Uncertainty and wave-trains My textbook and the following extract from feynman's lectures present the same idea regarding wavetrains and uncertainty in their wavelengths. Why is it that a wavetrain confined to some space has an uncertainty in its wavelength or the wave number? Is not a confined wave-train equivalent to a burst of successive pulses which can have a definite wavelength according to their origin or nature of origin. 
Next, i understand that the De Broglie relation relates the uncertainty in wavelength to the uncertainty in momentum but what links the finiteness of the wave-train to the uncertainty in position?
 A: To find out what frequencies a signal contains you Fourier transform it.
If you have an infinite plane wave, $\psi$, then when you Fourier transform it you get a delta function centred on the wave frequency i.e. it contains a single frequency.
However a wave packet of the type shown is the product of an infinite plane wave, $\psi$, and some envelope function, $E$, e.g. a Gaussian. When you Fourier transform this product the frequency distribution you get is the Fourier transform of $\psi$ convolved with the Fourier transform of $E$. $\psi$ tranforms to a delta function but, assuming it's a Gaussian, $E$ transforms to  another Gaussian. The convolution of the two gives a Gaussian i.e. the packet contains a range of frequencies not just a single frequency.
It's because the packet contains a distribution of frequencies that you can't assign it a definite frequency/wavelength. The best you can do is assign an average frequency.
Re the uncertainty in position: the position is given by the envelope function, $E$, as described above. In effect $E$ tells you the probability of finding the particle at some position/time. To get a precise position you need to shrink $E$ to a delta function, but if you do that you find it's Fourier transform goes to a constant i.e. the packet now has an infinitely uncertain frequency and therefore momentum.
A: John is right.
I just want to try to give a more intuitive explanation.
Suppose you plot the sine and cosine of a wave that has a wavelength of exactly 1.
Then of course the peaks and zero-crossings would be exactly 1 unit apart.
Now suppose you put an envelope on that plot, such as multiplying it by $x$.
Here's that plot:

Notice the zero-crossings of the sine curve,
and notice the peaks of the cosine curve.
They are not 1 unit apart. They are shorter, because of the slope of the envelope.
It's not a big effect, for this simple envelope, but you can see it.
So if the envelope is not a straight line, but a curve, the wavelength also varies along the curve.
