The displacement of two interfering light waves are `y_(1)=4 sin omega t" and "y_(2)= 3 cos (omega t)`. The amplitude of the resultant waves is `(y_(1)" and "y_(2)` are in CGS system)

Options

A. A. 5 cm
B. B. 7 cm
C. C. 1 cm
D. D. zero

Answer

To find the amplitude of the resultant wave formed by the interference of the two given waves, we can follow these steps:

### Step 1: Identify the given waves
The two waves are:
- \( y_1 = 4 \sin(\omega t) \)
- \( y_2 = 3 \cos(\omega t) \)

### Step 2: Convert the cosine wave to sine form
To combine these two waves, we can express \( y_2 \) in terms of sine. We know that:
\[ \cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_2 \) as:
\[ y_2 = 3 \cos(\omega t) = 3 \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Step 3: Represent the waves in a common form
Now we can represent both waves in terms of sine:
- \( y_1 = 4 \sin(\omega t) \)
- \( y_2 = 3 \sin\left(\omega t + \frac{\pi}{2}\right) \)

### Step 4: Use the formula for resultant amplitude
The resultant amplitude \( A \) of two waves can be calculated using the formula:
\[ A = \sqrt{A_1^2 + A_2^2} \]
where \( A_1 \) and \( A_2 \) are the amplitudes of the individual waves.

In our case:
- \( A_1 = 4 \)
- \( A_2 = 3 \)

### Step 5: Calculate the resultant amplitude
Substituting the values into the formula:
\[ A = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 6: State the final answer
The amplitude of the resultant wave is \( 5 \) cm.