The frequencies of two sound sources are 256 Hz and 260 Hz, At `t=0` the intesinty of sound is maximum. Then the phase difference at the time `t=1//16` sec will be

Options

A. zero
B. `pi`
C. `pi/2`
D. `pi/4`

Answer

To solve the problem, we need to determine the phase difference between two sound waves at a given time, given their frequencies. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the frequencies
We have two sound sources with frequencies:
- \( f_1 = 256 \, \text{Hz} \)
- \( f_2 = 260 \, \text{Hz} \)

### Step 2: Calculate the beat frequency
The beat frequency \( f_b \) is calculated as the absolute difference between the two frequencies:
\[ f_b = |f_1 - f_2| = |256 - 260| = 4 \, \text{Hz} \]

### Step 3: Determine the frequency of temporal interference
The frequency of temporal interference \( f \) is half of the beat frequency:
\[ f = \frac{f_b}{2} = \frac{4}{2} = 2 \, \text{Hz} \]

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) at time \( t \) can be calculated using the formula:
\[ \Delta \phi = 2 \pi f t \]
Substituting the values:
- \( f = 2 \, \text{Hz} \)
- \( t = \frac{1}{16} \, \text{s} \)

We get:
\[ \Delta \phi = 2 \pi \times 2 \times \frac{1}{16} \]
\[ \Delta \phi = \frac{4\pi}{16} = \frac{\pi}{4} \, \text{radians} \]

### Conclusion
Thus, the phase difference at \( t = \frac{1}{16} \, \text{s} \) is:
\[ \Delta \phi = \frac{\pi}{4} \, \text{radians} \]