations, we can follow these steps:

1. **Identify the Amplitudes and Phase Difference**:
The two waves are given as:
- \( y_1 = 4 \sin(\omega t) \) (Amplitude \( A_1 = 4 \))
- \( y_2 = 3 \sin(\omega t + \frac{\pi}{2}) \) (Amplitude \( A_2 = 3 \))

The phase difference between \( y_1 \) and \( y_2 \) is \( \frac{\pi}{2} \) radians.

2. **Use the Resultant Amplitude Formula**:
When two waves interfere, the resultant amplitude \( R \) can be calculated using the formula:
\[ R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\theta)} \]
where \( \theta \) is the phase difference between the two waves.

3. **Substituting the Values**:
Since the phase difference \( \theta = \frac{\pi}{2} \), we know that:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]
Therefore, the formula simplifies to:
\[ R = \sqrt{A_1^2 + A_2^2} \]

4. **Calculating the Resultant Amplitude**:
Substitute \( A_1 = 4 \) and \( A_2 = 3 \):
\[ R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

5. **Conclusion**:
The amplitude of the resultant wave is \( R = 5 \).