Respuesta :

Distance between the Tulsa, OK observatory and the center of this fictional earthquake: 20 kilometers.

An earthquake generates two major seismic waves that travel through the crest at different speeds:

  • Primary (P) waves travel at between 6 and 7 [tex]\text{km} \cdot \text{s}^{-1}[/tex]
  • Secondary (S) waves travel at approximately 3.5 [tex]\text{km} \cdot \text{s}^{-1}[/tex].

Let [tex]v_\text{P}[/tex] and [tex]v_{\text{S}}[/tex] denote the speed of the two waves, respectively. Let [tex]t_\text{P}[/tex], [tex]t_\text{S}[/tex] resembles the time required for each waves to reach the observatory.

Both [tex]t_\text{P} \cdot v_\text{P}[/tex] and [tex]t_\text{S} \cdot v_\text{S}[/tex] would represent the separation between the source and the observatory. Thus

[tex]t_\text{P} \cdot v_\text{P} = d\\t_\text{S} \cdot v_\text{S} = d\\[/tex]

[tex]t_\text{P} = d / v_\text{P}\\t_\text{S} = d / v_\text{P}[/tex]

Given the difference between  [tex]t_\text{P}[/tex] and [tex]t_\text{S}[/tex]:

[tex]t_\text{S} - t_\text{P} = 2.5 \; \text{minutes} = 150 \; \text{seconds}[/tex]

([tex]t_\text{P} < t_\text{S}[/tex] given that [tex]v_\text{P} > v_\text{S}[/tex] and  [tex]1/ v_\text{P} < 1/ v_\text{S}[/tex])

[tex]d / v_\text{S} - d / v_\text{P} = t_\text{S} - t_\text{P} = 150 \; \text{seconds}\\d \cdot (1/v_\text{S} - 1/v_\text{P}) = 150 \; \text{seconds}[/tex]

[tex]d = 150 \; \text{s} / ((1/(3.5 \; \text{km} \cdot \text{s}^{-1}) - 1/(6.5 \; \text{km} \cdot \text{s}^{-1}))\\\phantom{d} =2.0 \times 10\; \text{km} = 20 \; \text{km}[/tex]

Note that wave speed data involved in calculations above came from an external source. You shall repeat those steps with speeds indicated on the worksheet if possible.