Why is Heisenberg uncertainty principle not valid in waves in string? We know from high school physics that when the incident wave is traveling from a  low density region (high wave speed) region towards a high density  (low wave speed) region on a string, the width of the transmitted wave is smaller than initial width of incoming wave.  
If we apply Heisenberg principle $\Delta X \Delta P \ge \hbar/2$ to transmitted wave, the width of the transmitted wave must be bigger than incoming wave because the velocity of the transmitted wave is less than incoming wave, as a result the uncertainty in momentum decreases, the uncertainty in position increases. So Heisenberg principle implies the opposite of the result stated in the first paragraph. Somebody could explain why this logic is wrong.
 A: I think you have misunderstood the meaning of the equation
$$\Delta X \Delta P \geq \hbar / 2 \, .$$
This is not surprising given that the notation used here is really, really misleading.
It should be written like this
$$\sigma_X \sigma_P \geq \hbar / 2 \, .$$
To understand this we have to explain what $\sigma_X$ and $\sigma_P$ mean.
Suppose you have a wave pulse at a specific fixed point in time.
You can describe this pulse as a function of position $f(x)$.
That pulse has some width; it can be very narrow or very sharp.
A common way of characterizing this width is with the variance defined as
$$\text{variance} \equiv \int f(x) (x - \mu)^2 \, dx $$
where $\mu$ is the mean value of $x$ as weighted by $f(x)$, defined as
$$\mu \equiv \int x f(x) dx \, .$$
From now on lets assume we set up the coordinates so that $\mu=0$ and we have
$$\text{variance} \equiv \int f(x)x^2 \, dx \, .$$
Again, the variance is just a characterization of how wide the pulse is.
We also define the "standard deviation" of the pulse as
$$\text{standard deviation} \equiv \sigma_x \equiv \sqrt{\text{variance}} \, .$$
The take-home message here is that $\sigma_x$ is just a measure of the width of the pulse.
See the diagram.
You can also think of this as the "uncertainty in the pulse's position", but that particular interpretation really makes more sense in the quantum case where you have a wave function which represents the probability amplitude for finding a particle at various positions.

The Heisenberg uncertainty principle relates the width of this pulse $\sigma_X$ to the uncertainty in the pulse's momentum (or velocity if you like) $\sigma_P$.
So you see now that the actual speed of the wave is not the thing involved in the uncertainty principle; rather it's the uncertainty in the speed that comes in.
Now to go a little further let's think more about what $\sigma_P$ actually means.
You can re-express the wave function $f(x)$ as a function of wave vector via the Fourier transform
$$\tilde{f}(k) \equiv \int_{-\infty}^\infty f(x) e^{-i k x} \, dx \, . $$
This function tells you how to break the pulse down into waves, each of which has a specific momentum $p = \hbar k$.
The Heisenberg uncertainty principle says precisely that the width of this new function, multiplied by the width of the original position wave function, must be equal to or greater than $\hbar/2$.
Important: If you forget about momentum and talk only about position and wave vector you get a relation which holds for any function $f$ and has absoolutely nothing to do with quantum mechanics:
$$ \sigma_x \sigma_k \geq 1/2 \, .$$
You can think of this as the classical limit of the Heisenberg uncertainty relation if you want, but again it's really just a mathematical statement about the shapes of waves.
A: Your situation is a little confusing since the velocity of the wave packet is not related to the momentum. If you want to speak about the uncertainty principle for classical waves, the 'momentum' is proportional to the inverse wavelength $\lambda^{-1}$. In this situation the speed of the wave does not depend on the wavelength, only things like the density of the medium. So even though the wave speed is slower that has nothing to do with the spread in wavelengths needed to construct a pulse (in a Fourier sense), and so nothing to do with the spread in 'momentum'.
In fact you can convince yourself intuitively that if the wave packet has a shorter $\Delta x$ you will need comparatively more short wavelengths to compose it. So in fact there is comparatatively higher momentum (and spread in momentum) in the shorter wavepacket and the classical version of the uncertainty principle holds.
Edit- What I mean is say you have a wavepacket that is a linear superposition of a bunch of plane waves with different wavelengths. Now say you scale this wavepacket to have half the linear extent $\Delta x$ (you can define this as standard deviation, or the support if it's finite, or however) but the same shape. It is intuitive that this will be the superposition of the same distribution of plane waves each with half of the previous wavelength. Since the wavelength is halved, the 'momentum' $\lambda^{-1}$ is doubled. So that's why momentum increases and a product $\Delta x \Delta p$ appears in the uncertainty relation.
