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## Physics library

### Course: Physics library>Unit 8

Lesson 5: The Doppler effect

# Doppler effect: reflection off a moving object

Does the doppler effect still apply when sound is being reflected off a moving object? Created by David SantoPietro.

## Want to join the conversation?

• At I don't understand where Vs/(Vs-Vd) came from. Why is Vs on top and Vs-Vd on the bottom?
(19 votes)
• (14 votes)
• @ He writes the formula for the observed frequency, why is this formula different from the doppler effect formula for an observed frequency as mentioned in the other doppler effect formula videos.
(14 votes)
• The reason why you can't view an observer that is approaching a speaker as the speaker that is approaching the observer, is because when a speaker moves towards the observer, the waves don't move faster towards the observer, the observer may get more waves in shorter time, but the speed of the waves themselves is not increasing as the speed of the speaker increasing.
(4 votes)
• I tried to work this out using this formula (which honestly makes a little more sense to me) http://slideplayer.com/slide/5861208/19/images/9/Doppler+Effect+General+Case.jpg

I just wanted to share this in case it helps anyone to have it in more general terms:

Fo = Fs (Vw + Vo)/(Vw - Vs) Because the two Vw cancel on the top and bottom.

The first part assumes the person is the source and the door is the observer. The second part assumes the person is the observer and the door is the source. In both cases they are getting closer.

That really hurt my brain to think about! Does that sound right to everyone?
(8 votes)
• nice summary

on top of your finding, this equation works not only for the case of two bodies getting closer but for the other cases too

if they are at rest, it reduces to Fo=Fs(Vw+0/Vw-0)=Fs*1=Fs
if they are taking apart, the frequency gets smaller as (Vw+Vo) < (Vw-Vs)

one more thing, if you define Vo and Vs as V_moving_body, you can generalize this further and avoid any confusion with observer in Fo
(1 vote)
• Do ships have the same sonar system? That the waves reflect off the enemy ships and tell how close they are?
(4 votes)
• Yes, its the same thing. There's a computer onboard that calculates the distance using these formulas
(8 votes)
• In why the second part of the formula of Fhear is inverted? (Vs/(Vs+-V))? The formula of frequence isn't Fobs = Fs ((Vw+Vs)/Vw)? And the Fdoor isn't just
replacing the Fs in the original formula?
(3 votes)
• the first part (fo=fs((vs+vd)/vs)) is for when the observer (door) is moving with respect to a stationary source of sound (screamer). The second part of the formula (fo=fs(vs/(vs-vd))) is for when the source of the sound (scream coming off door) is moving towards a stationary observer.
(7 votes)
• Does wind affect the velocity of a sound source? and if it does, how to calculate the net velocity of the sound affected by the wind if the wind blows in the same direction with the sound?
(4 votes)
• good question.

You could use the same equations as for the Doppler effect (which usually assume no wind) but
assume that the observer is moving towards the source at the same speed as the wind.
(3 votes)
• I didn't get why he considered the frequency of the scream again while looking for the frequency the person screaming will hear after it being reflected by the door. I thought we could only consider the frequency heard by the door and the fact that the door is moving towards the person screaming.
(4 votes)
• Normally you would think the formula is f(hear) = f(door) * Vs/(Vs-Vd). Now just replace f(door) with the first formula that david gave for f(door). What you are doing in doppler effect formulas is multiplying one frequency with a factor that gives more or less frequency. Its like half of 10 is 5 and half of 5 is 2.5. You can do it all in one by picking half of half of 10 means 1/4 of 10. With variables:
Initial value * a = b (say) , b*c= d thus initial value*a*c=d
again d*e =f thus initial value *a*c*e=f. and so on.
Here you can think of b,d,f as the intermediate frequencies (like fdoor) and a,c,e as the factors (like (vs/vs-vd).
(2 votes)
• I have a question if the frequency observed by the door is v(door)=v(screem)*{(speed(screem) + speed(door))/speed(screem)} =>equation 1
That's okay !
but, Will the speed of the wave emitted by the door be equal to the speed of wave received by the door of the screem?
let's say that the door emitted a wave of speed =speed(door-wave)=s(dw)
then the frequency received by us will be
= v(door)*s(dw)/(s(dw)-speed(door))
so in all it will be (from equation 1)
v(screem)*{(speed(screem) + speed(door))/speed(screem)}*s(dw)/(s(dw)-speed(door))
But in the video it assumed as if the speed of the wave emitted by the door is equal to the speed of the wave received by the door = speed(screem)
So does speed(screem)=s(dw) ?
Does the speed attained by the wave when it strikes a moving object equals the wave's initial speed.
It would be if the obeject was at rest, but it does not seem intuitive to me that the speed will be constant even if the object(on which the wave strikes) is moving.
Thanks...
(2 votes)
• Why f sub hear is equal to f sub scream * (Vs+Vd)/(Vs-Vd)?
I mean why can we just merge them by multiplying?
(2 votes)
• Normally you would think the formula is f(hear) = f(door) * Vs/(Vs-Vd). Now just replace f(door) with the first formula that david gave for f(door). What you are doing in doppler effect formulas is multiplying one frequency with a factor that gives more or less frequency. Its like half of 10 is 5 and half of 5 is 2.5. You can do it all in one by picking half of half of 10 means 1/4 of 10. With variables:
Initial value * a = b (say) , b*c= d thus initial value*a*c=d
again d*e =f thus initial value *a*c*e=f. and so on.
Here you can think of b,d,f as the intermediate frequencies (like fdoor) and a,c,e as the factors (like (vs/vs-vd).
(1 vote)
• Why does the sound reflect instead of propagating through the door like David shows in this video: https://www.khanacademy.org/science/physics/mechanical-waves-and-sound/sound-topic/v/relative-speed-of-sound-in-solids-liquids-and-gases ?
(1 vote)
• When a wave encounters a material of a different stiffness than what it is currently traveling in there is a boundary that will have some amount of reflection and transmission. When a wave is traveling from a a material to a material that has a much greater stiffness the majority of the waves energy will be reflected.
(2 votes)

## Video transcript

-Imagine, you are at the gym. But, you got there at a really bad time. Because there's this big, beefy bodybuilder. He just got told his membership is revoked. Because he's never wiping down his sweat off all the equipment. And he's mad. He's so mad, he takes the door, he rips the door off. Here's the door, and he throws it. He doesn't know where he's throwing it, but he ends up throwing it straight at you. At a speed, we'll call it Vd. You're looking at this door coming towards you. What do you do? I'm not ashamed to say, I might start screaming. I'm like, "Oh, no!" I start screaming at a certain frequency. And I'm going to call that frequency of the scream. So, my question is, here's the thing. This sound is gonna come over to here, it's gonna reflect off the door. Then it's gonna come back towards you. What sound would you hear, after this sound wave reflected off the door and got back to you? Would you hear the same frequency that you're screaming at? Would you hear a higher frequency? Would you hear a lower frequency? Exactly what frequency would you hear? Let's figure it out, it's a good problem. It's tricky. I wanna know, what frequency would I hear? So I'm gonna have to use the Doppler Effect formula. This could be best in moving stuff going on here. So the frequency that I hear-- I'm gonna have to do this in two steps actually. First of all, the sound gets over here to the door. I'm gonna think of the door as an observer first. This door is gonna act as a moving observer. It's gonna hear. I wanna know what sound this door would hear if it was moving this way. Now, doors can't hear anything, but if this door was a person, what frequency would it hear? It wouldn't be f-scream. It'd be something else. So I'm gonna call this f-door. Because I wanna know what the door hears. And if we use the Doppler Effect formula. f-door is gonna equal the frequency that the wave actually has. So, frequency of the scream. Times, the factor of speed of sound up top, plus or minus for a moving observer. The formula was plus or minus, speed of the observer. And my door, is the observer here. Pretending that my door can observe. I wanna know what frequency it receives, that it receives. That it hears. Divided by the speed of sound. Then I have to decide whether I want the plus or the minus. So I ask myself this-- I could never remember this as a student. I always forget, there's too many plus and minuses here. And I have to say to myself. If I'm moving toward the source, or if the source is moving toward me. I know that means frequency that I hear goes up. So the frequency this door experiences should be bigger than the frequency that's actually emitted by the person over here, by me. So if I want it to be bigger. I wanna have plus on top. I want a big numerator. I want the plus sign, because a big numerator gives me a larger frequency. That's the way I remember whether I should add or subtract it here. Alright, so I'm not done yet. You might think we're done. If that's the frequency the door hears. Well, that just reflects off the door. And then I should hear that frequency too. But no, because this is the frequency the door hears. But now the door re-radiates that sound back at you. This door is acting like a speaker, and it's acting a like moving speaker, because the door's moving. We know what happens with moving speakers. There's another Doppler Shift. So there's two Doppler Shifts that go on here. One, by the fact that this door is observing, experiencing a higher frequency. Then, when it re-radiates that sound outward. When it reflects that sound outward. It acts as a moving speaker. So I wanna know the frequency that I will actually hear. So, what frequency when it gets back to me. What frequency will I actually hear? Alright, we gotta do another Doppler Shift. Here we go. The frequency I hear will be the frequency the speaker was trying to radiate, which was this. Remember, this was the frequency, the door was receiving. This was the frequency right here, f-door. So, it's gonna be that times another shift. So, let me just rewrite this over here. But times--I need to multiply this by another factor. I need to multiply this by another factor that takes into account, the fact that the speaker-- this door's not a speaker, but it's acting as a moving speaker, because it's moving. The fact that this door is moving. And that looks like this. That factor, if you remember, speed of sound on top. And on the bottom it's speed of sound plus or minus the speed of the actual speaker. Which is the door here, so I'm gonna put Vd. Now again, we're gonna have to pick plus or minus. Well, same thing. I go through the same thought process. The only thing I like memorizing is the fact that if things come toward each other, the frequency should be increasing. So again, speaker, this door moving toward me. The Doppler Shift should increase the frequency that I hear. So, to increase the frequency. Let's see, this is the denominator here. I want to increase the total amount of frequency. I need to make the denominator smaller. So it seems weird, but I need to have a negative sign down here. I wanna subtract. I wanna subtract because subtracting gives me a smaller denominator. Dividing by a smaller denominator gives me a larger total amount. And I know that I should be getting a larger total amount for the frequency I hear. Because the frequency should get increased by this Doppler Shift. So, this is it. The frequency that I would hear. I would hear a frequency of the amount I scream times, these two factors. Because there's two Doppler Shifts happening here. And, you can clean things up if you want. You can cancel off these Vs's, if you wanted to. Which is fine. But, the important thing is that there's two Doppler Shifts. And, you might think well this is just stupid. I mean, this was a really stupid example. When am I ever gonna find myself in a gym, and some beefy, bodybuilder throws a door at me? Well, it is kinda stupid But, here's a real life example. Instead of a door, imagine this is blood, blood flowing. No, the bodybuilder didn't explode or something. Imagine, instead of a gym this is a vein inside of your body. And, instead of me. This is some sensitive piece of equipment that wants to image, that wants to scan the blood flow and know how fast is the blood flowing. Well, here's a way to do it. You send sound waves in. Those sound waves reflect off of moving objects, and come back at you. If you can tell how much the frequency changed. You know--this is dependent on the speed of the thing that was moving. So, I called it speed of door. But in this case it would be the speed of the blood. You can figure out how fast that blood's flowing in there. That's useful. Or, if you're not a doctor. And you wanted to know--if you're a police officer. There's a car chase, and there's a car heading toward you. You wanna know how fast it goes. This car's coming toward you at some speed. And you wanna know how fast it's going. You could be parked over here. You could use radar, shoot that radar. It wouldn't be sound, this time it's electromagnetic waves. This radar would hit the car. It would bounce back at you, that frequency would shift. It would shift by a certain amount that depends on the speed of the car. This is how they can figure how fast you're going. And maybe give you a ticket. So, slow down out there. Make sure you don't get caught by the radar.