Two out-of-phase radio antennas at x=±300 m on the x-axis are ...

Question

Two out-of-phase radio antennas at x=±300 m on the x-axis are emitting 3.0 MHz radio waves. Is the point (x, y) =(300 m, 800 m) a point of maximum constructive interference, maximum destructive interference, or something in between?

Accepted Answer

Hey, everyone in this problem, two speakers are placed along the X axis. One at X equals negative 100 m. The other at X is equal to positive 100 m. The speakers emit sound waves with a frequency of 127.5 Hertz. A person stands at the point X comma Y is equal to 100 m, comma 600 m. And we're asked to determine whether there will be perfectly constructive interference, perfectly destructive interference or something in between at that point, the phase difference between the two speakers is two pi divided by three radiant. And we're told to take the speed of sound and air as 340 m per second. For simplicity, we're asked to only consider one decimal place. We'll get our answer. So we have four answer choices. Option A, it cannot be determined. Option B, partially constructive. Option C perfectly constructive and option D perfectly destructive. So let's start by drawing out what we have going on. OK. So we can think about our X axis and we know that we have two speakers. So we have a speaker at X equals negative 100 m. OK? And then we have a speaker at X equals positive 100 m. So that means that the distance between these two speakers will call the one on the left speaker, one, the one on the right speaker two. That means that the distance between these two speakers, let me do this in another color. So it's clear is going to be 200 m. Hey, we have a person, they are located at the 0.100 m, comma 600 m. OK. So they're directly above speaker two in our diagram, 600 m. And they're gonna be some distance away from speaker one as well. And we'll call that distance X one. All right. So this is what we have going on and we wanna figure out at this point where a person is, is it, what kind of interference do we have? All right. So in order to think about the interference, we wanna find the path or sorry, the phase difference in this case. So the phase difference at this point. And so the phase difference Delta Phi recall is gonna be equal to two pi multiplied by delta X where delta X is a path difference divided by lambda. OK. Lambda is the wavelength plus the source phase difference. OK. So this is the phase difference we were given in the problem that delta Phi not of two pi divided by three. OK. So we wanna look at this phase difference and look at whether it's an even or odd multiple of pi in order to figure out if it's perfectly destructive, perfectly constructive. OK? Or if we have something in between. Now, let's start with some of these values. We're gonna need to calculate the path difference. Delta X. We're gonna need to calculate the wavelength lambda. And we've already said that we have delta phi, not from the question. Let's calculate some of these values. OK. So the power difference delta X, that's just gonna be the difference in the distance from person, the person to the further away speaker. So in this case, speaker one, OK to speaker two. So delta X in this case is going to be X one minus X two. Now looking at our diagram and we can calculate X one using Pythagorean theorem, we have a right angle triangle. We know the two side lines, we can use that to find the hypotenuse. So X one squared, it's gonna be equal to 200 m squared plus 600 m squared. And this tells us that X one squared, it's gonna be equal to 400,000 or we can approximate X one as approximately 632.4555 3 2 m. And this 400,000 squared that should be meters squared. OK. So we have an approximation for X one. We know from our diagram that X two is equal to 600 in the distance between the person and speaker two. And so our path difference, delta X is gonna be about 632.455532 m minus 600 m, which gives us a path difference of about 32 0.455532 m. OK. So we have our path difference. Delta X. Now we can think about the wavelength lambda. OK. We're called that lambda is gonna be related to the speed and the frequency. And we're told the frequency of these sound waves, we're told the speed of sound and air. OK. So we want to recall that that speed is gonna be equal to the wavelength lambda multiplied by the frequency F. So we can say that the wavelength is gonna be the speed divided by the frequency and this is gonna be equal to 340 m per second divided by 127.5 Hertz. And which gives us a wavelength of 2.66 repeated meters, right? We have our wavelength, we have our path difference, we have the phase difference of the source. So we can get back to this calculation for the phase difference. OK. So the phase difference delta Phi is now going to be equal to two pie multiplied by 32.455532 m divided by 2.66 repeated meters plus two pi divided by three if we simplify this. OK? Two multiplied by 32.45532 divided by 2.66 repeated what we're gonna get. OK? Is that delta phi is approximately equal to 25.0 PP. And so there are gonna be some additional decimal places. But this question is telling us to just use a single decimal point to evaluate the answer. So 25.0 pi hey, what this means is that it is an odd multiple of pie. OK. So at this point because our phase difference is an odd multiple of pi we're gonna have perfectly destructive interference. Bye. And that's what the question was asking. And it was really asking about this type of interference that we would get. And we were able to find that looking at that phase difference. So we go back to our answer choices. We can see that the correct answer is going to be option D we found that at the point the person is standing, we have perfectly destructive interference. Thanks everyone for watching. I hope this video helped see you in the next one.

Two out-of-phase radio antennas at x=±300 m on the x-axis are emitting 3.0 MHz radio waves. Is the point (x, y) =(300 m, 800 m) a point of maximum constructive interference, maximum destructive interference, or something in between?

Key Concepts

Interference of Waves

Path Difference

Wavelength and Frequency

Watch next