To determine the nature of interference when two sound waves meet, we first need to analyze the situation using the information you've provided. We have a source vibrating at a frequency of 1.0 kHz, and one sound wave travels 83 cm longer than the other. The speed of sound in air is given as 332 m/s. Let's break this down step-by-step to find out whether the interference will be constructive or destructive.
Step 1: Calculate the Wavelength
The wavelength (λ) of a sound wave can be calculated using the formula:
λ = v / f
Where:
- v is the speed of sound (332 m/s)
- f is the frequency (1.0 kHz or 1000 Hz)
Substituting in the values:
λ = 332 m/s / 1000 Hz = 0.332 m
Step 2: Determine the Path Difference
Given that one path is 83 cm longer than the other, we convert 83 cm to meters:
83 cm = 0.83 m
This path difference will affect how the waves interact when they meet.
Step 3: Analyze the Path Difference Relative to the Wavelength
Now, we need to see how this path difference relates to the wavelength:
The wavelength is 0.332 m, and the path difference is 0.83 m. To determine the nature of the interference, we need to calculate how many wavelengths fit into the path difference:
Number of wavelengths = Path difference / Wavelength
Number of wavelengths = 0.83 m / 0.332 m ≈ 2.5
Step 4: Determine the Type of Interference
The key point here is that the path difference corresponds to 2.5 wavelengths. Since this is not an integer multiple of the wavelength (like 0, 1, 2, etc.), but rather 2.5, this means that the two waves will not be perfectly in phase or perfectly out of phase when they meet.
Specifically, a path difference of 2.5 wavelengths indicates that one wave is a half wavelength (0.5 λ) out of phase with the other wave at the point they meet. When two waves are half a wavelength out of phase, they interfere destructively.
Final Analysis
In summary, when the two sound waves originating from the same source meet after traveling different paths—one being 83 cm longer than the other—the interference will be destructive due to the half-wavelength phase difference. This means that the waves will partially cancel each other out, which can result in a reduction in sound intensity at that point.
