Question:medium

Given the Rayleigh wave velocity \(V_r\), shear wave velocity \(V_s\) and the P-wave velocity \(V_p\), which of the following relationships is/are CORRECT?

Show Hint

Rayleigh waves must decay evanescently with depth relative to both P and S potentials, forcing Vr below the smaller of the two body-wave speeds; for a Poisson solid, Vr is about 0.92 Vs.
Updated On: Jul 21, 2026
  • \(V_r < V_s < V_p\)
  • \(V_s < V_r < V_p\)
  • \(V_s < V_p < V_r\)
  • \(V_s = V_r < V_p\)
Show Solution

The Correct Option is A

Solution and Explanation

An alternative, non-algebraic way to fix the ordering is to use the physical requirement that a surface wave must be evanescent (its amplitude must decay with depth away from the free surface) with respect to BOTH the P and S wavefields it is built from.

A wave of horizontal phase velocity \(c\) has a vertical wavenumber \(k_z = k\sqrt{(V/c)^2 - 1}\) for a body wave of velocity \(V\). For this vertical wavenumber to be imaginary (giving exponential decay with depth, which is what makes it a surface wave rather than a wave that radiates energy into the half-space), we need \(c < V\) for that wave type. Since the Rayleigh wave must decay evanescently in terms of both its P-wave and SV-wave potentials, its phase velocity must be less than the SLOWER of the two body-wave velocities, i.e. less than \(V_s\) (because \(V_s < V_p\) always). Hence \(V_r < V_s\) is not a coincidence of the Poisson-solid root -- it is a structural requirement for the wave to exist as a surface wave at all.

Combined with the universal elastic-solid inequality \(V_s < V_p\) (shear modulus alone is always less than the P-wave modulus \(K + 4\mu/3\)), the only consistent ordering is

\[ V_r < V_s < V_p \]

which is option (A). Options (B), (C) and (D) all place \(V_r\) at or above \(V_s\), which would make the Rayleigh wave a radiating (non-evanescent, non-surface) wave -- physically inconsistent with it being a surface wave at all.

\(\boxed{\text{Answer: (A)}}\)

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