A travelling sound wave is described by the equation y = 6.0 (600 t...

The given equation of the wave is $y = 6.0 \times 10^{-5} \sin(600 t - 1.8 x)$.

Comparing with the standard wave equation $y = A \sin(\omega t - k x)$, we get:
$A = 6.0 \times 10^{-5} \text{ m}$
$\omega = 600 \text{ rad/s}$
$k = 1.8 \text{ m}^{-1}$

The velocity amplitude of the particles is given by $v_m = A \omega$.

The wave speed is given by $v = \dfrac{\omega}{k}$.

The ratio of the velocity amplitude of the particles to the wave speed is:
$\text{Ratio} = \dfrac{v_m}{v} = \dfrac{A \omega}{\omega / k} = A k$

Substituting the values:
$\text{Ratio} = (6.0 \times 10^{-5}) \times 1.8 = 10.8 \times 10^{-5} = 1.08 \times 10^{-4} \approx 1.1 \times 10^{-4}$