## Doppler Effect

### Transcript

Now, we'll look at the Doppler Effect. The Doppler Effect tells us that motion changes the perceived frequency of a sound. Now, the perceived frequency is given by f' = f * (v +/- vD)/(v -/+ vS), where f is the emitted frequency, v is the velocity of sound in the medium, vs is the velocity of the source of the sound and vD is the velocity of the detector of the sound.

So if we had a detector for a person standing around, and an emitter or a source, like an ambulance, the velocity of the source is how fast it's moving. The velocity of sound is just whatever might be. So most likely, its the velocity sound in air and then this detector has a velocity vD. Now, if something is moving towards you use the top operator.

If something is moving away you used the bottom operator. So, for example in this drawing here our detector is moving away from the sound. The source of moving towards the detector. So if we were looking at the effect on the perceived frequency we would use the bottom operator, for the detector but the top operator for the source. Because the detector is moving away, we used the bottom.

The source is moving towards, so we use the top. Let's do a few practice problems to master this concept. Let's do our Doppler problems. Suppose a lawyer is chasing an ambulance. What happens to the frequency of the siren he hears if he is following at the same speed?

Now we always want to remember our Doppler equation, f' = f * (v +/- vD)/(v -/+ vS). Now in problems like this, it's often helpful to have a little diagram. So if we have our ambulance, Moving at some speed, vS, this is being followed by our lawyer, Who is moving at some speed, vD.

Now the ambulance is moving away from the detector. But the detector is moving towards the source. So in this problem, we would use the top operator for the detector, and the bottom operator for the source. We can then rewrite the equation, f' = f * (v + vD)/(v + vS).

And we said here that he's following at the same speed. That means that the velocity of the detector equals the velocity of the source. Well, the velocity of sound plus vD over the velocity of sound plus vS, where vD equals vS, would reduce to 1. So the perceived frequency will equal the actual emitted frequency.

This means there will be no change in the frequency of the siren as he's following it. Now if the ambulance stops to pick up someone in need and the lawyer continues his pursuit, what happens to the observed frequency? Again, remember that we want to use the Doppler equation f' = f * (v +/- vD)/(v -/+ vS).

Here, the ambulance has stopped, So its velocity equals 0. The lawyer, however, is still moving at some velocity, vD. The ambulance isn't moving either toward or away, but the lawyer is still moving toward. That means we use the top operator for the detector, and the bottom operator since vS = 0, we can cancel this portion out.

So, we can re write the equation f' = f * (v + vD)/v, Now since v + vD is greater than v, f' or the observed frequency is going to go up. It will be greater than the actual emitted frequency. Finally, the police arrive at the scene and begin to pursue the lawyer. If he races away at twice the speed of the police, what happens to the observed frequency of the police siren?

Again we write out the Doppler equation, f' = f * (v +/- vD)/(v -/+ vS). I'll draw this up. Now the lawyer is moving away and the police car is moving towards. We also know that vD is 2*vS because he's moving away at twice the speed of the police.

We can now look at our original equation, if the lawyer is moving away that makes the detector, the bottom operator. The police car's coming towards the lawyer, so we'd use the top operator on the source. We can rewrite our equation f' = f * (v- vD)/(v- vS), We also know that vD = 2*vS.

So if we look at the top and bottom here v- 2*vS is less than v- vS because we're subtracting a larger number from this velocity of sound. Since the numerator of this fraction is less than the denominator, the perceived frequency is going to go down. So as the lawyer runs away as fast as he can, the police siren will trail off in the distance.

So with Doppler problems just think of what happens in real life when you hear a passing police car or as you approach the scene of an accident.

Read full transcriptSo if we had a detector for a person standing around, and an emitter or a source, like an ambulance, the velocity of the source is how fast it's moving. The velocity of sound is just whatever might be. So most likely, its the velocity sound in air and then this detector has a velocity vD. Now, if something is moving towards you use the top operator.

If something is moving away you used the bottom operator. So, for example in this drawing here our detector is moving away from the sound. The source of moving towards the detector. So if we were looking at the effect on the perceived frequency we would use the bottom operator, for the detector but the top operator for the source. Because the detector is moving away, we used the bottom.

The source is moving towards, so we use the top. Let's do a few practice problems to master this concept. Let's do our Doppler problems. Suppose a lawyer is chasing an ambulance. What happens to the frequency of the siren he hears if he is following at the same speed?

Now we always want to remember our Doppler equation, f' = f * (v +/- vD)/(v -/+ vS). Now in problems like this, it's often helpful to have a little diagram. So if we have our ambulance, Moving at some speed, vS, this is being followed by our lawyer, Who is moving at some speed, vD.

Now the ambulance is moving away from the detector. But the detector is moving towards the source. So in this problem, we would use the top operator for the detector, and the bottom operator for the source. We can then rewrite the equation, f' = f * (v + vD)/(v + vS).

And we said here that he's following at the same speed. That means that the velocity of the detector equals the velocity of the source. Well, the velocity of sound plus vD over the velocity of sound plus vS, where vD equals vS, would reduce to 1. So the perceived frequency will equal the actual emitted frequency.

This means there will be no change in the frequency of the siren as he's following it. Now if the ambulance stops to pick up someone in need and the lawyer continues his pursuit, what happens to the observed frequency? Again, remember that we want to use the Doppler equation f' = f * (v +/- vD)/(v -/+ vS).

Here, the ambulance has stopped, So its velocity equals 0. The lawyer, however, is still moving at some velocity, vD. The ambulance isn't moving either toward or away, but the lawyer is still moving toward. That means we use the top operator for the detector, and the bottom operator since vS = 0, we can cancel this portion out.

So, we can re write the equation f' = f * (v + vD)/v, Now since v + vD is greater than v, f' or the observed frequency is going to go up. It will be greater than the actual emitted frequency. Finally, the police arrive at the scene and begin to pursue the lawyer. If he races away at twice the speed of the police, what happens to the observed frequency of the police siren?

Again we write out the Doppler equation, f' = f * (v +/- vD)/(v -/+ vS). I'll draw this up. Now the lawyer is moving away and the police car is moving towards. We also know that vD is 2*vS because he's moving away at twice the speed of the police.

We can now look at our original equation, if the lawyer is moving away that makes the detector, the bottom operator. The police car's coming towards the lawyer, so we'd use the top operator on the source. We can rewrite our equation f' = f * (v- vD)/(v- vS), We also know that vD = 2*vS.

So if we look at the top and bottom here v- 2*vS is less than v- vS because we're subtracting a larger number from this velocity of sound. Since the numerator of this fraction is less than the denominator, the perceived frequency is going to go down. So as the lawyer runs away as fast as he can, the police siren will trail off in the distance.

So with Doppler problems just think of what happens in real life when you hear a passing police car or as you approach the scene of an accident.