Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air` = 320 ms^-1`

Explanation

To solve the problem of finding the frequencies for which there is destructive interference at the listener's position, we will follow these steps:

### Step 1: Determine the Path Difference
The listener is located 6.0 m from speaker S1 and 6.4 m from speaker S2. The path difference (Δd) can be calculated as:
\[ \Delta d = d_2 - d_1 = 6.4 \, \text{m} - 6.0 \, \text{m} = 0.4 \, \text{m} \]

### Step 2: Condition for Destructive Interference
For destructive interference to occur, the path difference must be an odd multiple of half the wavelength (λ):
\[ \Delta d = \left(2n + 1\right) \frac{\lambda}{2} \]
where \(n\) is an integer (0, 1, 2, ...).

### Step 3: Express Wavelength in Terms of Frequency
The wavelength (λ) can be expressed using the speed of sound (v) and frequency (f):
\[ \lambda = \frac{v}{f} \]
Given that the speed of sound in air is \(v = 320 \, \text{m/s}\), we can substitute this into the equation for path difference:
\[ 0.4 = \left(2n + 1\right) \frac{v}{2f} \]
Substituting \(v\):
\[ 0.4 = \left(2n + 1\right) \frac{320}{2f} \]
This simplifies to:
\[ 0.4 = \left(2n + 1\right) \frac{160}{f} \]

### Step 4: Rearranging to Find Frequency
Rearranging the equation to solve for frequency \(f\):
\[ f = \left(2n + 1\right) \cdot 400 \]

### Step 5: Finding Frequencies within the Given Range
We need to find the values of \(f\) for \(n = 0, 1, 2, ...\) that lie within the range of 500 Hz to 5000 Hz.

1. For \(n = 0\):
\[ f = (2 \cdot 0 + 1) \cdot 400 = 400 \, \text{Hz} \quad \text{(not in range)} \]

2. For \(n = 1\):
\[ f = (2 \cdot 1 + 1) \cdot 400 = 1200 \, \text{Hz} \quad \text{(in range)} \]

3. For \(n = 2\):
\[ f = (2 \cdot 2 + 1) \cdot 400 = 2800 \, \text{Hz} \quad \text{(in range)} \]

4. For \(n = 3\):
\[ f = (2 \cdot 3 + 1) \cdot 400 = 3600 \, \text{Hz} \quad \text{(in range)} \]

5. For \(n = 4\):
\[ f = (2 \cdot 4 + 1) \cdot 400 = 4400 \, \text{Hz} \quad \text{(in range)} \]

6. For \(n = 5\):
\[ f = (2 \cdot 5 + 1) \cdot 400 = 5200 \, \text{Hz} \quad \text{(not in range)} \]

### Conclusion
The frequencies for which there is destructive interference at the listener's position are:
- 1200 Hz
- 2800 Hz
- 3600 Hz
- 4400 Hz