Question:medium

If \(A_b\) and \(A_s\) denote the amplitudes of the body and surface waves, respectively, at a distance, \(r\), from the source, then the relation between them is given by

Show Hint

Body waves spread over an expanding sphere (amplitude ~1/r); surface waves spread over an expanding circle near the surface (amplitude ~1/√r).
Updated On: Jul 21, 2026
  • \( \dfrac{A_b}{A_s} \propto \sqrt{r} \)
  • \( \dfrac{A_b}{A_s} \propto r \)
  • \( \dfrac{A_b}{A_s} \propto \dfrac{1}{r} \)
  • \( \dfrac{A_b}{A_s} \propto \dfrac{1}{\sqrt{r}} \)
Show Solution

The Correct Option is D

Solution and Explanation

Alternate route using geometrical-spreading exponents directly. For an isotropic point source, amplitude decay with distance follows \(A \propto r^{-n}\), where the exponent \(n\) is fixed by the dimensionality of the spreading wavefront: \(n=1\) for a spherical (3-D, body-wave) wavefront and \(n=1/2\) for a cylindrical (2-D, surface-wave) wavefront. This is a standard result from ray theory / energy conservation over expanding wavefronts of different geometry, and does not need to be re-derived from Green's functions each time — it is tabulated for body waves (spherical), surface waves (cylindrical) and guided/channel waves (planar, \(n=0\)).

So directly, \(A_b \propto r^{-1}\) and \(A_s \propto r^{-1/2}\), giving

\[ \frac{A_b}{A_s}\propto r^{-1}\big/r^{-1/2} = r^{-1/2} = \frac{1}{\sqrt r} \]

confirming option (D) without needing to separately invoke energy-density arguments.

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