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### Course: MCAT > Unit 9

Lesson 15: Sound- Sound questions
- Sound is a longitudinal wave
- Production of sound
- Sound Properties: Amplitude, period, frequency, wavelength
- Speed of Sound
- Relative speed of sound in solids, liquids, and gases
- Decibel Scale
- Why do sounds get softer?
- Ultrasound medical imaging
- Standing waves in open tubes
- Standing waves in closed tubes
- Doppler effect introduction
- Doppler effect formula for observed frequency
- Doppler effect formula when source is moving away
- When the source and the wave move at the same velocity
- Doppler effect for a moving observer
- Doppler effect: reflection off a moving object

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# Doppler effect formula when source is moving away

Continue your exploration of the Doppler effect by looking at the case where the source is moving away from the listener. Understand the formulas for observed frequency when the source is moving towards or away from the observer. Apply these concepts to real-world scenarios like the changing pitch of a moving train's horn. Created by Sal Khan.

## Want to join the conversation?

- so, what will the formula be if the observer is in an angle from the source's direction of motion ( not exactly in front of or behind the source)?(17 votes)
- I would think that it would be cos of angle that the velocity of the source makes with the observer.(4 votes)

- For both of the equations, why must we divide by Vw ?

To = Ts ((Vw - Vs) / Vw)(7 votes)- In this case, Ts(Vw-Vs) is the distance between the source and the observer after the source moved. Mathematically speaking, Time=Distance/Speed; which corresponds with this situation because To is the time, Ts(Vw-Vs), and Vw is the speed of the sound wave.(3 votes)

- What happens if the observer is traveling at the exact same speed as the wave from the source or the source itself? What does the observer experience?(4 votes)
- yeah does the observer seem as if he remains the same as in position or does his angle change with the motion(2 votes)

- Can the observer sitting in the source emitting sound, if traveling faster than the speed of sound, hear the source sound normally?(6 votes)
- No, if the observer is moving faster than the speed of sound, then the wave being emitted will never reach him, assuming his speed is constant. lets say the wave moves at a speed of 8m/s for example but if the source moves at 10m/s the source would be 2 meters in front of the wave within the time interval of 1 second.(2 votes)

- Can this change in frequency also be observed in electromagnetic waves?(3 votes)
- Yes, it's called red-shift or blue-shift, depending on if the frequency is lower or higher.(4 votes)

- Why speed of the source doesn't affect speed of the wave ?

And this is a law or an assumption ?(2 votes)- The speed of a wave depends on the medium it is travelling in. This is a law.

It is a good question though, because the speed of the source is important for other things.

For instance, since the speed of the wave depends on the medium, it is constant if the medium does not change, but the source can speed up. Some aircraft are capable of going faster than the speed of sound. When they do this the actually catch up to the sound waves they emitted earlier, and emit more sound waves at the same time, eventually they can break through its own sound waves, and when this happens a very loud sonic boom can be heard. If the airplane is not strong enough then this sonic boom can even break the plane.(4 votes)

- Why does the frequency of an approaching object seem to be increasing gradually, and then decrease gradually? From the diagram above it seems that frequency suddenly drops as the source passes you by.(2 votes)
- It does suddenly drop. Listen to what it sounds like as a car goes by you. The change is very gradual until the car is close. As it goes by, the change is very fast.(3 votes)

- Will the same formula hold true if the observer is not exactly in the path of source but sideways?(2 votes)
- If you mean the source is moving straight and the observer is stationed at an angle from it, this was an answer given by Andrew M for a similar question:

The same concept holds true but now you just have to use trigonometry to get the velocity component that is parallel to the movement of the source given the angle between the source's path and the listener's location.(3 votes)

- at 8.00 I don't understand why this expression (10ms / 10ms-5ms) isn't evaluated in a way that the 10 m/s over 10 m/s doesn't jcancel out leaving you with 1 / 1-5.

Instead he seems to do the denominator first 10ms-5ms giving him 10ms / 5 ms = 2 cycles. Is their a rule I've missed out when evaluating in fractions? because I always assumed that multiplication and division took priority over addition and subtraction.(1 vote)- If you look at what is written the division line is over both the 10 - 5 so it is 10/(10 -5) not 10/10-5.(5 votes)

- What happens when the source is moving away(2 votes)
- The wave coming from it takes time to travel, so the wavelength to the person observing it appears lower.(2 votes)

## Video transcript

In the last video, we figured
out the formulas for the observed period and frequency
for an observer sitting in the path of the source. So the source is moving
towards the observer. So this is the example where
the train is moving towards you, and you perceive the
train's horn as having a higher pitch or a higher
frequency. And we were able to do that by
doing a thought experiment. Saying, OK, my object
starts here. After one period-- a period is
just a measure of time, but it's the measure of time over
which the source emits a cycle, so it emits a
cycle every period. But after one period, we said,
OK, where is that first wave front, or that first pulse,
or that first crest? And where is the source? Because exactly one period has
passed by, and the source will be ready to emit another
crest or another cycle. So the distance between where
the source is and that front of the crest, or that first
crest, that is going to be the wavelength. Because this next thing that
emits is going to be traveling at the exact same velocity, and
it's going to be separated by that distance, which we
saw is this expression. We said how long will it take
it to travel that distance? Well, it's traveling at
a speed of v sub w. That'll tell you what the
observed period would be for this dude over here. We calculated it right here,
and then the observed frequency is just the
inverse of that. Now let's think about the
situation where the observer is over here. So these equations, or these
formulas that we came up with right here, this is observer--
or let me say source traveling in direction of observer. Now let's think about the
opposite case, where the source is traveling away from
the observer, and in this case, the observer is
that guy over there. Maybe I'll do it in
a different color. He'll be blue. So this is the observer. So when we started off, our
source was right here. After exactly one period from
the source's point of view, that first crest emitted
has traveled radially outward that far. This is the distance. The velocity of the wave times
the amount of time that passed-- velocity times time
is going to give you distance-- and where the source
of the wave will have traveled to the right exactly
this distance. It's velocity times the amount
of time that's gone by. Now, in the last video, we said,
OK, that wave is just passing this guy. How long will it take for that
pulse that's being emitted right then to also reach him? And then that tells us the
period between two pulses or between two crests. Now let's think about that exact
same situation here. That first crest is just
passing this guy. And a period, or the t sub s,
which is a period of the emitted wave has
just passed by. So this guy is just about to
emit another wave, So? That other wave is going
to be right here. So what is the distance between
the crest, or the cycle, or however you want to
think about it, the pulse that is passing him by right now,
what is the distance between that and the pulse, the front
edge of that pulse that is being emitted right
at that moment? Right at that moment. Well, it's going to be this
radius, which is this value. It'll be v sub w times
the period. That is that distance plus the
amount of distance that our source has traveled away
from this guy. So plus v sub s times
the period. So that's what this distance is,
and this is how far apart this wave pulse is going to be
from that one or this crest is going to be from that one. So if he's seeing this first
crest right now, right at this moment, how long will it take
him to see the crest that's being emitted right now, that's
this far away from him? Well, it's that far
away from him. So let me write this down. So the amount of time it takes
for him to see the next crest or the same point in the next
cycle, that's the period. That's the observed period. That's going to be equal
to this distance. The velocity of the wave times
the period from the perspective of the source plus
the velocity of the source, because the source has gotten
that much further away from him, velocity of the source
times the period of the source, so that's how far
the next crest is. And then you divide it by the
speed of the wave, by the speed of each of the crests,
which is just the velocity of the wave. And we can just factor
out the t sub s's here and say this is t sub s times
v sub w, the velocity of the wave plus the velocity of the
source divided by the velocity the wave. So this will be a
larger observed period than if this guy was stationary and
especially if the observer was in the path of the guy. That make sense, because every
time this guy issues a cycle, he is moving a little
bit further away. So every crest the same point
in the cycle is going to be further and further apart, so
you're going to have longer wavelengths, longer periods. And if you want the observed
frequency for that guy over there-- I'll do it in
the same color. The observed frequency for
a guy where the source is traveling away from him is
just the inverse of that. So one over the period, one over
the period, same argument we did there, one over the
period from the point of view of the source is the frequency
of the source. Let me Color code it. So 1 over t sub s is equal to
the frequency of the source. This is the inverse of that. So I'm just taking one
over everything here. So 1 over t sub s is the
frequency of the source, and then we take the inverse of this
over here: the velocity of the wave divided by the
velocity of the wave plus the velocity of the source. So we're done, at least
for the simple cases. Obviously, it becomes a little
more interesting when someone isn't exactly in the direction
of the source or exactly being moved away from, but these are
kind of the two extreme cases. So this is the situation when
it is moving away from you. Now, just to check our math and
maybe make it a little bit concrete in relation to the
video we did where we introduced to the idea of the
Doppler effect, let's actually apply those numbers. So on that video, two videos
ago, we had a situation where the velocity of our source was
5 meters per second to the right, and the velocity of the
wave was 10 meters per second radially outward, and the period
of our wave-- I'll call it the period of our-- let me do
it in another color-- from the point of view of the source
was 1 second per cycle, and the frequency was just
the inverse of that. So 1 cycle per second,
or 1 Hertz, which is a cycle per second. So using those numbers, let's
see if we get to the exact same answer we got in that first
video where we first learned about the
Doppler effect. So let's look at the frequency
from the point of view of this gentleman right here. So the frequency of the source
is going to be 1 cycle per second, 1 Hertz. The velocity of the wave is
10 meters per second. Let me write this. 1 cycle per second. The velocity of the wave is
10 meters per second. The velocity of the wave is 10
meters per second minus the velocity of the source is
5 meters per second. So what's this going
to be equal to? The observed frequency for this
guy right there, is going to be 1 cycle per second times
10 over 10 minus 5, and the meters per second cancel out. Meters per second in the
numerator, meters per second in the denominator. So 10 divided by 10 minus
5, or 10 divided by 5, is going to be 2. So it's going to be 2
cycles per second. And if you want the observed
period for this guy, its going to be the inverse of that,
or it's going to be 1/2 second per cycle. And this is exactly what we got
in the previous video, or actually, it was
two videos ago. Now what about the guy who this
guy's running away from? Well, we'll do the
exact same thing. You have 1 cycle per
second, or 1 Hertz. That's the emitted frequency
from the point of view of the source times the velocity of
the wave divided by the velocity of the wave plus the
velocity of the source, because it's moving
away from him. So it's 10 over 10 plus 5. That's 10 over 15. That is 2/3, so this
is equal to 2/3. The units over here
all cancel out. This was cycles per second. So 2/3 cycle per second, which
confirms the numbers we got in that first video, so that should
make us feel good and it also makes a lot of sense. This guy is going to see the
wave crest more frequently. He's going to observe
a higher frequency. If this is a sound,
a higher pitch. This guy, since each crest or
the cycles are getting spread out, he's going to see them less
frequently, and if this is sound, he's going to
observe a lower pitch.