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Ch 15: Mechanical Waves

Chapter 15, Problem 35

Two radio antennas A and B radiate in phase. Antenna B is 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied. (b) What is the longest wavelength for which there will be constructive interference at point Q?

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Hey everyone in this problem we have mobile phone network towers P and Q. That are separated by 375 m. A tower P is located to the left of tower que their antennas are radiating in phase at identical frequencies, a mobile phone is located to the left of antenna P at a point R. Which is 1.2 kilometers from that tower P. On the axis connecting the two Antennas. We're told that mobile networks have various bands with different frequencies and wavelengths were told to assume that the bands can have any frequency value. And we are asked to determine the largest wavelength that produces constructive interference at that point. Our. Alright. So let's start by drawing this out. Okay, so we have our two towers P and Q. Okay, so we have our two towers. We're told that P is located to the left of Q. Okay, so this tower on the left is P. tower on the right is cute and they're separated by m. So the distance here is 375 m. Okay. And then we have a mobile phone located to the left of P. Okay. And it's going to be on the same axis connecting these two And tests. Okay. So we have over here a mobile phone, we're just drawing everything as points. Um just for simplicity in our diagram. Okay. So this is gonna be point are where we have our mobile phone and we're told that it is 1.2 km from Tower P. So this is 1.2 km. All right. So, we have our diagram here. Now we're asked to determine the largest wavelength that produces constructive interference. Now, when we're talking about constructive interference, what we want to find, what's important to find is the path difference. Okay? And the path difference is going to tell us the difference in the distance that the wave has to travel from the further point, which in this case is going to be Q to R. Okay, versus the wave that has to travel from P to R. So we're gonna start by finding the distances from our two towers P. R and Q two R. Okay. And then use those to find our path difference. So we're gonna call D. P. The distance from point P. Two point are. Okay, so from this tower P to our cell phone at point R. And we know that this is 1.2 km. Now we want to convert this into meters. So this is gonna be 1.2 km times m per kilometer. The unit of kilometer cancels, we get 1200 m. Okay? So to go from kilometers two m, we multiply by 1000. Okay, so this is the distance from tower P. Two point are now we're going to do the same for Q. So we have D. Q. Is gonna be the distance from point Q to that point where the cell phone is our, what is this distance gonna be, well this is going to be the distance from Q. Two point P which is 375 m plus the distance from P two R. Okay. Because they're all along the same axis, we can add up these separate distances and we found that the distance between P and r is 1200 m. Okay, 1.2 kilometers we can write us 1200 m and this gives us a value for the distance between Q and R. Of 1575 m. Alright, so we found our two differences. Let's go ahead and calculate the path difference. So the path difference, we're gonna call it D and it's gonna be the difference between these two distances and we're gonna take the absolute value D q minus dp and in this case D Q is bigger. Um So we can drop the absolute values, it's going to be positive. You can flip this, you could have dp minus D Q. K. We just care about the magnitude of the difference. So we get 15 75 minus 1200 m which gives us a path difference of 375 m. Alright, so we have our path difference d remember we're looking for the largest wavelength that produces constructive interference. Okay, so let's think about when constructive interference is going to occur, constructive interference is going to occur when the path difference, D is some integer multiple of the wavelength. So it's either zero lambda two lambda dot dot dot. Okay. Or we can write this as D is equal to n. Lambda? Where N is a positive integer 012 dot dot dot. Now, why is that if the difference in the distance that one wave has to travel over the other to get the point R is a multiple of the wavelength can enter multiple of the wavelength. That means that when those waves get to point, are there going to be in phase and if they're in phase, Okay, the waves are going to kind of add to each other. Okay, that's construction constructive interference. Okay. If we had the opposite case, okay, Where the path difference? Wasn't a multiple. Okay. We would have them being out of phase when they get to point. Are those two waves out of phase when they get to point are okay. And they would kind of cancel each other. Okay. So when we're looking at constructive interference, we want an integer multiple of the wavelength for a path difference. Okay, well remember we're trying to find lambda. Okay, so let's isolate for lambda and we can write that lambda. And we're gonna say lambda subscript N because it's going to depend on the value of N is equal to the path difference. D divided by ed. Now, when we're looking at the possible values of N. Ok, let's just right. For our particular case we cannot have an equal zero. Okay. Why? Well, when N is zero, that means that D A zero. And we know our path difference is not zero. Okay, So we know that we don't have N is equal to zero. Okay, So we just have N is equal to 123. Okay. The natural numbers. Alright, when are we going to get the largest wavelength? The largest lambda is gonna occur when N is the smallest? Okay, we're dividing by end. So we want end to be as small as possible so that we're dividing by a smaller number. That's going to give us a bigger wavelength lambda. Okay, well, the smallest value of N we can have is one. Okay, so lambda max is gonna be equal to lambda one. She's gonna be The path difference D which we calculated to be 375 m Divided by our end value of one, Which gives our maximum value for the wavelength of m. So, if we go back up to our answer choices, We see that the answer is going to be be the largest wavelength that produces constructive interference is going to be m. Thanks everyone for watching. I hope this video helped see you in the next one
Related Practice
Textbook Question
Two speakers that are 15.0 m apart produce in-phase sound waves of frequency 250.0 Hz in a room where the speed of sound is 340.0 m>s. A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (a) What does she hear: constructive or destructive interference? Why?
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Textbook Question
Two speakers that are 15.0 m apart produce in-phase sound waves of frequency 250.0 Hz in a room where the speed of sound is 340.0 m>s. A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (c) How far from the center must she walk before she first hears the sound maximally enhanced?
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Textbook Question
Two radio antennas A and B radiate in phase. Antenna B is 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point Q?
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Textbook Question
Two speakers, emitting identical sound waves of wavelength 2.0 m in phase with each other, and an observer are located as shown in

Fig. E35.5. (a) At the observer's location, what is the path difference for waves from the two speakers?

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Textbook Question
Two small stereo speakers A and B that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point P starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker B

(Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced. Limit your solution to the cases where x … 1.50 m
542
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Textbook Question
Two small stereo speakers A and B that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point P starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker B

(Fig. E35.1). For what values of x will the sound this person hears be (b) cancelled? Limit your solution to the cases where x … 1.50 m
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