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# Doppler effect: reflection off a moving object

Does the doppler effect still apply when sound is being reflected off a moving object? Explore the Doppler Effect with two shifts in frequency. Understand how sound waves reflect off moving objects and change frequency, using the Doppler Effect formula. This knowledge is crucial in medical imaging and radar speed detection. Created by David SantoPietro.

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• At I don't understand where Vs/(Vs-Vd) came from. Why is Vs on top and Vs-Vd on the bottom?
• @ 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.
• 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.
• 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?
• 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?
• Yes, its the same thing. There's a computer onboard that calculates the distance using these formulas
• 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?
• 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.
• 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?
• 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.
• 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.
• 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).
• im just wondering does it not depend what the door is made of??
• 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...