Question

Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a particular seismic station detects the arrival of these two types of waves 1.2 min apart

Answer 1

The earthquake occurred at a distance of 1122 km

Step-by-step explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ = 5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

`t_1 = (72v_2)/(v_1-v_2)\\\\t_1 = (72*5.5)/(8.5-5.5)\\\\t_1 = 132 \ s`

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km