Two speakers are 2m apart. An observer is 3m Infront of one speaker.
a) if both are driven by a single oscillator at 300Hz, what is the phase difference at the observer
b) what is the frequency closest to 300Hz for which the observer hear minimal sound?
To answer these questions, we need to consider the principles of wave interference and phase difference.
a) To find the phase difference at the observer, we need to understand how distance affects phase difference in wave propagation. The phase difference between two points along a wave is determined by the path length difference between those points.
In this case, the two speakers are 2 meters apart. Since they are driven by a single oscillator, they will generate identical waves with the same frequency and wavelength. The wave from the first speaker travels directly to the observer, while the wave from the second speaker has to travel an additional 2 meters to reach the observer.
Since the wavelength (λ) is the distance between corresponding points on a wave, we can use the formula:
λ = v / f
Where:
λ = Wavelength
v = Velocity
f = Frequency
For sound waves in air, the velocity (v) is approximately 343 meters per second. Therefore, using the equation above, we can calculate the wavelength:
λ = 343 m/s / 300 Hz
λ ≈ 1.14 meters
Since the two speakers are 2 meters apart, the path length difference between them is 2 meters. To find the corresponding phase difference, we need to calculate how many wavelengths fit into this path length difference:
Δphase = (2 meters) / (1.14 meters/wavelength)
Δphase ≈ 1.75 wavelengths
Therefore, the phase difference at the observer will be approximately 1.75 wavelengths.
b) To determine the frequency closest to 300 Hz for which the observer hears minimal sound, we need to understand the concept of constructive and destructive interference.
When two waves arrive at the same location, their amplitudes add up. If the waves are in phase (with the same phase), they will add constructively, resulting in an increased amplitude and a louder sound. If the waves are out of phase (with opposite phases), they will add destructively, resulting in a decreased amplitude and a softer sound.
In this case, the observer is positioned 3 meters in front of the first speaker. So, for the observer to hear minimal sound, there needs to be destructive interference between the waves from the two speakers.
Since the path length difference between the two speakers is 2 meters, we can use the formula:
Δphase = Δpath length / λ
To achieve destructive interference, the phase difference should be an odd multiple of half a wavelength (λ/2), which corresponds to 180 degrees out of phase.
Therefore, we can write the equation:
Δphase = (2 meters) / λ = (λ/2)
Rearranging the equation and solving for λ:
λ ≈ 4 meters
Now, we can determine the frequency closest to 300 Hz for which the observer hears minimal sound. Using the formula for wavelength:
λ = v / f
Rearranging the equation and substituting the velocity:
f ≈ v / λ ≈ 343 m/s / 4 meters
f ≈ 85.75 Hz
Therefore, the frequency closest to 300 Hz for which the observer hears minimal sound is approximately 85.75 Hz.