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## Class 11 Physics (India)

### Course: Class 11 Physics (India) > Unit 19

Lesson 7: The Doppler effect- Doppler effect introduction
- Doppler effect: Wavefront diagrams and word problems
- Doppler effect review
- 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 for observed frequency

The Doppler effect is the change in frequency of a wave as the source moves relative to an observer, and explains why the pitch of a sound sometimes changes as it moves closer or further to or from an observer. Learn how to derive the formula for perceived frequency of a sound using an equation that accounts for a sound and observer’s velocities relative to one another. Created by Sal Khan.

## Want to join the conversation?

- But, isn't the velocity of the wave for moving particle the sum Vw and Vs, because the speed of the wave would also include the speed of the object emitting the light?(25 votes)
- The speed of the sound wave is not the speed of the particles moving, but the speed at which the peaks (compressed air) propagate through the medium that the sound is travelling in. It is not related to the speed of the emitting source. If you think this is unclear, you should watch the earlier videos in this series.(4 votes)

- What happens if the wave source is moving faster than the waves themselves?(11 votes)
- Since the source is moving faster than the sound waves it creates, it actually leads the advancing wavefront. In the case of Sound, The sound source will pass by a stationary observer before the observer actually hears the sound it creates. " SONIC BOOM" is heard and seen when source or object breaks the speed of sound(15 votes)

- im still really confused... can someone explain the Doppler effect in simple terms??(17 votes)
- When the sound source is approaching the observer or when the observer is approaching the sound source, the apparent frequency of the sound source increases. You notice this when a train whistling is approaching you. The apparent frequency of the whistling train is larger compared to the frequency of the whistling when both you and the train are stationary. However if the train is approaching you, then you'll notice an apparent increase in frequency, which is perceived by the higher pitch of the train's whistling sound, keeping in mind that higher pitch is associated with higher frequency sound.(5 votes)

- What would happen if the source travels faster than the wave?(6 votes)
- this can only happen with sound waves in which one experiences a sonic boom by the pressure difference at the front and rear of a aircraft. since other electromagnetic waves travel at the speed of light,no source can travel faster thus making such a situation impossible(7 votes)

- What happens when things approach the speed of light(4 votes)
- I didn't understand how the speed of light is slowed down? Can you explain in more detail.(0 votes)

- Can't we write To = Ts [ ( Vw - Vs ) / Vw ] as

To = Ts [ ( 1 - Vs / Vw ) ] ?(4 votes)- In fact, this is the best way to express this equation(1 vote)

- I didn't exactly understand how Sal represented the crest of the waves in the form of circles. Please help..(3 votes)
- It's like when you throw a stone in a pond. It makes circular ripples that spread out from the stone.(6 votes)

- Is Sal saying that the time taken by source and time taken by wave both are equal? because in calculating its distance he has used the same time(T sub S). And if the time is different how is the calculation going to differ?(4 votes)
- He just calculated the distance after a fixed time to both the source and the wave.(1 vote)

- Sal says that in the last video he showed the formula for doppler effect when the object is moving towards you. Where is that video(2 votes)
- Why do we divide (VwTs - VsTs) by Vw? I didn't understand that in the vidoe.(2 votes)
- Well the time equals the distance divided by the speed, so if (VwTs - VsTs) = the distance then Vw = the speed and T, the period equals the time… Do you have any other questions?(4 votes)

## Video transcript

I've got this source of a wave
right here that's moving to the right at some velocity. So let's just say that the
velocity of the source-- let's call it v sub s to the right--
so we're really going to do what we do in the last video,
but we're going to do it in more abstract terms so we can
come up with a generalized formula for the observed
frequency. So that's how fast he's moving
to the right, and he's emitting a wave. Let's say the
wave that he's emitting-- so the velocity of wave--
let's call that v sub w radially outward. We've got to give a magnitude
and a direction. So radially outward. That's the velocity of the wave,
and that wave is going to have a period and a
frequency, but it's going to have a period and a frequency
associated from the point of view of the source. And we're going to
do everything. This is all classical
mechanics. We're not going to be talking
about relativistic speed, so we don't have to worry about all
of the strange things that happen as things approach
the speed of light. So let's just say it has
some period of-- let me write it this way. The source period, which is the
period of the wave from the perspective of the source,
so the source period, we'll call it t sub source, And the
source frequency, which would just be-- we've learned,
hopefully it's intuitive now-- would be the inverse of this. So the source frequency would
be-- we'll call it f sub s. And these two things are the
inverse of each other. The inverse of the period
of a wave is its frequency, vice versa. So let's think about what's
going to happen. Let's say at time equal zero,
he emits that first crest, that first pulse, so he's
just emitted it. You can't even see it because
it just got emitted. And now let's fast forward
t seconds. Let's say that this is in
seconds, so every t seconds, it emits a new pulse. First of all, where is
that first pulse after t sub s seconds? Well, you multiply the velocity
of that first pulse times the time. Velocity times time is going
to give you a distance. If you don't believe me, I'll
show you an example. If I tell you the velocity is 5
meters per second, and let's say that this period is 2
seconds, that's going to give you 10 meters. The seconds cancel out. So to figure out how far that
wave will have gone after t sub s seconds, you just multiply
t sub s times the velocity of the wave. And let's say it's
gotten over here. It's radially outward. So, I'll draw it radially
outward. That's my best attempt
at a circle. And this distance right here,
this radius right there, that is equal to velocity
times time. The velocity of that first
pulse, v sub w, that's actually the speed. I'm saying it's v sub
w radially outward. This isn't a vector quantity. This is just a number
you can imagine. v sub w times the period,
times t of s. I know it's abstract, but just
think, this is just the distance times the time. If this was moving at 10 meters
per second and if the period is 2 seconds,
this is how far. It will have gone 10 meters
after 2 seconds. Now, this thing we said
at the beginning of the video is moving. So although this is radially
outward from the point at which it was emitted, this thing
isn't standing still. We saw this in the last video. This thing has also moved. How far? Well, we do the same thing. We multiply its velocity times
the same number of time. Remember, we're saying what does
this look like after t sub s seconds, or some period
of time t sub s. Well, this thing is moving
to the right. Let's say it's here. Let's say it's moved
right over here. In this video, we're assuming
that the velocity of our source is strictly less than the
velocity of the wave. Some pretty interesting things happen
right when they're equal, and, obviously, when
it goes the other way. But we're going to assume that
it's strictly less than. The source is traveling slower
than the actual wave. But what is this distance? Remember, we're talking
about-- let me do it in orange as well. This orange reality is what's
happened after t sub s seconds, you can say. So this distance right here. That distance right there--
I'll do it in a different color-- is going to be the
velocity of the source. It's going to be v sub
s times the amount of time that's gone by. And I said at the beginning,
that amount of time is the period of the wave. That's
the time in question. So period of the wave t sub s. So after one period of the wave,
if that's 5 seconds, then we'll say, after 5 seconds,
the source has moved this far, v sub s times t sub s,
and that first crest of our wave has moved that far,
V sub w times t sub s. Now, the time that we're talking
about, that's the period of the wave
being emitted. So exactly after that amount of
time, this guy is ready to emit the next crest.
He has gone through exactly one cycle. So he is going to emit
something right now. So it's just getting emitted
right at that point. So what is the distance between
the crest that he emitted t sub s seconds ago or
hours ago or microseconds ago, we don't know. What's the distance between this
crest and the one that he's just emitting? Well, they're going to move at
the same velocity, but this guy is already out here, while
this guy is starting off from the source's position. So the difference in their
distance, at least when you look at it this way, is the
distance between the source here and this crest. So what is this distance
right here? What is that distance
right there? Well, this whole radial
distance, we already said, this whole radial distance is
v sub w, the velocity of the wave, times the period of the
wave from the perspective of the source, and we're going to
subtract out how far the source itself has moved. The source has moved in the
direction, in this case, if we're looking at it from
this point of view, of that wave front. So it's going to be minus v
sub s, the velocity of the source, times the period of the
wave from the perspective of the source. So let me ask you a question. If you're sitting right here, if
you're the observer, you're this guy right here, you're
sitting right over there, and you've just had that first
crest, at that exact moment that first crest has passed you
by, how long are you going to have to wait for
the next crest? How long until this one that
this guy's emitting right now is going to pass you by? Well, it's going to have
to cover this distance. It's going to have to
cover that distance. Let me write this down. So the question I'm asking is
what is the period from the point of view of this observer
that's right in the direction of the movement of the source? So the period from the point
of view of the observer is going to be equal to the
distance that the next pulse has to travel, which is that
business up there. So let me copy and paste that. So it's going to be that. Let me get rid of that. It shouldn't look like an equal
sign, so I can delete that right over there. Or a negative sign. So it's going to be this
distance that the next pulse is going to travel, that one
that's going to be emitted right at that moment, divided by
the speed of that pulse, or the speed of the wave, or the
velocity the wave, and we know what that is. That is v sub w. Now this gives us the period
of the observation. Now. If we wanted the frequency-- and
we can manipulate this a little bit. Let's do that a little bit. So we can also write this. We could factor out the
period of the source. So t sub s we could
factor out. So it becomes t sub s times the
velocity of the wave minus the velocity of the source, all
of that over the velocity of the wave. And so just like
that, we've gotten our formula for the observed period for this
observer who's sitting right in the path of this moving
object as a function of the actual period of this wave
source, the wave's velocity and the velocity
of the source. Now, if we wanted the frequency,
we just take the inverse of this. So let's do that. So the frequency of the
observer-- so this is how many seconds it takes for him
to see the next cycle. If you want cycles per second,
you take the inverse. So the frequency of the observer
is just going to be the inverse of this. So if we take the inverse of
this whole expression, we're going to get 1 over t sub
s times v sub w over the velocity of the wave minus
the velocity the source. And of course, 1 over the period
from the point of view of the source, this
is the same thing. This right here is the
same thing as the frequency of the source. So there you have it. We have our two relations. At least if you are in the path,
if the velocity of the source is going in
your direction, then we have our formulas. And I'll rewrite them, just
because the observed period of the observer is going to be the
period from the point of view of the source times the
velocity of the wave minus the velocity of the source-- that's
the velocity of the source-- divided by the velocity
of the wave itself. The frequency, from the point
of view of this observer, is just the inverse of that,
which is the frequency. The inverse of the period is the
frequency from the point of view of the source times
the velocity of the wave divided by the velocity
of the wave minus the velocity of the source. In the next video, I'll do the
exact same exercise, but I'll just think about what happens to
the observer that's sitting right there.