Fig 1.3.5.1 The ambulance moves from A →B →C →D. The wave front emitted at A has spread to A1 and A2; B has spread to B1 and B2; C has spread to C1 and C2; D has spread to D1 and D2.

For reference, the wave as emitted by the siren (yellow)...

Consider an ambulance with the siren emitting at a frequency of 660HZ moving at 33ms^{1} on a level road between two people standing 660m m apart.

What does each person hear?

The person in the ambulance...

660 wave fronts are emitted every second. Since the person is moving with the same speed as the wave fronts he will experience a frequency of 660Hz, with a wavelength λ of 330ms^{1}/660ms^{1}.

f = 660 Hz and

λ = ^{v}/_{f}

λ = 0.5m and

T = ^{1}/_{f}

T = 1.52 X 10^{-3} s.

The person behind the ambulance...

660 wave fronts are emitted every second. Since the source is moving away with a speed of 33ms^{1} he will experience a delay in the wave fronts. The speed of the sound in that direction is [330 - 33] ms^{1}. The wavelength is extended by the distance covered by the ambulance in one period of the source.

Distance covered by the ambulance in one period = [v_{ambulance}] t at constant velocity.

Distance covered by the ambulance in one period =^{[33 ms1]}/_{[660Hz]}

Distance covered by the ambulance in one period = 0.05m

λ_{behind} = 0.5 + 0.05m

λ_{behind} = 0.55m

f_{behind} = [v_{sound}] / [ λ_{behind}]

f_{behind} = [330] / 0.55

f_{behind} = 600Hz.

The person in front of the ambulance...

660 wave fronts are emitted every second. Since the source is moving towards him with a speed of 33ms^{1} he will experience a increase in the wave fronts. The speed of the sound in that direction is [330 + 33] ms^{1}. The wavelength is extended by the distance covered by the ambulance in one period of the 'normal' wave.

Distance covered by the ambulance in one period = [v_{ambulance}] t at constant velocity.

Distance covered by the ambulance in one period = [33 ms^{1}] [660Hz]

Distance covered by the ambulance in one period = 0.05m

λ_{behind} = 0.5 - 0.05m

λ_{behind} = 0.45m

f_{behind} = [v_{sound}] / [ λ_{behind}]

f_{behind} = [330] / 0.45

f_{behind} = 733Hz.

Deriving a formula...

Source moving and observer stationary.

Consider observers at rest at X and Y and the source S moving to the right. If the source is moving directly towards the observer then the new wave front is displaced to the right of the where the previous wave originated... (the yellow dots in fig 1.3.5.2). The wave fronts are heard more quickly than if the source was at rest. the wavelength observed λ is shorter than the wavelength of the source.

Fig 1.3.5.2

During this time the source movesV_{s}T or ^{Vs}/_{f' }which is the amount that the wavelength is shortened.

λ' = λ - Δλ

λ' = λ - ^{Vs}/_{f }

The general formula for frequency is f = ^{V}/_{λ}

So the new frequency f' = ^{V}/_{λ'}

f' = ^{V}/_{[λ-(Vs/f)]}

f' = ^{V}/_{[(v/f)-(Vs/f)]}

f' = f[^{V}/_{[V -Vs]} (source moving toward stationary observer)

If the source is moving directly away from the observer then the new wave front is displaced to the left of the where the previous wave originated... (the yellow dots in fig 1.3.5.2). The wave fronts are heard more slowly than if the source was at rest. the wavelength observed λ is longer than the wavelength of the source.

During this time the source moves V_{s}T or ^{Vs}/_{f' }which is the amount that the wavelength is lengthened.

λ' = λ + Δλ

λ' = λ + ^{Vs}/_{f }

The general formula for frequency is f = ^{V}/_{λ}

So the new frequency f' = ^{V}/_{λ'}

f' = ^{V}/_{[λ+(Vs/f)]}

f' = ^{V}/_{[(v/f)+(Vs/f)]}

f' = f[^{V}/_{[V+Vs]} (source moving away from stationary observer)

Source stationary and observer moving.

Consider a source at rest the observer moving to the left. If the observer was stationary the frequency observed would be f. However, by moving towards the source the relative speed v' of the source is Vs + Vo.

Fig 1.3.5.2

V' = Vs + Vo

Recall V = fλ

f' = ^{V'}/_{λ}

f' = ^{[V + Vo]}/_{λ}

f' = ^{[V + Vo]}/_{[V/f]}

f' = f {^{[V + Vo]}/_{V}} (observer moving toward stationary source)

If the observer is moving away from the source the speed is

V' = Vs - Vo

f' = ^{V'}/_{λ}

f' = ^{[V - Vo]}/_{λ}

f' = ^{[V - Vo]}/_{[V/f]}

f' = f {^{[V - Vo]}/_{V}} (observer moving toward stationary source)

Although the average person will observe the Doppler effect with sound waves, all waves display this phenomenon. For example, astronomers use the effect to determine the speeds of celestial objects relative to the Earth. The Doppler effect is used in police radar systems to measure the speeds of motor vehicles.

We can combine both sets of equations to get:

f' = f {^{[V ± Vo]}/_{[V - Vs]}}

A positive value is used for motion of the observer or the source toward the other (result an increase in observed frequency), and a negative value is used for motion of one away from the other (result a decrease in observed frequency).

The Doppler Effect and Sonic Booms

Stationary Sound Source

Source moving with v_{s} < v_{sound}

Source moving with v_{s} = v_{sound} ( Mach 1 - breaking the sound barrier )

Source moving with v_{s} = v_{sound} (Mach >1 - supersonic)

Sound waves are produced at a constant frequency.

Wavefronts are produced with the same frequency as before.

Wavefronts are produced with the same frequency as before.

Wavefronts are produced with the same frequency as before.

Wavefronts propagate symmetrically away from the source at a constant speed, which is the speed of sound in the medium.

The center of each new wavefront is now slightly displaced to the right.

Jet pilots flying at Mach 1 report that there is a noticeable "wall" or "barrier" which must be penetrated before achieving supersonic speeds. This "wall" is due to the intense pressure front, and flying within this pressure front produces a very turbulent and bouncy ride.

The sound source has now broken through the sound speed barrier. Intense pressure front on the Mach cone that causes the shock wave known as a sonic boom as a supersonic aircraft passes overhead. The shock wave advances at the speed of sound, and since it is built up from all of the combined wave fronts, the sound heard by an observer will be quite intense. A supersonic aircraft usually produces two sonic booms, one from the aircraft's nose and the other from its tail,

Distance between wavefronts is the wavelength.

Wavefronts begin to bunch up in front of source (shorter wavelength than source) and spread further apart behind the source (shorter wavelength than source).

Wavefronts in front of the source are now all bunched up at the same point.

Source is moving faster than the sound waves it creates, it actually leads the advancing wavefront.

All stationary observers will hear the same frequency, which will be equal to the actual frequency of the source.

Stationary observer in front of the source will hear a higher frequency f' > f_{source}, and an observer behind the source will hear a lower frequency f'' f_{source}.

Stationary observer in front of the source will detect nothing until the source arrives. The pressure front will be quite intense (a shock wave), due to all the wavefronts adding together, and will not be percieved as a pitch but as a boom of sound as the pressure wall passes by.

Sound source will pass by a stationary observer before the observer actually hears the sound it creates.

Stationary Sound Source

Sound source is moving to the right with a speed v_{s} = 0.4 v_{sound} (Mach 0.4)

Sound source is moving at the speed of sound in the medium (Mach 1.0). The speed of sound in air at sea level is about 340 ms^{-1} or about 750 mph.

Sound source is moving faster than the speed of sound in the medium (> Mach 1.0). The speed of sound in air at sea level is about 340 ms^{-1} or about 750 mph.

Examples

1. (a) What frequency will you hear if you are at
rest and a police car moves toward you at 25.0 ms^{1} with the siren on at a frequency of 1600 Hz?

(b) What would the frequency be if it moved away from you?

(a) Answer = .....

(b) Answer = .....

2. (a) What frequency will you hear if you are moving at 25.0 ms^{1} towards a police car at
rest with the siren on at a frequency of 1600 Hz?

(b) What would the frequency be if you moved away from the car?

(a) Answer = .....

(b) Answer = .....

3. A bat flying at 5.0 ms^{1} emits a frequency of 20000Hz towards a wall. What will be the shift in frequency as the echo returns to the bat.

Answer = .....

4. You are driving at a speed of ^{}25.0 ms^{1 }and encounter an ambulance at a speed of 30ms^{1 }with siren emitting a sound of frequency 800Hz. What frequency would you hear when

(a) both vehicles are moving directly towards each other?

(b) both vehicles are moving directly away from each other?

Answer = .....

Answer = .....

5. A bat flying at 5.0 ms^{1} emits a frequency of 20000Hz towards a car moving at a speed of 30 ms^{1} directly towards it. What will be the shift in frequency as the echo returns to the bat?