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Ch 37: Special Relativity
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All textbooksYoung & Freedman Calc 14th EditionCh 37: Special RelativityProblem 37

Chapter 36, Problem 37

A source of electromagnetic radiation is moving in a radial direction relative to you. The frequency you measure is 1.25 times the frequency measured in the rest frame of the source. What is the speed of the source relative to you? Is the source moving toward you or away from you?

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Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem, an electromagnetic waves emitter moves in a straight line relative to a detector. The detector records a frequency that is 3/ times the frequency measured in the emitter's rest frame. I determine the speed of the emitter relative to the detector. I I, what is the direction of motion toward or away relative to the detector? OK. So we have two goals that we're trying to accomplish here. We need to determine the speed of the emitter relative to the detector. And we need to find out what the direction of motion relative to the detector is. OK. So we're given some multiple choice answers. Let's read them off to see what our final answer set might be. A is I 0.28 ci I away from the detector B is I 0.28 ci I toward the detector C I 0. I is 0.14 ci I away from the detector and D is I 0.14 ci I toward the detector. OK. So to solve for part I, first off, let us recall the equation for the relativistic Doppler effect of light. And also note that the source of light is moving towards the observer. So let's call it equation one. So the Doppler the relativistic Doppler effect of light states that the frequency is equal to the speed of light multiplied by the square root of the speed of light plus the velocity all divided by C minus the velocity where C is the speed of light all multiplied by the frequency F subscript zero F zero. So f in this case is the frequency of the observed or is the frequency observed and F subscript zero is the frequency of the source. OK. So our goal is to find the speed of the emitter. So we need to use equation one to solve for the velocity. Also we are given wavelength. So we need to relate the frequency and the wavelength to one another. OK. So let's start off by rearranging equation one using algebra and start by squaring both sides to work towards isolating U the velocity. So when we do that, we should get equation two, which states that the velocity is equal to the speed of light multiplied by F divided by F zero. So the frequency observed divided by the frequency of the source squared minus one divided by the frequency of the observed divided by the frequency of the source squared plus one. So now we need to substitute in the wavelength value into our frequency equation. So we need to note that the frequency is equal to the speed of light divided by the wavelength. And that, so this is the frequency of the observed is equal to the speed of light divided by the wavelength. So the frequency of the source is equal to the speed of light divided by the wavelength of the source. OK. So we can state that the frequency of the source divided by the frequency of the observed is F zero, divided by F is equal to C divided by wavelength divided by C divided by wavelength zero. Which when we simplify means that the wavelength of the source divided by the wavelength of the observed. So the observed wavelength, OK. That was a lot of math and a lot of relating but we did it. So now we need to plug this value the value of wavelength of the ob of the source. So the source wavelength divided by the observed wavelength back into equation two. So let's do that. So the velocity is equal to the speed of light multiplied by the wavelength of the source divided by the wavelength observed squared minus one divided by the wavelength of the source divided by the wavelength of the deer observed divide like squared. So the wavelength of the source divided by the wavelength observed squared plus one. So now at this stage, we can plug in all of our known values and solve for you the velocity. So let's do that. So the velocity is equal to the speed of light multiplied by. So the wavelength of, of the source is given to us as 0. F zero because it as it was given to us in the problem. So the 0.75 is the frequency is, the frequency is 3/4 times the frequency observed. OK. So 0.75 frequency observed divided by the frequency observed squared minus one, divided by 0. frequency observed divided by frequency observed plus one. So when we plug that into a calculator and simplify, we get that the velocity is zero point 28 C. And that is our answer for part I, OK. So now to solve for part I I, we need to consider that F equals zero point mark is I I that F equals 0.75 F zero. The frequency observed is less than the frequency of the observed. So 0.75 multiplied by the frequency observed is less than the frequency observed. So the object is moving away from the detector. So the emitter moves away from the detector. So that means our final answer has to be that it moves away from the detector. So moves away from detector. OK. So let's go back up to look at our multiple choice answers. So that means the correct answer has to be the letter A I is 0.28 C and I I is away from the detector. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.
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