Question

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

• A. \( \dfrac{A_b}{A_s} \propto \sqrt{r} \)
• B. \( \dfrac{A_b}{A_s} \propto r \)
• C. \( \dfrac{A_b}{A_s} \propto \dfrac{1}{r} \)
• D. \( \dfrac{A_b}{A_s} \propto \dfrac{1}{\sqrt{r}} \)

Hint:

Body waves spread over an expanding sphere (amplitude ~1/r); surface waves spread over an expanding circle near the surface (amplitude ~1/√r).

Answer 1

The Correct Option is D

Body waves radiate outward from a point source through the full 3-D volume of the Earth, so at distance \(r\) their energy is spread over the surface of an expanding sphere, whose area grows as \(4\pi r^2 \propto r^2\). By conservation of energy, the energy flux (energy per unit area) falls off as \(1/r^2\):

\[ E_b \propto \frac{1}{r^2} \]

Since seismic energy density is proportional to the square of the displacement amplitude, \(E_b \propto A_b^2\), so

\[ A_b \propto \frac{1}{r} \]

Surface waves, in contrast, are trapped within roughly one wavelength of the free surface and spread out only along an expanding circle (cylindrical/2-D spreading); the "area" over which their energy is shared grows only as the circumference, \(2\pi r \propto r\):

\[ E_s \propto \frac{1}{r} \;\Rightarrow\; A_s \propto \frac{1}{\sqrt{r}} \]

Taking the ratio of the two amplitude decay laws:

\[ \frac{A_b}{A_s} \propto \frac{1/r}{1/\sqrt{r}} = \frac{\sqrt{r}}{r} = \frac{1}{\sqrt{r}} \]

\(\boxed{\dfrac{A_b}{A_s}\propto \dfrac{1}{\sqrt{r}}}\) — this is exactly why, far from an earthquake source, surface waves (which decay more slowly) dominate the seismogram even though body waves arrive first.