1
concept
The Doppler Effect
7m
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Hey, guys, in this video, we're going to talk about the Doppler effect, which is an interesting thing that happens toe waves when the source and the listening to the source or the detector are moving relative to one another. All right, let's get to it. The measurement of the frequency of the sound will always be the same whenever you measure it at rest. If the source is at rest and you are at rest, no matter where you are in relation to the source, the frequency of the sound admitted will always be the same. The frequency will not be the same if there's a relative motion between you and the source when you measure it. Okay, The Doppler effect is the changing of frequencies of sound in motion when in motion relative to a source, the frequency increases when moving towards the source. Okay, when you're coming together, which is sometimes called approaching, and the frequency decreases when moving away from the source. Sometimes when you two are moving away, sometimes called receding. All right. I want to show two figures here, and I'm gonna get out of the way for this. At the left, we have a source at rest. Okay, this isn't physical space. Each of these circles is a wave front. It's a peak of a sound wave. Okay? And they are emitted with the distance being the wavelength. And they're admitted at different times. A pulse, a pulse, a pulse, beat, beat, beat. But they are emitted at particular intervals in time. And no matter where you happen to be, you're going to hear the same frequency of sound. Because no matter where you are, the wavelength between these sounds is the same. Okay, you guys see that That wavelength is the same no matter where you measure it. But what if it's in motion? This outermost ring? His measure, It is produced at the initial position off the source. Then the source moves to the right and the second ring is admitted at the second position of the source. Then it moves to the right, and this third ring is emitted at the third position of the source. Then it moves to the right again, and this little list ring is emitted where it is right now. Okay, So this is what the source moving to the right. Look at what this does to the frequency. If on observers over here it sees a huge wave length the right. Okay, these waves air very, very, very far apart. Over here, these wavelengths stack up. It's too small for even for me to even draw. So if you see when they stack up, it hears Wavefront Wavefront, Wavefront, Wavefront wavefront. Very, very, very fast. That means that the frequency is up on this side because of how rapidly those wave fronts air coming. And the frequency is down on this side because of how slowly it takes the wavelengths to reach them because of how far apart those wavelengths are. Okay, so the frequency increases win moving towards the source or when the source is moving to you when you guys are approaching one another and the frequency decreases when moving away from the source, or when the source is moving away from you when you are receding. Okay. Now, the equation for the Doppler effect gives us the frequency detected due to the Doppler effect due to the relative motion between the source and detector, and it's given by the plus V D, which is the speed of the detector divided by the plus ve s, which is the speed of the source times. The frequency of the source emits, sometimes called the natural frequency. All right. And now, in order to use this equation with these particular signs, you need to use a very particular coordinate system where your detectors at the origin and the positive access is drawn from the detector to the source. Okay, let's do a quick example. A speaker emits a sound at 20 hertz while at rest. If the speaker removing away from you at 45 m per second, What frequency of sound would you hear? Consider the speed of sound to be 343 m per second. Okay, so first, what's the natural frequency of the speaker? 20 hertz? It says That's what it admits, while at rest that's it's natural frequency. Next, what's the speed of the source? It's 45 m per second, but what signs should I give it? Notice that it's moving away from you. You are the detector, so the source is moving away from you. So the source is moving in the positive direction. So this is positive. 45 m per second. What about your speed. It doesn't mention anything about you moving, so don't assume you are assuming zero. And the speed of sound is 343 m per second. So all we have to do is plug and chug with the Doppler effect equation. So the frequency detected is V plus V D over V plus V s times F s, which is 3 43 plus Sorry, not 2020 is the frequency. This is 45 divided by what's the detector zero divided by 3. 43 plus The source is moving at and the source frequency is 20 hertz and this holding becomes at 17.7 hurts and it should be lower. Okay, It should have decreased because the source is moving away from you. So the frequency detective should definitely be lower than the natural frequency. Alright, guys, that wraps up our discussion on the Doppler effect. Thanks for watching
2
Problem
A police siren emits a sound somewhere around 700 Hz. If you are waiting at a red light, and a police car approaches you from behind and passes you, moving at a constant 30 m/s, what is the frequency you hear from the siren as it approaches you from behind? What about once it’s passed you? Assume the air temperature to be 20°C.
A
fD,towards=639 m/s, fD,away=761 m/s
B
fD,towards=644 m/s, fD,away=767 m/s
C
fD,towards=761 m/s, fD,away=638 m/s
D
fD,towards=767 m/s, fD,away=644 m/s
3
example
Two Submarines Approaching One Another
5m
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Hey, guys, let's do an example. Two submarines approach one. Another underwater sub one emits a sonar pulse at one times 10 to the three. Hertz, which travels from sub one towards sub two, is then reflected off of sub to traveling back to sub one. If the speed of sound of water is 1500 m per second, sub one is approaching sub two at a speed of 18 m per second and sub two is approaching sub one at a speed of 15 m per second. What frequency to sub to detect the sonar polls at what frequency to sub one detector reflected sonar pulse app. Okay, so I'll break this up into questions A and questions be question A is what? Frequency to sub to detect the pulse emitted by sub one act. Okay, so here's one sub your second sub. My crappy subs. Sub one is moving towards sub two at 18 m per second sub twos, moving towards sub one at 15 m per second and sub one is emitting sound towards it sub to Okay, so we have the source. We have the detector, the sources sub one, the detectors sub to so the detected frequency is going to be V plus V D over V plus V s ffs. Okay, now the speed of sound is just gonna be 1500. Just like the problem says, 1500 m per second. The question is, what are the signs going to be? And it's emitted at 1000 hertz. What are the signs? And value is gonna be for our velocities for the detector and source. Okay, The sources sub one. So clearly the source should be 18. But is it plus or minus the detectors? Sub two. So includes should be 15 for the detective speed, but is it plus or minus? Don't forget the, uh, corden system. The detectors at the origin and you draw a line from the detector to the source, and that's the positive direction. So the detector is moving towards the source, so that's positive. The source is moving towards the detector, so that is negative. Okay. And all of this equals a detected frequency of 10 hurts. So this is the frequency that sub to detects the sonar pulse from sub one act. Okay, now, part B. Same problem. But now some to someone this one's moving at 18. Still, this one's moving at 15. Still, But now sub two is acting like the source because those sonar be sonar waves are bouncing off of south of sub two and heading back towards sub one. Remember that sound that any wave is reflected off of any circus at the same frequency? So this is exactly like the sub emitting a sound at 1000 at 10. 22 hurts. That's exactly the set up to the problem. So the detected sound is V plus V D the plus V s s. But our detector on our source have flipped. Now, Sub two is the source, Right? And sub one is the detector. So this is still gonna be 1500 plus some speed, some velocity. This is still gonna be 1500 plus some velocity, but now the source frequency is 10. 22 hurts, okay? And our coordinate system has sub two on the right sub one on the left. And remember that the corn systems from the detector to the source is positive. So the detectors velocity is positive. This is plus 15 right? Oh, sorry. The This is the source. I got these backwards. I had also gotten this one backwards, but it still worked because the mirror image looks the same. In this case, the detector is truly on the right and the source is on the left. If I were to flip this around, you would see sub two is the detector. Sub one is the source. This is positive. They're moving towards each other. The detector speed is still positive. The source speed is still negative. Okay, so work out the exact same way. The detectors moving towards the source source moving towards the detector. So you get that same thing, but now the detector sub one. So this is plus 18 the sources sub too. So this is minus 15 and plugging this all in. We get 10 45 hertz. This is how sonar works. Sub one admitted a pulse at 1000 hertz and the received the pulse back at 1000 45 Hertz. Because that sub knows its own speed, it can use this exact equation to find the speed of the other sub based on how large that Doppler shift is. Okay, So is this how sonar for submarines work? Alright, guys, that wraps up this problem. Thanks for watching
Additional resources for The Doppler Effect
PRACTICE PROBLEMS AND ACTIVITIES (5)
- The shock-wave cone created by a space shuttle at one instant during its reentry into the atmosphere makes an ...
- The siren of a fire engine that is driving northward at 30.0 m>s emits a sound of frequency 2000 Hz. A truc...
- The siren of a fire engine that is driving northward at 30.0 m>s emits a sound of frequency 2000 Hz. A truc...
- A railroad train is traveling at 30.0 m>s in still air. The frequency of the note emitted by the train whis...
- A railroad train is traveling at 30.0 m>s in still air. The frequency of the note emitted by the train whis...