Question

A P-wave of frequency 20 Hz is travelling through a non-dispersive medium with a velocity of 5 km/s. The amplitude retained at a distance of 10 km from source is _____________ % (rounded off to one decimal place). (Use quality factor, \(Q = 80\))

Hint:

Use A/A0 = exp(-pi*f*x/(Q*v)); with f=20 Hz, x=10 km, v=5 km/s, Q=80 the exponent is pi/2, giving about 20.8% amplitude retained.

Answer 1

Correct Answer: 20

Intrinsic (anelastic) attenuation of a seismic wave's amplitude as it travels a distance \(x\) through a medium of quality factor \(Q\) is given by:

\[ \frac{A}{A_0} = \exp\left(-\frac{\pi f x}{Q v}\right) \]

where \(f\) is the frequency, \(v\) is the (non-dispersive) wave velocity, and \(x\) is the propagation distance. This form follows from writing the travel time \(t = x/v\) and using the standard temporal attenuation law \(A/A_0 = e^{-\pi f t/Q}\).

Substitute the given values: \(f = 20\ \text{Hz}\), \(x = 10\ \text{km}\), \(Q = 80\), \(v = 5\ \text{km/s}\):

\[ \frac{\pi f x}{Qv} = \frac{\pi \times 20 \times 10}{80 \times 5} = \frac{200\pi}{400} = \frac{\pi}{2} = 1.5708 \]

So:

\[ \frac{A}{A_0} = e^{-1.5708} = 0.2079 \]

Converting to a percentage:

\[ \frac{A}{A_0} \times 100\% = 20.8\% \]

This value of 20.8% lies within the official accepted range of 20 to 21%. Physically, this shows how strongly a 20 Hz P-wave is attenuated by intrinsic damping (Q = 80) over just 10 km -- only about a fifth of the original amplitude survives, which is why higher-frequency seismic signals lose amplitude much faster with distance than lower-frequency ones for the same Q.

\(\boxed{\dfrac{A}{A_0} \approx 20.8\%}\)