Here's a 2-minute, mathematically accurate explanation of the Uncertainty Principle:
Particles are singular points, which means that they have one and only one location is space (a Dirac delta function, if you know about that sort of thing). Idealized waves are infinite in extent (think about the graph of a sine function carrying on towards infinity). In quantum mechanics, all of the information about the system is contained within its wave function (which is usually more like a sine wave than a Dirac delta function).
However, the wave function can take on a range of shapes between a sine wave and a Dirac delta function. The more DDF-like the wave function is the better we "know" its position. The more sine-like the wave function is the better we "know" its momentum. Now, in truth, we don't actually know the particle's position or momentum in either case, and the only way to find out is to do a measurement and thus collapse the wave function. But in the more localized cases (DDF-like) the range of possible positions is small while the range of possible momenta is big. Conversely, in the sine-like cases the range of possible momenta is small while the range of possible positions is big.
In fact, these ranges are closely linked through a process called the Fourier Transform of the wave function. And from the mathematical analysis of Fourier Transforms it can be deduced that the "width" of the position wave function times the "width" of its Fourier Transform (i.e. the momentum wave function) must be no less than $\frac{\hbar}{2}$. Thus,
$$
\sigma_x\sigma_p\geq\frac{\hbar}{2}
$$
where the sigmas represent the standard deviations of their respective wave functions.
EDIT: To address @ZeroTheHero from the comments. I agree that the way I've presented this material is fairly tightly focused towards the position-momentum uncertainty relation, but that's what OP was asking about and it is the best example for developing a meaningful intuition about this property of quantum mechanics.
That said, there is nothing analogical about the Fourier Transform explanation. It is exact, and it is easy to show its exactness. Here is the proof:
To show the inequality, it suffices to show that the minimum uncertainty state satisfies the equality (i.e. $\sigma_x\sigma_p=\frac {\hbar}{2}$). The minimum uncertainty state is known to be a Gaussian with arbitrary mean and standard deviation (see Shankar or Griffiths for that proof). Without loss of generality, let's examine the normalized position-space state
$$\psi (x)=\frac{1}{\pi^{\frac{1}{4}}}e^{-\frac{x^2}{2}}
$$
which has a corresponding Fourier Transform (that also happens to be its momentum-space wave function)
$$\phi (p)=\frac{1}{\pi^{\frac{1}{4}} \sqrt{\hbar}} e^{-\frac{p^2}{2\hbar^2}}
$$
From here,
$$\sigma_x^2=\frac{1}{\sqrt {\pi}}\int^{\infty}_{-\infty} x^2 e^{-x^2} dx =\frac {1}{2}\\ \ \ \\ \sigma_p^2=\frac{1}{\hbar\sqrt {\pi}}\int^{\infty}_{-\infty} p^2 e^{-\frac{p^2}{\hbar^2}} dp=\frac{\hbar^2}{2} \\ \ \ \\ \therefore \sigma_x\sigma_p=\frac {\hbar}{2}
$$
