# Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both...

## Question:

Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of the S wave is about 8.0 km/s, and that of the P wave is 4.0 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in a straight line, at what distance does the earthquake occur?

## Source of Earthquake:

If an object has speed {eq}v {/eq} then the time required to travel distance {eq}d {/eq} is given by, $$\begin{align} t=\dfrac dv \end{align} $$

This implies it will take different time to travel the same distance if two objects have a different speed. The situation in the problem is quite similar.

## Answer and Explanation: 1

Let the source is at a distance {eq}x {/eq} and let the velocity of the transverse wave is {eq}v_s {/eq} and the longitudinal wave is {eq}v_p {/eq}.

Then time is taken by the transverse wave to reach the seismograph {eq}t_1=\dfrac{x}{v_s} {/eq}

And time is taken by the longitudinal wave {eq}t_2=\dfrac{x}{v_p} {/eq}.

From the question we have,

- {eq}v_s=8.0\ \rm km/s {/eq}

- {eq}v_p=4\ \rm km/h {/eq}

- {eq}t_2-t_1=4\ \rm min=240\ \rm s {/eq}

Hence we get

$$\begin{align} &t_2-t_1=240\ \rm s\\[.4 cm] &\dfrac{x}{v_p}-\dfrac{x}{v_s}=240\ \rm s\\[.4 cm] &x\left(\dfrac1{v_p}-\dfrac1{v_s}\right)=240\ \rm s\\[.4 cm] &x=\dfrac{240\ \rm s}{\dfrac1{v_p}-\dfrac1{v_s}}\\[.4 cm] &x=\dfrac{240v_sv_p}{v_s-v_p}\\[.4 cm] &x=\dfrac{240\times 8\times4}{8-4}\\[.4 cm] &x=1,920\ \rm km \end{align} $$

Hence, the center of the earthquake is {eq}1,920\ \rm km {/eq} away.

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