Two speakers are driven by the same oscillator with frequency of 140 Hz. They are located 4.00 m apart on a vertical pole. A man walks straight toward the lower speaker in a direction perpendicular to the pole. He will hear the minimum in sound intensity twice.
How far is he from the pole at these moments?
When is the difference in distances equal to half a wavelength or 1.5 wavelengths?
Look up velocity of sound call it c
then period = 1/f = L/c where L is wavelength
Calculate L from that
say man is x meters from pole
distance to lower speaker is x
distance from higher speaker is
sqrt (x^2+4^2)
difference = sqrt(x^2+4^2) - x
when is that equal to L/2 ??
To find the distance from the pole at the moments when the man hears the minimum in sound intensity, we need to consider the interference pattern created by the two speakers.
Given:
- The speakers are driven by the same oscillator with a frequency of 140 Hz.
- The speakers are located 4.00 m apart on a vertical pole.
When a sound wave reaches the man, it can interfere constructively (maximum intensity) or destructively (minimum intensity) depending on the path length difference between the two speakers.
To find the distance from the pole at the moments of minimum sound intensity heard by the man, we need to consider the destructive interference condition. Destructive interference occurs when the path length difference between the two speakers is equal to an odd multiple of half the wavelength.
First, we need to find the wavelength of the sound wave using the given frequency. The formula for wavelength (λ) is:
λ = v / f
Where:
λ = Wavelength of the sound wave
v = Speed of sound in air (approximately 343 m/s at room temperature)
f = Frequency of the sound wave (given as 140 Hz)
Substituting the values into the formula:
λ = 343 m/s / 140 Hz
λ ≈ 2.45 m
Now that we know the wavelength of the sound wave, we can determine the path length difference between the two speakers needed for destructive interference to occur.
For the first minimum in intensity, the path length difference should be half a wavelength (λ/2). This means that the difference in the distances traveled by the sound waves from the two speakers to the man should be half a wavelength.
Let's assume the distance of the man from the lower speaker (directly in front of it) is x meters. The distance to the upper speaker would then be (4.00 - x) meters since the speakers are 4.00 m apart.
For the first minimum in intensity, the path length difference is:
λ/2 = 2.45 m / 2
λ/2 ≈ 1.22 m
Based on the path length difference, we can set up the following equation:
(4.00 - x) - x = 1.22
Simplifying the equation:
4.00 - 2x = 1.22
Rearranging the equation:
2x = 4.00 - 1.22
2x = 2.78
Solving for x:
x = 2.78 / 2
x ≈ 1.39 meters
Therefore, the man is approximately 1.39 meters away from the pole at the moment of the first minimum in sound intensity.
To find the distance from the pole at the second minimum in sound intensity, we can use the same process but this time when the path length difference is equal to one and a half wavelengths (3λ/2).
Following the steps:
3λ/2 = 3 * 2.45 m / 2
3λ/2 ≈ 3.68 m
Setting up the equation:
(4.00 - x) - x = 3.68
Simplifying:
4.00 - 2x = 3.68
Rearranging:
2x = 4.00 - 3.68
2x = 0.32
Solving for x:
x = 0.32 / 2
x = 0.16 meters
Therefore, the man is approximately 0.16 meters away from the pole at the moment of the second minimum in sound intensity.
In conclusion, the man is approximately 1.39 meters away from the pole at the first moment of minimum sound intensity and approximately 0.16 meters away from the pole at the second moment of minimum sound intensity.