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Chapter 11: Problem 61

An underground explosion sends out both transverse (S waves) and longitudinal (P waves) mechanical wave pulses (seismic waves) through the crust of the Earth. Suppose the speed of transverse waves is \(8.0 \mathrm{km} / \mathrm{s}\) and that of longitudinal waves is \(10.0 \mathrm{km} / \mathrm{s} .\) On one occasion, both waves follow the same path from a source to a detector (a seismograph); the longitudinal pulse arrives 2.0 s before the transverse pulse. What is the distance between the source and the detector?

Short Answer

Expert verified
Question: Calculate the distance between the source and the detector of seismic waves if the given time difference between the arrival of transverse waves and longitudinal waves at the detector is 2.0 seconds. The speed of transverse waves is 8.0 km/s and the speed of longitudinal waves is 10.0 km/s. Answer: The distance between the source and the detector is 80.0 km.

Step by step solution

01

Define variables

Let's define the variables we need: - \(V_T\): the speed of transverse waves (\(8.0 \mathrm{km} / \mathrm{s}\)) - \(V_L\): the speed of longitudinal waves (\(10.0 \mathrm{km} / \mathrm{s}\)) - \(\Delta t\): time difference between the arrival times of P and S waves at the detector (\(2.0 \mathrm{s}\)) - \(T_T\): time taken by transverse waves to reach the detector - \(T_L\): time taken by longitudinal waves to reach the detector - \(D\): the distance between the source and the detector
02

Write the equation for the time difference

The given time difference between the arrival times of the P and S waves at the detector is 2.0 seconds. In terms of the variables we defined above, we can write the equation for the time difference as: $$ T_T - T_L = \Delta t $$
03

Calculate the time taken by each type of wave

To find the time taken by each type of wave to reach the detector, we'll use the formula for distance: $$ D = V_T \cdot T_T = V_L \cdot T_L $$ From this equation, we can express \(T_T\) in terms of \(D\) and \(V_T\), and \(T_L\) in terms of \(D\) and \(V_L\): $$ T_T = \frac{D}{V_T} $$$$ T_L = \frac{D}{V_L} $$
04

Substitute the expressions for \(T_T\) and \(T_L\) into the equation for time difference

Replace the calculated values of \(T_T\) and \(T_L\) in the equation for time difference: $$ \frac{D}{V_T} - \frac{D}{V_L} = \Delta t $$
05

Solve for the distance

Now we can solve the equation for \(D\). First, let's express the left side of the equation with a common denominator: $$ \frac{D(V_L - V_T)}{V_T \cdot V_L} = \Delta t $$ Next, multiply both sides by \(V_T \cdot V_L\) to eliminate the fraction: $$ D (V_L - V_T) = \Delta t \cdot V_T \cdot V_L $$ Finally, divide both sides by \((V_L - V_T)\) to find \(D\): $$ D = \frac{\Delta t \cdot V_T \cdot V_L}{V_L - V_T} $$
06

Substitute the given values and calculate the distance

Now we can plug in the given values for the speeds of transverse and longitudinal waves and the time difference: $$ D = \frac{2.0 \mathrm{s} \cdot 8.0 \mathrm{km} / \mathrm{s} \cdot 10.0 \mathrm{km} / \mathrm{s}}{10.0 \mathrm{km} / \mathrm{s} - 8.0 \mathrm{km} / \mathrm{s}} $$ Simplify and calculate \(D\): $$ D = \frac{160.0 \mathrm{km}}{2.0 \mathrm{s}} = 80.0 \mathrm{km} $$ So the distance between the source and the detector is 80.0 km.

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