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Topic: Changing frequency by changing speed
Replies: 7   Last Post: Feb 3, 2005 3:28 PM

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Posts: 6
Registered: 12/13/04
Changing frequency by changing speed
Posted: Feb 3, 2005 4:58 AM
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I would like to see if anyone can comment on these two
opposing points argued below. The first argument denoted
by double arrows `>>` states that a shorter wavelength
can have its frequency adjusted to equal the frequency
of a slightly longer wavelength simply by slowing
down the speed of the shorter wavelength. Implicit in
this argument is the fact that it is possible to reduce
the observed frequency of a wavelength by reducing
its speed.

The second argument at the bottom denoted by single
`>`claims that this is not true and that it is impossible
to adjust the speed of a shorter wavelength so that its
observed frequency matches that observed from a slightly
longer wavelength.
I would like to see if anyone can say which argument is

>> Here look at this example.
>> lets say we have the shorter wavelength as .8 meter
>> and the longer wavelength at 1 meters
>> so we then get this visual below. A is longer
>> than B, X is the observor point. Direction
>> of movement is r-l
>> X meters
>> A 0 1 2 3 4 <<
>> a b c d e
>> B 0 .8 1.6 2.4 3.2 4 <<
>> f g h i j k
>> If they both travel at the same speed lets
>> say 1 meter a second then they are not at the same
>> frequency as B is observed to have 5 peaks per sec
>> and A is observed to have 4 per peaks per second.
>> However if B travels at .8 meters a second then
>> they will both have the same frequency observed at
>> X. So now, every second, 1 peak from A passes X and
>> every second 1 peak from B passes X.
>> So, where A`s speed is 1 m/s and B`s speed
>> is .8 m/s, we get the following using the letters a-k
>> under each wave denoted above,...
>> at 0 seconds a and f pass X
>> at 1 second b and g pass X
>> at 2 seconds c and h pass X
>> at 3 seconds d and i pass X etc etc...
>> Both wavelengths now have the same frequency!
>> You are wrong to say this is not possible.

>Let me try a sketch too. 'a' is the series
>of peaks at frquency A, is those for B. The
>horizontal scale is time, not distance.

>As emitted:

> time ->
> a a a a a a
> b b b b b b b

>As received - equal speeds:

> |
> |---D/v-->|
> |
> a a a a a a
> b b b b b b b

>As received - different speeds:

> | | |
> | | |---D/va->|
> | | |
> | |---D/va->| |
> | | |
> |---D/va->| | |
> | | | |
> | a a a a a a
> | b b b b b b b
> |
> |---D/vb-->|
> |---D/vb-->|
> |---D/vb-->|
>Each letter 'b' in this diagram is delayed by
>the same amount, D/vb from the time when it
>was transmitted.

>In the above, D is the distance from the source,
>va is the speed for frequency A and vb is the
>speed for frequency B.

>I have drawn in three of the delays for A and B
>in the bottom diagram and you could add them
>for each peak but it would get too cluttered
>in this post. Notice how the delay has
>aligned one pair of peaks that would otherwise
>have missed but thereafter they diverge again.

thanks ahead


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