Hey, guys. If you've ever stood on the side of a road while an ambulance blaring its siren has moved past you, you might have noticed that the sound changes once it passes by. It sounds like eeew, and that's what the Doppler effect is. So in this video, I'm going to show you what causes the Doppler effect, but more importantly, I'm going to show you the one equation that you need to solve any kind of problem that deals with it. Let's check this out here. Basically, what the Doppler effect is is a shift in the frequency that you hear from the frequency of the sound source. So we call the frequency that you hear \(f_{\text{listener}}\) or \(f_L\) and the frequency of the sound source \(f_{\text{sound}}\) or \(f_S\). So it's not as if the ambulance actually produces a different sound as it passes by. What happens is that the frequency you hear gets shifted, so we call it sometimes a Doppler shift. Now this Doppler effect happens whenever the sound source, meaning the siren, or the sound listener, meaning you, are moving relative to each other. So you have to have some relative motion of the two objects in order for you to have a Doppler shift. Now I'm going to get back to the equation in just a second here, but I want to show you sort of visually what's going on here and why the Doppler effect gets produced. So if there's no relative motion, there's no Doppler effect. So if you are standing on the side of the road, your velocity of the listener is going to be 0, and if the ambulance is parked on the side of the road, the velocity of the sound source is going to be 0. So what happens is if this sound wave, or sorry, if this ambulance is producing waves at a rate of 5 hertz, so it's 5 waves per second, then what happens is that these waves are traveling towards you at the speed of sound. The ambulance is producing 5 waves every second, then what happens is that later on 5 waves are going to pass through you every second. So if nothing's moving, what's going to happen is that the frequency that you hear is going to be equal to the frequency of the sound source. So you're just going to hear 5 hertz. Right? Nothing gets shifted, nothing like that because 5 waves are going to pass through you. Now things get a little bit trickier whenever you do have some relative motion, and there's actually four different scenarios. You could have you, the listener, moving towards or away from the source or the source that's moving towards or away from you. So I'm going to show you these really where you see that there's only actually two scenarios. So now imagine that you're actually moving, running towards the sound source with some \(v_L\), and the sound source is basically just standing still. What happens is that the ambulance is still producing 5 waves a second, still producing a sound of 5 hertz. So what happens here though is that if you're moving towards the sound source, you're going to be able to cover more waves every second. You're going to be able to move through more waves as they're passing by you. So in general, what happens is you're going to hear something that's greater than 5 hertz. You're going to hear basically more waves per second. The exact same thing happens if the ambulance were actually moving towards you with some sound source like this. Basically, what happens is that these waves sort of get crammed up together in front of the sound source, and they sort of get stretched out or elongated behind it. But the effect is the same. You're going to hear more of these waves every second because the thing is actually moving towards you. So you're going to hear something that's greater than 5 Hertz. So, generally, what happens is that the listener and the sound source are moving towards each other, then the frequency you hear is going to be greater than the frequency of the source. That's the general rule. Now if you reverse everything, it's basically opposite. If you're moving away from this source or if the source is moving away from you, the exact opposite happens. Basically, you're going to hear fewer waves per second, so you're going to hear something that's less than 5 Hertz in both of these situations. So in if the listener and the source are moving away from each other, you're going to hear fewer waves per second and the frequency that you hear is going to be less than the source. So that's why what happens is that when the ambulance is moving towards you, you're going to hear a higher pitch or a higher frequency. It's going to sound like eeep, and then once it passes you, it's going to be moving away from you, and you're going to hear something that is a lower frequency. It just sounds like eeo as it passes by. So let me show you the equation real quick. Remember it's a shift between \(f_L\) and \(f_S\), and what goes inside here is really just a ratio. It's going to be \(\frac{v + v_L}{v + v_S}\). Now unfortunately, there are three \(v\)'s inside this equation, so I'm going to go through them very quickly. This \(v\) here with no subscripts is going to be the speed of sound. It's always going to be positive at 343 meters per second, no matter what. This \(v_L\) here is going to be the velocity of the listener, which is you, and this, the \(s\) here is going to be the velocity of the source, basically the velocity of the siren or the ambulance or whatever it is. Right? So that's just the one equation. It works for any kind of these situations here. Let's go ahead and take a look at our practice problem here. So we have the alarm of a car that's at rest and it produces sound waves of frequency 550. So this is actually the frequency that is produced by the sound source. It's going to be 550. Now you're on a motorcycle and you're traveling directly towards it. So what happens here is you have this \(f_S\), but you're going to be moving towards this sound source with some \(v_L\). And you're hearing an observed frequency of 600, so this is the frequency that you're actually hearing. So this is \(f_L\) at 600. So we want to calculate how fast must you be traveling. All right? So that's basically going to be this \(v_L\) over here. We have the velocity of the sound source, but remember the car is actually going to be at rest. So this velocity of the sound source is actually going to be 0. All right. So basically, now let's go ahead and take a look at our equation. So we have this equation here. This is \(f_L = \frac{v + v_L}{v + v_S} \times f_S\). So let's go through each one of our variables here. So I've got \(f_L\), right, that's just the 6600. Now I've got my \(v\) because it's always going to be positive, 343. What about this \(v_L\)? That's actually what I'm looking for here. We also know that \(v_S\) is going to be 0, so we just can cancel that out and we also have what the frequency of the sound source is. So you have one unknown. The problem here is that if you'll notice, what happens is this ambulance produces sound waves and remember these sound waves travel at the speed of sound that's going towards the listener, towards you. So what happens is we have these two arrows that are actually pointing in different directions. So in these kinds of problems, remember, velocity is a vector, we're going to have to establish a direction of positive here. And basically, here is the rule. The rule is that the direction of positive is always going to be from the listener to the source. So what happens here is that this is the listener and this is the source, so the direction of positive is always going to be from \(L\) to \(S\), from listener to source. So here is our direction of positive, and what that means is that this \(v_L\) is actually going to be positive. You're always going to write it like this, but depending on your sign, you actually might have to pick up a negative sign if that happens. All right. So let's take a look here. So we're going to plug in some numbers. Our \(f_L\) is 600, this equals 343. Now what happens is, remember, this 343 is always going to be positive. Even though the direction of positive is this way and your velocity, the speed of sound moves to the right, these sound waves, you're still going to actually write the 343 as positive. All right? So we have 343 plus \(v_L\), and then we're going to divide this by 343, and now we're going to multiply this by 550. So let's just go ahead and move this 550 over to the other side. Once you divide this 550, you're going to get 1.091, which equals \(\frac{343 + v_L}{343}\). So what happens is when you move this 343 up over here, you're going to get 374.2 equals 343 plus \(v_L\). All right. So all we have to do is just move this 343 over. We're going to get 374.2 minus 343, and this equals \(v_L\). And if you go ahead and work this out, you're going to get 31.2 meters per second. So that is the speed that you have while you are traveling towards the sound source. All right. So that's it for this one, guys. Let me know if you have any questions.

18. Waves & Sound

The Doppler Effect

18. Waves & Sound

# The Doppler Effect - Online Tutor, Practice Problems & Exam Prep

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### The Doppler Effect

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PRACTICE PROBLEMS AND ACTIVITIES (22)

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- The siren of a fire engine that is driving northward at 30.0 m>s emits a sound of frequency 2000 Hz. A truc...
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- A railroad train is traveling at 30.0 m>s in still air. The frequency of the note emitted by the train whis...
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- A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and...
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- (II) A bat at rest sends out ultrasonic sound waves at 50.0 kHz and receives them returned from an object movi...
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- (II) Two trains emit 508-Hz whistles. One train is stationary. The conductor on the stationary train hears a 4...
- Two loudspeakers face each other at opposite ends of a long corridor. They are connected to the same source wh...
- (II) A police car sounding a siren with a frequency of 1580 Hz is traveling at 120.0 km/h . ...
- (III) A factory whistle emits sound of frequency 770 Hz. The wind velocity is 15.0 m/s from the north (heading...
- (III) A factory whistle emits sound of frequency 770 Hz. The wind velocity is 15.0 m/s from the north (heading...
- (III) A factory whistle emits sound of frequency 770 Hz. The wind velocity is 15.0 m/s from the north (heading...
- (II) A police car sounding a siren with a frequency of 1580 Hz is traveling at 120.0 km/h .(c) The police car ...