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 correct.
>> 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.
> 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.