# earthquake

• Mar 1st 2008, 07:33 PM
lovinhockey26
earthquake
When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 8.03 km/s and the secondary, or S, wave has a speed of about 5.63 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 77.9 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far is the seismograph from the earthquake?
• Mar 2nd 2008, 12:46 AM
earboth
Quote:

Originally Posted by lovinhockey26
When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 8.03 km/s and the secondary, or S, wave has a speed of about 5.63 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 77.9 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far is the seismograph from the earthquake?

You know that the distance is calculated by:

$d = v \cdot t$

Both waves travel the same distance

$d = v_P \cdot t_P$ and

$d = v_S \cdot t_S$. Use the fact that $t_S - t_p = 77.9\ s$ :

$v_P \cdot t_P = v_S \cdot (t_P + 77.9)$ . After a few steps you'll get:

$t_P=\frac{v_S \cdot 77.9}{v_P - v_S}\approx 182.7\ s$

And now you can calculate the distance to the earthquake.
• Mar 2nd 2008, 04:20 AM
Soroban
Hello, lovinhockey26!

Quote:

When an earthquake occurs, two types of sound waves are generated
and travel through the earth.
The primary or $P$ wave has a speed of about 8.03 km/s
and the secondary or $S$ wave has a speed of about 5.63 km/s.

A seismograph, located some distance away, records the arrival of the $P$ wave
and then, 77.9 s later, records the arrival of the $S$ wave.

Assuming that the waves travel in a straight line,
how far is the seismograph from the earthquake?

The $P$ wave travels for t seconds at 8.03 km/sec.
. . Its distance is: . $8.03t$ feet.

The $S$ wave travels for $(t+77.9)$ seconds at 5.03 km/sec.
. . Its distance is: . $5.03(t+77.9)$ feet.

These distances are equal: . $8.03t \;=\;5.63(t + 77.9)$

Go fot it!