The problem involves interference from two loudspeakers, considering destructive and constructive interference. Let's break it down:

Given:

• $d_1 = 2.70 \, \text{m}$ (distance between speakers)
• $d_2 = 3.81 \, \text{m}$ (distance from the listener to one speaker)
• Speed of sound, $v = 343 \, \text{m/s}$
• Audible frequency range: $20 \, \text{Hz}$ to $20,000 \, \text{Hz}$

Part (a): Lowest frequency for destructive interference (minimum signal)

1. Path difference: $\Delta d = d_2 - d_1$ $\Delta d = 3.81 \, \text{m} - 2.70 \, \text{m} = 1.11 \, \text{m}$

2. For destructive interference: \Delta d = \left( m + \frac{1}{2} \right) \lambda, \, \text{where m = 0, 1, 2, \ldots} For the lowest frequency, we take $m = 0$, so: $\lambda = 2 \Delta d$ $\lambda = 2 \times 1.11 \, \text{m} = 2.22 \, \text{m}$

3. Frequency is given by: $f = \frac{v}{\lambda}$ $f = \frac{343}{2.22} \, \text{Hz} \approx 154.5 \, \text{Hz}$

Part (b): Lowest frequency for constructive interference (maximum signal)

1. For constructive interference: \Delta d = m \lambda, \, \text{where m = 1, 2, \ldots} For the lowest frequency ($m = 1$): $\lambda = \Delta d$ $\lambda = 1.11 \, \text{m}$

2. Frequency is: $f = \frac{v}{\lambda}$ $f = \frac{343}{1.11} \, \text{Hz} \approx 308.1 \, \text{Hz}$

Final Answers:

• (a) Destructive interference: $154.5 \, \text{Hz}$
• (b) Constructive interference: $308.1 \, \text{Hz}$

Would you like more detailed steps or clarification?

Related Questions:

1. How does path difference affect interference patterns?
2. What happens if $m$ is higher in the constructive/destructive formulas?
3. What would change if the listener's position $d_2$ changed?
4. Why does interference depend on wavelength and path difference?
5. What are practical applications of interference in sound engineering?

Tip: Remember that the frequency of a wave depends directly on its speed and inversely on its wavelength!