Two sound waves (expressed in CGS units) given by ` y_(1)=0.3 sin (2 pi)/(lambda)(vt-x) and y_(2)=0.4 sin (2 pi)/(lambda)(vt-x+ theta)` interfere. The resultant amplitude at a place where phase difference is ` pi //2 ` will be

Answer

To find the resultant amplitude of two sound waves that interfere with each other, we can follow these steps:

### Step 1: Identify the Amplitudes and Phase Difference
From the given equations of the sound waves:
- \( y_1 = 0.3 \sin\left(\frac{2\pi}{\lambda}(vt - x)\right) \)
- \( y_2 = 0.4 \sin\left(\frac{2\pi}{\lambda}(vt - x + \theta)\right) \)

We can identify:
- Amplitude of wave 1, \( a_1 = 0.3 \, \text{cm} \)
- Amplitude of wave 2, \( a_2 = 0.4 \, \text{cm} \)
- Phase difference, \( \phi = \frac{\pi}{2} \)

### Step 2: Use the Formula for Resultant Amplitude
The formula for the resultant amplitude \( R \) when two waves interfere is given by:
\[ R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos(\phi)} \]

### Step 3: Substitute the Values into the Formula
Substituting the values we have:
\[ R = \sqrt{(0.3)^2 + (0.4)^2 + 2 \cdot (0.3) \cdot (0.4) \cdot \cos\left(\frac{\pi}{2}\right)} \]

### Step 4: Calculate the Cosine of the Phase Difference
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), the equation simplifies to:
\[ R = \sqrt{(0.3)^2 + (0.4)^2 + 0} \]

### Step 5: Calculate the Squares of the Amplitudes
Calculating the squares:
\[ R = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} \]

### Step 6: Add the Values Inside the Square Root
\[ R = \sqrt{0.25} \]

### Step 7: Take the Square Root
\[ R = 0.5 \, \text{cm} \]

### Final Result
The resultant amplitude at the point where the phase difference is \( \frac{\pi}{2} \) is \( R = 0.5 \, \text{cm} \).

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