# Doppler

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### Physics

#### Everything Waves: Sound

• Wave Phenomena: (True for all waves)
• Pipes:

#### Strong Nuclear Force

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We all know the "Doppler 4000", the weather tracking program, but how does Doppler work with sound?

Imagine an ambulance driving in your direction, siren blaring. Its pitch gets higher as it moves toward you, then immediately drops in pitch once it's passed you. Why is that? Like many of life's deep questions, this one can be answered by a quick physics explanation.

Ok, so we have a sound source, emitting sound that looks like ripples in a pond (image above). To any observer the wavelength and the frequency of the waves are the same (think of it as the radius of a circle, as the source is in the center of the circle; the point is an equal distance from a person standing anywhere). What happens when it moves? Once the source starts moving, the waves start to shift in that direction (see below).

As the source travels away from the observer, the sound has a longer wavelength and a low frequency, producing a lower pitch (this is called a red shift). As the source travels towards an observer, the wavelength becomes shorter and the frequency, higher (called a blue shift). The equation for measuring the frequency of a moving source is as follows:

```                  $f1=(1-(u/v))f$
```

Where:

```     f=the actual frequency
f1=the observed frequency
v=the wave speed
u=the observer's speed
```

Note: If observer and the wave are moving in opposite directions, v and u will have the opposite sign. Likewise, if the observer and the wave are moving in the same direction, they will have the same sign.

Therefore, as the ambulence drives towards you, the wavelengths are getting shorter and the frequency is getting higher, so the pitch is higher. Once it passes you, the wavelengths become immediately longer and the frequency lower, and so the pitch drops immediately.

Try solving this problem for practice: A street musician is playing a violin and plays a 440-Hz A. How does this note sound to an observer moving towards him at 5m/s?

Solution: $f1=(1-(u/v))$ $f1=(1+(5m/s / 330m/s))440Hz = 446.666Hz = 450Hz$

A thought experiment: An ambulance comes towards you at 5m/s, and its 440Hz siren thus sounds to be 450Hz to you. You listen for 2 seconds, and thus 900 wavefronts reach your ear. But in that 2 seconds, the siren only produced 880 waverfronts. Where did the extra 20 wavefronts come from?