Two sound waves, originating from the same sound source travel along different paths in air and than meet at a point . The speed of the sound is `330m//s` . If the source vibrates at a frequency of 500 Hz and one path is 33 cm longer than the other, then the nature of interference is -

Options

A. Destructive
B. Constructive
C. Neither destructive nor constructive
D. Nothing can be predicted

Answer

To solve the problem, we need to determine the nature of interference between two sound waves that have traveled different paths and meet at a point. We will follow these steps:

### Step 1: Identify the given data
- Speed of sound, \( V = 330 \, \text{m/s} \)
- Frequency of the source, \( f = 500 \, \text{Hz} \)
- Path difference, \( \Delta x = 33 \, \text{cm} = 0.33 \, \text{m} \)

### Step 2: Calculate the wavelength of the sound wave
Using the formula for the speed of sound:
\[ V = f \lambda \]
we can rearrange it to find the wavelength \( \lambda \):
\[ \lambda = \frac{V}{f} = \frac{330 \, \text{m/s}}{500 \, \text{Hz}} = 0.66 \, \text{m} \]

### Step 3: Calculate the phase difference
The phase difference \( \Delta \phi \) due to the path difference can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting the values we have:
\[ \Delta \phi = \frac{2\pi}{0.66} \times 0.33 \]

### Step 4: Simplify the phase difference calculation
Calculating the above expression:
\[ \Delta \phi = \frac{2\pi \times 0.33}{0.66} = \frac{2\pi}{2} = \pi \]

### Step 5: Determine the nature of interference
The phase difference \( \Delta \phi = \pi \) corresponds to an odd multiple of \( \pi \) (i.e., \( (2n + 1)\frac{\pi}{2} \) where \( n \) is an integer). This indicates that the waves are out of phase by half a wavelength, leading to destructive interference.

### Conclusion
Since the phase difference is an odd multiple of \( \pi \), the nature of interference is **destructive interference**.