I read the following from wikipedia:
When a string is plucked normally, the ear tends to hear the fundamental frequency most prominently, but the overall sound is also colored by the presence of various overtones (frequencies greater than the fundamental frequency).
When I pluck a string, I just notice a node at each end and an antinode at the middle. How can we have overtones in addition to the fundamental frequency? It seems counterintuitive for me.
The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular.
If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The Fourier components correspond to the overtones.
Eigenmodes of a string have sinusoidal spatial form $f_m(x) = C_m \sin(\pi m x/L)$, where $x$ is the parallel coordinate and $L$ is the length of the string. Plucking a string at a fixed location $x_0$ means giving it a non-sinusoidal initial perturbation, e.g., something like a piece-wise linear function, $f(x) = A x/x_0$ for $x \le x_0$ and $f(x) = A (L-x)/(L-x_0)$ for $x \ge x_0$. Expanding the initial perturbation $f(x)$ in eigenmodes $f_m(x)$ shows how much each harmonic is excited initially, in general it would be a full spectrum of eigenmodes.
You start with a triangular form, which has its fourier series. Let's say the initial shape is $f(x)$:
$$f(x)=\sum_n a_n \sin \frac{\pi n x}{L}$$ where $n$ counts the modes ($1$ is fundamental). So initial shape determines the harmonic content $a_n$ of certain modes. If you pluck in the middle, you will put more of the fundamental into the initial spectrum, and in every odd mode, but no even modes at all. If you pluck near the end of the string (like on a guitar), you get all modes, with higher modes still less prominent.
This shape then evolves in time. Every mode has a frequency, related to the wavelength:
$$f(x,t)=\sum_n a_n \sin \frac{\pi n x}{L}\cos \omega_n t$$ $\omega_n$ is proportional to $n$ (determined by $\omega=kc$, where $k=2\pi/\lambda$ the wavevector of the mode and $c$ is the speed of propagation of waves on the string).
This would make motion completely periodic and the shape would return to initial form after one cycle of the fundamental frequency. However, vibration is always damped: some small amount by radiating sound into the air, and mostly, by conversion into heat (the string not perfectly elastic). Higher frequency modes are always much more damped: if you demand very fine-detailed curly vibrations, this will get damped a lot because curvature of the string is high. Usually, you get something like that: $$f(x,t)=\sum_n a_n e^{-t/\tau_n} \sin \frac{\pi n x}{L}\cos \omega_n t$$
where $\tau_n$ is the characteristic damping time of $n$th mode. You could say that the intensity of spectral components falls off faster for higher overtones and the tone is getting more sinusoidal. Very high harmonics associated by the sharp triangular kink at the plucking position just produce a "plunk" sound and disappear almost instantly.
Damping higher frequencies always has a smoothing effect on the shape: with time, the shape gets rounded, without sharp corners.
You are not hearing the string but the resonating body and the string. An acoustic guitar resonates with its plucked strings. The colour of the sound is determined by its formant, and what you hear as the fundamental frequency may in fact be some multiple of that, with both under and overtones, depending on the type of resonating body. The resonating bodies of a guitar and a cello (for example) have different formants and therefore result in a different spectral envelope and a different timbre, for the same fundamental frequency.
