Topic 4.4 Extended C – The Doppler effect Consider the ice-cream truck that cruises through the neighborhood playing that interminable music. We’ll

Topic 4.4 ExtendedC – The Doppler effect


Consider the ice-cream truck that cruises through the neighborhood playing that interminable music.

We’ll use a bell for our sound source, and assume it is ringing at a fixed frequency.The sound waves radiate spherically from the bell, as shown in the planar cross-section.Since the ice cream truck is not yet moving, the wave fronts are spherically symmetric.

Planar cross-section

Symmetric cross-section


We can represent a sound wave as a sine wave - just what you’d see on water if you dropped in a stone.

Whether Dobson is in front of the truck, or behind it, he hears the same frequency.

Dobson

FYI: Sound waves are not, strictly speaking, crests and troughs, like water waves. Sound waves are really alternating continuums of compressed and rarefied air.


Suppose the ice-cream truck is now moving, but ringing the bell at the same rate as before.

Note how the wave fronts bunch up in the front, and separate in the back.The reason this happens is that the truck moves forward a little bit during each successive spherical wave emission.


If Dobson is in front of the moving truck, he will hear a different frequency than if he is behind it.

In fact, as the truck approaches him he will hear a higher than actual frequency.As the truck recedes, he will hear a lower than actual frequency.We call this phenomenon the Doppler effect.


Now we want to look at TWO successive pulses emitted by a source that is MOVING at speed vs:

VS

Pulse 1t = 0

Pulse 1t = T

Pulse 2t = T

The distance between Pulse 1 at t = 0 and Pulse 1 at t = T is given by s = vT. The distance the Source has traveled in time T is given by d = vsT. This is the distance shown here:

s = vT

d = vsT

The wavelength of the sound heard by the observer Dobson is represented with the symbol o.

o

ThusNote that o = s - d.

o = vT - vsT


From o = vT - vsT we can do a little manipulation to get the following result:

o = vT - vsT

o = (v - vs)T

But the frequency perceived by the observer, fo is related to the speed of sound by v = ofo so that

o =v

fo

. But fs = 1/T so that

o = (v - vs)T

v

fo

= (v - vs)1

fs

1

fo

= (v - vs)v

1

fs

fo =v

v - vs

fs

fo = 11 - vs

v

fs

Doppler EffectSource moving TOWARD a stationary observer

FYI: The denominator is less than the numerator so that fo > fs as expected.


If the source is moving AWAY from the observer, a little thought should convince you that

fo =v

v + vs

fs

fo = 11 + vs

v

fs

Doppler EffectSource moving AWAY FROM a stationary observer

FYI: The denominator is more than the numerator so that fo < fs as expected.


If the source is stationary, and the observer is moving toward the source, the effective speed of the emitted wavefronts will be ve = v + vo. Then the frequency heard by the observer is given by

fo =ve

s

= v + vo

s

= v + vo

vfs

If the source is stationary, and the observer is moving away from the source, the effective speed of the emitted wavefronts will be ve = v - vo. Then the frequency heard by the observer is given by

fo =ve

s

= v - vo

s

= v - vo

vfs

fo =v + vo

vfs = 1 + vo

vfs

Doppler EffectObserver moving TOWARD a stationary source

fo =v - vo

vfs = 1 - vo

vfs

Doppler EffectObserver moving AWAY FROM a stationary source

FYI: The multiplier is greater than 1 so that fo > fs as expected.FYI: The multiplier is less than 1 so that fo < fs as expected.


Of course, there is always the situation where both source and observer are moving. Dare we talk about this case? Well, it turns out that the following formula works for ALL situations:

fo =v vo

v vs+fs source and observer moving

The Doppler Effect

(toward means greater)

If vo = 0 the formula reduces to fo = fs.v

v vs+

If vs = 0 the formula reduces to fo =v vo

vfs.


Suppose the ice cream truck is ringing the bell with a frequency of 120 Hz (about the frequency of a fire alarm) and is moving along at 25 m/s. (a) What frequency will Dobson hear if he is standing in front of the truck? Assume the speed of sound is 345 m/s.

We can use our general formula, with vo = 0:

fo =v vo

v vs+fs source and observer moving

The Doppler Effect

(toward means greater)

fo = fs

v

v vs+

FYI: The source is moving TOWARD the observer, so we want a GREATER frequency: USE THE NEGATIVE.

= fs

v

v - vs

= ·120345

345 - 25

= 129 Hz


Suppose the ice cream truck is ringing the bell with a frequency of 120 Hz (about the frequency of a fire alarm) and is moving along at 25 m/s. (b) What frequency will Dobson hear if he is standing behind the truck? Assume the speed of sound is 345 m/s.

We can use our general formula, with vo = 0:

fo =v vo

v vs+fs = fs

v

v vs+

FYI: The source is moving AWAY FROM the observer, so we want a SMALLER frequency: USE THE POSITIVE.

= fs

v

v + vs

= ·120345

345 + 25

= 112 Hz


Suppose the ice cream truck is ringing the bell with a frequency of 120 Hz (about the frequency of a fire alarm) and is parked on the side of the road. What frequency will Dobson hear if he is approaching the truck at 25 m/s on his motorcycle? Assume the speed of sound is 345 m/s.

We can use our general formula, with vs = 0:

fo = fs

FYI: The source is moving TOWARD the observer, so we want a HIGHER frequency: USE THE POSITIVE.

= fs

= ·120345 + 25

345

= 129 Hz

v + vo

vv vo

v


Suppose the ice cream truck is ringing the bell with a frequency of 120 Hz (about the frequency of a fire alarm) and is traveling at 25 m/s. What frequency will Dobson hear if he is approaching the truck from the front at 25 m/s on his motorcycle? Assume the speed of sound is 345 m/s.

We can use our general formula:

FYI: The source and observer are moving TOWARD each other, so we want a HIGHER frequency: USE THE POSITIVE in the top and the NEGATIVE in the bottom.

fo =v vo

v vs+fs

v + vo

v - vs

fs=

= ·120345 + 25345 - 25

= 139 Hz


MACH SPEEDS If the sound source moves faster than the speed of sound, the source passes up the wave front, producing what is called a sonic boom.

vs = 0 vs < v vs = v

Subsonic Mach 1

vs > v

Supersonic

FYI: At subsonic speeds, the wave fronts never cross each other, and thus cannot constructively interfere.

FYI: At supersonic speeds, the wave fronts DO cross each other, and thus CAN constructively interfere.


The vapor cone is explained by the Prandtl-Glauert singularity - a mathematical singularity that occurs in aerodynamics.

MACH SPEEDS

The transonic wave front is rounded rather than pointed


MACH SPEEDS The Prandtl-Glauert singularity is the point at which a sudden drop in air pressure occurs, and is generally accepted as the cause of the visible condensation cloud that often surrounds an aircraft traveling at transonic speeds.One view of this phenomenon is that it exhibits the effect of compressibility and the so-called "N-wave". The N-wave is the time variant pressure profile seen by a static observer as a sonic compression wave passes. The overall three-dimensional shock wave is in the form of a cone with its apex at the supersonic aircraft. This wave follows the aircraft. The pressure profile of the wave is composed of a leading compression component (the initial upward stroke of the "N"), followed by a pressure descent forming a rarefaction of the air (the downward diagonal of the "N"), followed by a return to the normal ambient pressure (the final upward stroke of the "N").


MACH SPEEDS These condensation clouds, also known as "shock-collars" or "shock eggs," are frequently seen during space-shuttle launches around 25 to 33 seconds after launch when the vehicle passes through the area of maximum dynamic air-pressure, or max Q. These effects are also visible in archival footage of some nuclear tests. The condensation marks the approximate location of the shock wave.


MACH SPEEDS Since heat does not leave the affected air mass, this change of pressure is adiabatic, with an associated change of temperature. In humid air, the drop in temperature in the most rarefied portion of the shock wave (close to the aircraft) can bring the air temperature below its dew point, at which moisture condenses to form a visible cloud of microscopic water droplets. Since the pressure effect of the wave is reduced by its expansion (the same pressure effect is spread over a larger radius), the vapor effect also has a limited radius. Such vapor can also be seen in low pressure regions during high–g subsonic maneuvers in humid conditions.


MACH SPEEDS vs > v Supersonic

vt

vst

sin = vtvst sin = v

vs

Line tangent to spherical supersonic wave fronts

FYI: The ratio M = vs / v is called the Mach number.Thus

sin = 1M

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Topic 4.4 Extended C – The Doppler effect Consider the ice-cream truck that cruises through the neighborhood playing that interminable music. We’ll