Longitudinal Wave Physics Example 4

Solution

1. 1
   Let $d$ be the distance. P-wave arrives at time $t_P = \frac{d}{6000}$ and S-wave at $t_S = \frac{d}{3500}$.
2. 2
   Time difference: $t_S - t_P = 20 \text{ s}$, so $\frac{d}{3500} - \frac{d}{6000} = 20$.
3. 3
   $$d\left(\frac{1}{3500} - \frac{1}{6000}\right) = 20 \implies d\left(\frac{6000 - 3500}{21000000}\right) = 20 \implies d \times \frac{2500}{21000000} = 20 \implies d = \frac{20 \times 21000000}{2500} = 168{,}000 \text{ m} = 168 \text{ km}$$