Two speakers, one directly behind the other, are each generating a 264-Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? Take the speed of sound to be 342 m/s.
The basic requirement for destructive interference is that the two waves be shifted by half a wavelength.
Now, as you know, v = λν, so just plug in your numbers.
To determine the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them, we need to consider the concept of phase difference.
In destructive interference, the crests of one wave align with the troughs of the other wave, causing the two waves to cancel each other out. This occurs when the phase difference between the waves is an odd multiple of π.
The phase difference between two waves can be calculated using the formula:
Δφ = 2πΔx / λ
where Δx is the separation distance between the speakers and λ is the wavelength of the sound wave.
To find the wavelength (λ) of the sound wave, we can use the formula:
v = fλ
where v is the speed of sound and f is the frequency of the sound wave.
Given:
Frequency (f) = 264 Hz
Speed of sound (v) = 342 m/s
First, let's find the wavelength (λ):
λ = v / f
λ = 342 / 264
λ ≈ 1.295 m
Now, we can calculate the phase difference (Δφ) required for destructive interference:
Δφ = 2πΔx / λ
Δφ = 2πΔx / 1.295
To achieve destructive interference, the phase difference (Δφ) should be an odd multiple of π (i.e., π, 3π, 5π, etc.).
Let's consider the smallest odd multiple of π, which is π:
π = 2πΔx / 1.295
Simplifying the equation:
Δx = (1.295 * π) / 2
Calculating the value:
Δx ≈ 2.03 m
Therefore, the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them is approximately 2.03 meters.