Yes — that’s a good intuitive way to express the Heisenberg uncertainty principle.

Formally, the uncertainty relation says:

$$\Delta x \, \Delta p \geq \frac{\hbar}{2}$$

where

• $\Delta x$ is the uncertainty (or spread) in position, and

• $\Delta p$ is the uncertainty (or spread) in momentum.

So, if a particle is very localized — meaning $\Delta x$ is small (you know where it is quite precisely) — then $\Delta p$ must be large (you know less about its momentum).
Conversely, if the particle’s momentum is sharply defined (small $\Delta p$), its position becomes very uncertain (large $\Delta x$).

Put differently:

• A particle spread out over space (like a wide wave) has a well-defined momentum (narrow wavelength distribution).

• A particle concentrated in one place (like a narrow wave packet) is made up of many different momentum components interfering together.

Would you like me to show how this emerges mathematically from the Fourier transform relationship between position and momentum wavefunctions?