Question

An earthquake produces longitudinal P waves that travel outward at 8000 m/s and transverse S waves that move at 4500 m/s. A seismograph at some distance from the earthquake records the arrival of the S waves 2.0 min after the arrival of the P waves. How far away was the earthquake? You can assume that the waves travel in straight lines, although actual seismic waves follow more complex routes.

Answer 1

Answer:

1234285.7 m or 1234.3 km

Explanation:

Let the distance be $d$, the time taken by P waves be $t_P$ and the time taken by the S waves be $t_S$.

$\text{Velocity}\dfrac{\text{Distance}}{\text{Time}}$

$\text{Time}\dfrac{\text{Distance}}{\text{Velocity}}$

For the P waves,

$t_P=\dfrac{d}{8000}$

$d=8000t_P$

For the S waves,

$t_S=\dfrac{d}{4500}$

$d=4500t_S$

Equating the $d$,

$8000t_P=4500t_S$

Divide both sides of the equation by 500 to reduce the terms.

$16t_P=9t_S$

Since S waves arrive 2 minutes (= 120 seconds) after P waves,

$t_S-t_P=120$

$t_S=120+t_P$

Substitute this in the equation of the distance.

$16t_P=9(t_P+120)$

$16t_P=9t_P+1080$

$7t_P=1080$

$t_P=\dfrac{1080}{7}$

Substitute this in the equation for $d$ involving $t_P$.

$d=8000t_P$

$d=8000\times\dfrac{1080}{7}$

$d=1234285.7 \text{ m }= 1234.3 \text{ km}$