To generate a Rayleigh wave in aluminum, the required incident angle is 57 degrees given Va = 3.1 x 10^5 cm/s and Vp = 2.6 x 10^5 cm/s.

• A. 37 degrees
• B. 57 degrees
• C. 75 degrees
• D. 48 degrees

Answer: (B)

A Rayleigh wave is excited when the refracted wave inside the solid travels along the surface, which happens at a grazing (nearly parallel) angle. Using Snell’s law at the interface, the tangential components must match, and in the grazing limit the transmitted P-wave inside the aluminum makes an angle of 90° with the normal. This leads to a simple relation between the incident angle and the velocities: sin(incident) = Vp / Va, where Va is the Rayleigh-wave velocity along the surface and Vp is the P-wave velocity in the metal.

Plugging in the given speeds, Vp / Va = 2.6 × 10^5 / 3.1 × 10^5 ≈ 0.8387. Taking the arcsin gives an incident angle of about 56.9°, which rounds to 57°. This angle ensures the P-wave is refracted along the surface, efficiently generating the Rayleigh wave rather than propagating as a bulk wave.