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Help...The Doppler Effect!

I dont really get it :confused:

How is it used to detect the speed of a moving vehicle (E.g. aeroplane) ? Why are pulses needed? Any help on this would be great :biggrin:

Reply 1

$Avatar for /V\ETALHED$

/V\ETALHED

http://galileo.phys.virginia.edu/classes/109N/more_stuff/flashlets/doppler.htm

Reply 2

MathematicalMind

OP

10

wow thanks..it helps and its fun lol. can any1 else help??

Imagine this: a (large enough) obejct is coming towards you. You have a scanner which sends out some sort of wave with large enough wave lenght (maybe some sort of microwave or radio wave). It can aslo detect the same types of waves.

You aim the device to send out a continuous wave of a particular, known, wavelenght towards the moving object. The waves will head towards the object, hit it and bounce back towards you, where they will be detected by the device.

The Doppler effect will mean the detected waves will be off different wavelengths to the ones sent out- you use the change to work out the speed.

But why is there a change. Well imagine the waves as a series of uniformly spaced out lines moving towards the moving object.As one line hits the moving object it immediately is reflected and moved back in the other direction. Next think about how far the next wave/line has to travel to hit the object. When the first wave hits the object, the second wave is one whole wavelenght distance from the object.

If the object was stationary, it would mean the next wave had one whole wave length in distance to move forward before it hit the object.

But because the object is moving forward, the distance the next wave has to travel is less than a whole wavelength. This is becuase the wave moves some of the distance while the object covers the rest. The second wave is then reflected.

Therefore the reflected waves have a shorter distance between them than they initally did.

A similar thing happens if the object is moving away, only this time, each successive wave has a distance larger than a wavelength to travel before being reflected as the object is moving away.

A similar thing can be described using the frequency instead of wavelength.

Theres is then a formula to help you work out the speed/velocity of the object based on the detected and sent wavelengths and frequencies.

Change in frequency

\Delta f=\frac{fv}{c}=\frac{v}{\lambda}

Observed frequency

f'=f + \frac{fv}{c}

where

f, is the transmitted frequency
v, is the velocity of the transmitter relative to the receiver in meters/second: positive when moving towards one another, negative when moving away
c, is the speed of light in a vacuum \left(3\cdot10^8\right) m/s

\lambda

, is the wavelength of the transmitted wave subject to change

Formulae taken from wikipedia.

From this you can derive the formulae I rememebr using at school for this and have just worked out now:

\frac{v}{c} = \frac{\lambda - \lambda '}{\lambda '}

and

\frac{v}{c} = \frac{f' - f}{f}

where
f = frequency sent out
f' = frequency detected

\lambda

= wavelength sent out

\lambda '

= wavelength detected
v = velocity of object (assuming the detector/emmitter is stationary and the object is moving at constant speed. Positive if moving towards you, negative if away.
c= speed of light.

This all works for waves moving at the speed of light.

Reply 4

MathematicalMind

OP

10

cheers! i mean i understand how to draw a diagram of it and stuff but just how u can use it to actually work out the speed. do you use v = f x lamda? or is there another more complicated formula lol. im not 100% if we need to know the formula for this AS exam but you can tell me anyway lol it will get me ready for A2! :cool:

edit: thanks for the formula...i knew it was something like that lol. i can just use that now if im asked how to work out the speed :smile:

Reply 5

Sidhe

Simply put in a medium such as sound as the source moves towards the observer the waves become more tightly packed ie wavelength is shortened, the more waves per second means the higher the frequency of sound so you can determine by the frequency of a sound the relative speed between you and the vehicle.

Unparseable latex formula:

$f' = \left( \frac{v}{v \pm v_s} \right) f$

v , is the speed of waves in the medium
v_s is the velocity of the sounds source

Speed of vehicle works out as

Unparseable latex formula:

$v\pm v_s=\frac{vf}{f'}$

This also depends on the angle of approach of the vehicle, although most simple examples assume the observer is directly in front and then behind.

Unparseable latex formula:

$ v_{s, r}=v_s\cdot \cos{\theta}$

v_s,r is the radial component.

As a moving ambulance approaches you may notice it's siren pitch is higher and as it passes you its pitch decreases. Using this increase and decrease you can determine the speed of the vehicle.

This effect is much more useful in astronomy where it can be used in light waves to distinguish whether a star is moving towards or away from you.

EDIT: Roger has supplied the equations for light above I notice :smile:

It's important to note as well the frequency of the waves nor the wavelength actually change, the guy in the ambulance wouldn't notice the sirens pitch changes. This is purely due to the relative motion of two objects.

MathematicalMind

cheers! i mean i understand how to draw a diagram of it and stuff but just how u can use it to actually work out the speed. do you use v = f x lamda? or is there another more complicated formula lol. im not 100% if we need to know the formula for this AS exam but you can tell me anyway lol it will get me ready for A2! :cool:

edit: thanks for the formula...i knew it was something like that lol. i can just use that now if im asked how to work out the speed :smile:

High again - I've edited my post above to work out the forumlae I used when at school for this. I couldn't find them anywhere, so spent the time from making that post to editting them to work then out for you :smile:

Reply 7

MathematicalMind

OP

10

Sidhe

This effect is much more useful in astronomy where it can be used in light waves to distinguish whether a star is moving towards or away from you.

Although I suspect that's beyond the scope of your question

cheers...We go onto this stuff in A2...sounds intrestin!

Roger Kirk

High again - I've edited my post above to work out the forumlae I used when at school for this. I couldn't find them anywhere, so spent the time from making that post to editting them to work then out for you :smile:

thanks u hav been a great help and answered so quickly! now i can get on with my revision......... :wink:

Reply 8

Sidhe

MathematicalMind

cheers...We go onto this stuff in A2...sounds intrestin!

Yeah although it's called red shift and blue shift in astronomy.

They use it to calculate the Hubble constant H₀- the rate at which the universe is expanding, and to determine if a star has planets orbiting around it, amongst other things.

Sidhe

Yeah although it's called red shift and blue shift in astronomy.

They use it to calculate the Hubble constant H₀- the rate at which the universe is expanding, and to determine if a star has planets orbiting around it, amongst other things.

Is it also used for looking at speed and direction of rotation in stars?

Reply 10

Morbo

17

Roger Kirk

Is it also used for looking at speed and direction of rotation in stars?

The Doppler shift can be used to measure the radial velocity of binary stars.

Reply 11

teachercol

9

Roger Kirk

Is it also used for looking at speed and direction of rotation in stars?

Yes - and doppler broadening of spectral lines (some molecules move toward and some move away from you so the line is broadened in frequency ) can be used to get a value for temperature of stars.

Reply 12

Morbo

17

teachercol

Yes - and doppler broadening of spectral lines (some molecules move toward and some move away from you so the line is broadened in frequency ) can be used to get a value for temperature of stars.

Stellar surface temperature is usually measured by fitting a blackbody distribution of frequencies to the received spectrum, taking into account any absorption between emission and detection. I've never heard of the Doppler broadening of the lines being used to measure the temperature: I get the feeling that you'd get a much bigger error on that small measurement than when you use the data from all the detected frequencies.

Reply 13

Mohit_C

10

Just a quick question, recessional velocity of a planet when using the equation of the Doppler Effect is the velocity of the planet relative to Earth right?

Reply 14

Morbo

17

Mohit_C

Just a quick question, recessional velocity of a planet when using the equation of the Doppler Effect is the velocity of the planet relative to Earth right?

Yes.

Reply 15

Mohit_C

10

This may have been asked elsewhere but how do you find the actual velocity of the planet? (just out of curiousity as I doubt that would be asked at A level)

Reply 16

Sidhe

Mohit_C

This may have been asked elsewhere but how do you find the actual velocity of the planet? (just out of curiousity as I doubt that would be asked at A level)

Copernicus worked out the orbital periods(siderial periods) of all the known planets from the time it took for a planet to move from the same position in the sky and back to the same position again(synodic period) Using a neat bit of maths.

E = the sidereal period of Earth (a sidereal year, not the same as a tropical year)
P = the sidereal period of the other planet
S = the synodic period of the other planet (as seen from Earth)

S = \frac1{\left|\frac1E-\frac1P\right|}

From knowing planets orbital periods you can calculate not only there distance from the sun but their speed using Keplers laws of planetary motion which he derived from Tyco Brahe's tireless observations. This can be used to calculate the orbital distances travelled by the planet so now you have both an orbital period and distance. Thus simply

speed=\frac {distance}{time taken}

The third law : "The squares of the orbital periods of planets are directly Proportionality to the cubes of the semi-major axis of the orbits."

P^2 \propto a^3

P

= orbital period of planet

a

= semimajor axis of orbit

Newton found a general law using calculus, which could be used to calculate the speed by knowing orbital period. Although kepler's laws on their own would be sufficient to show velocity without a more general rule.

\left({\frac{P}{2\pi}}\right)^2 = {a^3 \over G (M+m)}.

G

= gravitational constant

M

= mass of sun

m

= mass of planet

In general as well you can use this equation.

v=[br]\sqrt{2\left({\mu\over{r}}+{\epsilon}\right)}

\mu

is the standard gravitational parameter

r

is the distance between the orbiting body and the central body

\epsilon

is the specific orbital energy

Reply 17

Mohit_C

10

Ok, looks a bit complicating.

Reply 18

Eau

15

Mohit_C

Ok, looks a bit complicating.

Lol, yeah...

Reply 19

Sidhe

Mohit_C

Ok, looks a bit complicating.

Hehe well that was how we did it in history.

Shows the whole standing on the shoulder of giants deal really from Copernicus to Newton.

It's actually pretty simple now a days because we know lovely little things like the mass of the Sun the mass of the Earth the gravitational constant etc.

Now a days you can work out the speed of a planet from plugging values into simple general equation, but then they didn't know the constants involved there so it wasn't so easy for our ancestors.

Help...The Doppler Effect!

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