Chapter 4.6, Problem 61E

a.

To determine

Express $T_1$ , $T_2$ and $T_3$ in terms of $D$ , $h$ , $c_1$ , $c_2$ and $\theta$ .

Answer to Problem 61E

  $T_1=\frac{D}{c_1},T_2=\frac{2h\sec \theta }{c_1}+\frac{D-2h\tan \theta }{c_2},T_3=\frac{\sqrt{4h^2+D^2}}{c_1}$

Explanation of Solution

Given information:

The speeds of sound $c_1$ in an upper layer and $c_2$ in a lower layer of rock and the thickness $h$ of the upper layer can be determined by seismic exploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamitic charge is detonated at a point $P$ and the transmitted signals are recorded at a point $Q$ , which is a distance $D$ from $P$ . The first signal to arrive at $Q$ travels along the surface and takes $T_1$ seconds. The next signal travels from $P$ to a point $R$ , from $R$ to $S$ in the lower layer, and then to $Q$ , taking $T_2$ seconds. The third signal is reflected off the lower layer at the midpoint $O$ of $RS$ and takes $T_3$ seconds to reach.

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}$

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Draw the diagram of the speeds of sound $c_1$ in an upper layer and $c_2$ in a lower layer of rock and the thickness $h$ of the upper layer can be determined by seismic exploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamitic charge is detonated at a point $P$ and the transmitted signals are recorded at a point $Q$ , which is a distance $D$ from $P$ . The first signal to arrive at $Q$ travels along the surface and takes $T_1$ seconds. The next signal travels from $P$ to a point $R$ , from $R$ to $S$ in the lower layer, and then to $Q$ , taking $T_2$ seconds. The third signal is reflected off the lower layer at the midpoint $O$ of $RS$ and takes $T_3$ seconds to reach

  

Distance the first sound travels is $D$ and at speed $c_1$

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $\begin{array}{l}{c}_1=\frac{D}{T_1}\\ \Rightarrow T_1=\frac{D}{c_1}\end{array}$

Now, the sum of the distance the first sound travels is $2PR$ at speed $c_1$ because it has not reached the second layer of the stone and $RS$ at speed $c_2$ because it is now traveling on the second layer of the stone

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $T_2=\frac{2PR}{c_1}+\frac{SQ}{c_2}$

Recall that,

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

And

Trigonometric ratio:

  $\tan \theta =\frac{p}{b}\text{ and }\cot \theta =\frac{b}{p}$

  $\begin{array}{l}PR=\sqrt{h^2+h^2{\tan }^2\theta }\\ =\sqrt{h^2\left(1+{\tan }^2\theta \right)}\\ =\sqrt{h^2{\sec }^2\theta }\\ =h\sec \theta \end{array}$

The maximum horizontal distance is $D$

So,

  $\begin{array}{l}RS=D-P{R}_x-S{Q}_x\\ \text{}\text{}\text{}\text{}=D-2h\tan \theta \end{array}$

Now substitute the values to get,

  $T_2=\frac{2h\sec \theta }{c_1}+\frac{D-2h\tan \theta }{c_2}$

The third sound travels two times the distance of $PO$ at speed $c_1$

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $T_3=\frac{2PO}{c_1}$

Recall that,

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

And

Trigonometric ratio:

  $\tan \theta =\frac{p}{b}\text{ and }\cot \theta =\frac{b}{p}$

  $\begin{array}{l}PO=\sqrt{h^2+{\left(\frac{D}{2}\right)}^2}\\ =\sqrt{h^2+{\frac{D}{4}}^4}\end{array}$

Substitute the value to get,

  $\begin{array}{l}{T}_3=\frac{2\sqrt{h^2+{\frac{D}{4}}^4}}{c_1}\\ T_3{}^2=\frac{4h^2+D^2}{c_1{}^2}\\ T_3=\frac{\sqrt{4h^2+D^2}}{c_1}\end{array}$

Conclusion:

Thus $T_1=\frac{D}{c_1},T_2=\frac{2h\sec \theta }{c_1}+\frac{D-2h\tan \theta }{c_2},T_3=\frac{\sqrt{4h^2+D^2}}{c_1}$

b..

To determine

Show that $T_2$ is minimum when $\sin \theta =\frac{c_1}{c_2}$ .

Explanation of Solution

Given information:

The speeds of sound $c_1$ in an upper layer and $c_2$ in a lower layer of rock and the thickness $h$ of the upper layer can be determined by seismic exploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamitic charge is detonated at a point $P$ and the transmitted signals are recorded at a point $Q$ , which is a distance $D$ from $P$ . The first signal to arrive at $Q$ travels along the surface and takes $T_1$ seconds. The next signal travels from $P$ to a point $R$ , from $R$ to $S$ in the lower layer, and then to $Q$ , taking $T_2$ seconds. The third signal is reflected off the lower layer at the midpoint $O$ of $RS$ and takes $T_3$ seconds to reach $Q$

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}$

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Draw the diagram of the speeds of sound $c_1$ in an upper layer and $c_2$ in a lower layer of rock and the thickness $h$ of the upper layer can be determined by seismic exploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamitic charge is detonated at a point $P$ and the transmitted signals are recorded at a point $Q$ , which is a distance $D$ from $P$ . The first signal to arrive at $Q$ travels along the surface and takes $T_1$ seconds. The next signal travels from $P$ to a point $R$ , from $R$ to $S$ in the lower layer, and then to $Q$ , taking $T_2$ seconds. The third signal is reflected off the lower layer at the midpoint $O$ of $RS$ and takes $T_3$ seconds to reach

  

Distance the first sound travels is $D$ and at speed $c_1$

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $\begin{array}{l}{c}_1=\frac{D}{T_1}\\ \Rightarrow T_1=\frac{D}{c_1}\end{array}$

Now, the sum of the distance the first sound travels is $2PR$ at speed $c_1$ because it has not reached the second layer of the stone and $RS$ at speed $c_2$ because it is now traveling on the second layer of the stone

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $T_2=\frac{2PR}{c_1}+\frac{SQ}{c_2}$

Recall that,

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

And

Trigonometric ratio:

  $\tan \theta =\frac{p}{b}\text{ and }\cot \theta =\frac{b}{p}$

  $\begin{array}{l}PR=\sqrt{h^2+h^2{\tan }^2\theta }\\ =\sqrt{h^2\left(1+{\tan }^2\theta \right)}\\ =\sqrt{h^2{\sec }^2\theta }\\ =h\sec \theta \end{array}$

The maximum horizontal distance is $D$

So,

  $\begin{array}{l}RS=D-P{R}_x-S{Q}_x\\ \text{}\text{}\text{}\text{}=D-2h\tan \theta \end{array}$

Now substitute the values to get,

  $T_2=\frac{2h\sec \theta }{c_1}+\frac{D-2h\tan \theta }{c_2}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate with respect to $\theta$

  $\begin{array}{l}{T}_2{}^{\prime }\left(\theta \right)=\frac{2h\sec \theta \tan \theta }{c_1}-\frac{2h{\sec }^2\theta }{c_2}\\ =\frac{2h{\sec }^2\theta \sin \theta }{c_1}-\frac{2h{\sec }^2\theta }{c_2}\\ =2h{\sec }^2\theta \left(\frac{\sin \theta }{c_1}-\frac{1}{c_2}\right)\end{array}$

Solve for $T_2{}^{\prime }\left(\theta \right)=0$ , and simplified

  $\begin{array}{l}2h{\sec }^2\theta \left(\frac{\sin \theta }{c_1}-\frac{1}{c_2}\right)=0\\ \Rightarrow \frac{\sin \theta }{c_1}-\frac{1}{c_2}=0\\ \Rightarrow \frac{\sin \theta }{c_1}=\frac{1}{c_2}\\ \Rightarrow \sin \theta =\frac{c_1}{c_2}\end{array}$

The critical number obtained the positive value which means that the function is minimum at this point.

Conclusion:

Thus $T_2$ is minimum at $\sin \theta =\frac{c_1}{c_2}$

c.

To determine

Suppose that $D=1\text{km}\text{,}{T}_1=0.26\text{s},T_2=0.32\text{s},\text{and }{T}_3=034\text{s find }{c}_1,c_2\text{and }h$ .

Answer to Problem 61E

  $c_1=3.8463\raisebox{1ex}{$\text{km}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.,c_2=7.6599\raisebox{1ex}{$\text{km}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\text{ and }h=0.4213\text{km}$

Explanation of Solution

Given information:

The speeds of sound $c_1$ in an upper layer and $c_2$ in a lower layer of rock and the thickness $h$ of the upper layer can be determined by seismic exploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamitic charge is detonated at a point $P$ and the transmitted signals are recorded at a point $Q$ , which is a distance $D$ from $P$ . The first signal to arrive at $Q$ travels along the surface and takes $T_1$ seconds. The next signal travels from $P$ to a point $R$ , from $R$ to $S$ in the lower layer, and then to $Q$ , taking $T_2$ seconds. The third signal is reflected off the lower layer at the midpoint $O$ of $RS$ and takes $T_3$ seconds to reach $Q$

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}$

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Draw the diagram of the speeds of sound $c_1$ in an upper layer and $c_2$ in a lower layer of rock and the thickness $h$ of the upper layer can be determined by seismic exploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamitic charge is detonated at a point $P$ and the transmitted signals are recorded at a point $Q$ , which is a distance $D$ from $P$ . The first signal to arrive at $Q$ travels along the surface and takes $T_1$ seconds. The next signal travels from $P$ to a point $R$ , from $R$ to $S$ in the lower layer, and then to $Q$ , taking $T_2$ seconds. The third signal is reflected off the lower layer at the midpoint $O$ of $RS$ and takes $T_3$ seconds to reach

  

Distance the first sound travels is $D$ and at speed $c_1$

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $\begin{array}{l}{c}_1=\frac{D}{T_1}\\ \Rightarrow T_1=\frac{D}{c_1}\end{array}$

Substitute $D=1\text{km}\text{,}{T}_1=0.26\text{s}$ to get

  $\begin{array}{l}0.26=\frac{1}{c_1}\\ \Rightarrow c_1=3.8463\raisebox{1ex}{$\text{km}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

Now, the sum of the distance the first sound travels is $2PR$ at speed $c_1$ because it has not reached the second layer of the stone and $RS$ at speed $c_2$ because it is now traveling on the second layer of the stone

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $T_2=\frac{2PR}{c_1}+\frac{SQ}{c_2}$

Recall that,

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

And

Trigonometric ratio:

  $\tan \theta =\frac{p}{b}\text{ and }\cot \theta =\frac{b}{p}$

  $\begin{array}{l}PR=\sqrt{h^2+h^2{\tan }^2\theta }\\ =\sqrt{h^2\left(1+{\tan }^2\theta \right)}\\ =\sqrt{h^2{\sec }^2\theta }\\ =h\sec \theta \end{array}$

The maximum horizontal distance is $D$

So,

  $\begin{array}{l}RS=D-P{R}_x-S{Q}_x\\ \text{}\text{}\text{}\text{}=D-2h\tan \theta \end{array}$

Now substitute the values to get,

  $T_2=\frac{2h\sec \theta }{c_1}+\frac{D-2h\tan \theta }{c_2}$

Substitute $D=1\text{km}\text{,}{T}_1=0.26\text{s},T_2=0.32\text{s},\text{and }{c}_1=3.8463$ to get

  $c_2=7.6599\raisebox{1ex}{$\text{km}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

The third sound travels two times the distance of $PO$ at speed $c_1$

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $T_3=\frac{2PO}{c_1}$

Recall that,

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

And

Trigonometric ratio:

  $\tan \theta =\frac{p}{b}\text{ and }\cot \theta =\frac{b}{p}$

  $\begin{array}{l}PO=\sqrt{h^2+{\left(\frac{D}{2}\right)}^2}\\ =\sqrt{h^2+{\frac{D}{4}}^4}\end{array}$

Substitute the value to get,

  $\begin{array}{l}{T}_3=\frac{2\sqrt{h^2+{\frac{D}{4}}^4}}{c_1}\\ T_3{}^2=\frac{4h^2+D^2}{c_1{}^2}\\ T_3=\frac{\sqrt{4h^2+D^2}}{c_1}\end{array}$

Substitute $D=1\text{km}\text{,}{T}_1=0.26\text{s},T_2=0.32\text{s},\text{and }{T}_3=034c_1=3.8462,c_2=7.6599$ to get

  $\begin{array}{l}{T}_3{}^2=\frac{4h^2+D^2}{c_1{}^2}\\ \Rightarrow h=\frac{\sqrt{T_3{}^2{c}_1{}^2-D^2}}{2}\\ =0.4213\text{km}\end{array}$

Conclusion:

Thus $c_1=3.8463\raisebox{1ex}{$\text{km}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.,c_2=7.6599\raisebox{1ex}{$\text{km}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\text{ and }h=0.4213\text{km}$