Question

Earthquakes produce several types of shock waves. The most well-known are the P-waves (P for primary or pressure) and the S-waves (S for secondary or shear). In the earth's crust, the P-waves travel at around 6.5 km/s while the S-waves move at about 3.5 km/s. The actual speeds vary depending on the type of material they are going through. The time delay between the arrival of these two waves at a seismic recording center tells geologists how far away the earthquake occurred. If the time delay is 73 s, how far from the seismic station did the earthquake occur?

Answer 1

Answer:

$d=166.075Km$

Explanation:

If we write the velocity of P-waves as $v_P=\frac{d_P}{t_P}$, where these variables are the distance covered by them and the time taken, and the velocity of S-waves in the same manner as $v_S=\frac{d_S}{t_S}$, and we know the value of $\Delta t=t_S-t_P$ (since the P-waves are faster the time they take is shorter), we only need to notice that the distance they travel up to the seismic recording center must be the same for both, so we have:

$d_P=d_S$

$v_Pt_P=v_St_S=v_S(\Delta t - t_P)=v_S \Delta t - v_S t_P$

$v_Pt_P+v_S t_P=v_S \Delta t$

$(v_P+v_S) t_P=v_S \Delta t$

$t_P=\frac{v_S \Delta t}{v_P+v_S}$

And then we can obtain the distance asked:

$d=d_P=v_Pt_P=\frac{v_Pv_S \Delta t}{v_P+v_S}$

Which substituting for our values is:

$d=\frac{(6.5km/s)(3.5km/s)(73s)}{(6.5km/s)+(3.5km/s)}=166.075Km$