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

The given equation of the travelling sound wave is $y = 6.0 \sin(600 t - 1.8 x)$.

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

The wavelength $\lambda$ of the wave is given by:
$\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{1.8} = \dfrac{\pi}{0.9} \text{ m}$

The ratio of the displacement amplitude to the wavelength is:
$\dfrac{A}{\lambda} = \dfrac{6.0 \times 10^{-5}}{\dfrac{\pi}{0.9}} = \dfrac{5.4 \times 10^{-5}}{\pi}$

Using $\pi \approx 3.14$, we get:
$\dfrac{A}{\lambda} = \dfrac{5.4 \times 10^{-5}}{3.14} \approx 1.7 \times 10^{-5}$