A wave travelling on a string has the equation y = (0.10 mm) [(31.4...

The standard equation of a travelling wave is given by $y = A \sin(kx + \omega t)$.

Comparing the given equation $y = (0.10 \text{ mm}) \sin[(31.4 \text{ m}^{-1})x + (314 \text{ s}^{-1})t]$ with the standard equation, we get:
Amplitude $A = 0.10 \text{ mm}$
Angular frequency $\omega = 314 \text{ s}^{-1}$

The maximum displacement of a portion of the string is its amplitude, which is $A = 0.10 \text{ mm}$.

The maximum speed of a portion of the string (maximum particle velocity) is given by v_{\text\max} = A\omega.

Substituting the values:
v_{\text\max} = (0.10 \text{ mm}) \times (314 \text{ s}^{-1}) = 31.4 \text{ mm s}^{-1}

Converting to $\text{cm s}^{-1}$:
v_{\text\max} = 3.14 \text{ cm s}^{-1}

Thus, the maximum displacement is $0.10 \text{ mm}$ and the maximum speed is $3.14 \text{ cm s}^{-1}$.