That's question 18 solve it in English

The relationship between wave speed ($v$), frequency ($f$), and wavelength ($\lambda$) is given by the formula:

$$v = f\lambda$$

We need to find $\lambda$, so we rearrange the formula:

$$\lambda = \frac{v}{f}$$

Substitute the given values: $v = 330 \, \text{m/s}$ and $f = 256 \, \text{Hz}$.

$$\lambda = \frac{330 \, \text{m/s}}{256 \, \text{Hz}}$$ $$\lambda \approx 1.2890625 \, \text{m}$$

The phase difference ($\Delta \phi$) between two points separated by a distance $\Delta x$ on a progressive wave is given by the formula:

$$\Delta \phi = \frac{2\pi}{\lambda} \Delta x$$

Substitute the calculated wavelength and the given distance $\Delta x = 0.16 \, \text{m}$:

$$\Delta \phi = \frac{2\pi}{1.2890625 \, \text{m}} (0.16 \, \text{m})$$ $$\Delta \phi = 2\pi \times \frac{0.16}{1.2890625}$$ $$\Delta \phi \approx 2\pi \times 0.12412$$ $$\Delta \phi \approx 0.24824 \pi \, \text{rad}$$

This value is very close to $0.25\pi$, which is $\frac{\pi}{4}$.

To be more precise, let's express the ratio $\frac{\Delta x}{\lambda}$ as a fraction:

$$\frac{\Delta x}{\lambda} = \frac{0.16}{330/256} = \frac{0.16 \times 256}{330} = \frac{40.96}{330}$$

This fraction is approximately 0.12412.
If we consider the options, $\frac{\pi}{4}$ corresponds to a ratio of $\frac{1}{8} = 0.125$. The calculated value is very close to this.

The final answer is $\text{C}$.