Waves and its Characteristics: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Waves and its Characteristics. Understanding the underlying formula is key to arriving at the correct answer below:

Wavelength \(\lambda = \frac{v}{f} = \frac{330}{500} = 0.66\text{ m}\). Path difference \(\Delta x = \frac{\lambda}{2\pi} \Delta \phi = \frac{0.66}{2\pi} \left(\frac{\pi}{6}\right) = 0.055\text{ m}\).