Two loudspeakers are placed 3.00m apart. They emit 485 Hz sounds, in phase. A microphone is placed 3.20m distant from a point midway between the two speakers, where an intensity maximum is recorded.
How far must the microphone be moved to the right to find the first intensity minimum?
Suppose the speakers are reconnected so that the 485 Hz sounds they emit are exactly out of phase. At what positions are the intensity maximum and minimum now?(2 answers)
To understand how to solve this problem, let's break it down step by step.
Step 1: Find the wavelength
First, we need to find the wavelength of the sound wave. We can use the formula:
wavelength = speed of sound / frequency
The speed of sound is approximately 343 meters per second. So, the wavelength is:
wavelength = 343 m/s / 485 Hz = 0.707 m
Step 2: Determine the path difference
The path difference is the difference in the distances between the two loudspeakers and the microphone. In this case, the microphone is placed at a point midway between the two speakers, so the path difference is zero.
Step 3: Find the distance to the microphone for the first intensity minimum
To find the distance to the microphone for the first intensity minimum, we need to take into account the path difference and the wavelength. Intensity minimum occurs when the path difference is equal to half the wavelength (0.5 * wavelength).
So, the distance to the microphone for the first intensity minimum is:
distance = (0.5 * wavelength) + (3.00 m / 2)
Plugging in the values, we get:
distance = (0.5 * 0.707 m) + (3.00 m / 2) = 1.8535 m
Therefore, the microphone must be moved 1.8535 meters to the right to find the first intensity minimum.
Now, let's move on to the second part of the question, where the speakers are out of phase.
When the speakers are out of phase, the path difference will change. In this case, as the path difference goes from zero to a full wavelength (1 * wavelength), we will have an intensity maximum. Similarly, when the path difference goes from a full wavelength to two full wavelengths (2 * wavelength), we will have another intensity maximum.
Therefore, the positions of the intensity maximums will be located at:
1. distance = 1 * wavelength + (3.00 m / 2)
2. distance = 2 * wavelength + (3.00 m / 2)
Using the calculated wavelength of 0.707 m, we can plug in these values to find the positions of the intensity maximums:
1. distance = 1 * 0.707 m + (3.00 m / 2) = 3.8535 m
2. distance = 2 * 0.707 m + (3.00 m / 2) = 4.5605 m
Therefore, the positions of the intensity maximums are 3.8535 meters and 4.5605 meters, respectively.