Chapter 17, Problem 17.18P

A cowboy stands on horizontal ground between two parallel, vertical clifTs. He is not midway between the cliffs. Me fires a shot and hears its echoes. The second echo arrives 1.92 s after the first and 1.47 s before the third. Consider only the sound traveling parallel to the ground and reflecting from the cliffs, (a) What is the distance between the cliffs? (b) What If? If he can hear a fourth echo, how long after the third echo does it arrive?

To determine

The distance between the cliffs’s.

Answer to Problem 17.18P

The distance between the cliff’s is $826\text{m}$.

Explanation of Solution

Given info: The time of the second echo is $1.92\text{s}$ after the first and $1.47\text{s}$ after the third echo.

The distance travelled by the echo is the sum of the distance travelled by the sound and the distance that is received after striking to the object to the transmitter.

The formula to calculate the time interval of the first echo is,

$t_1=\frac{2\left(PQ\right)}{v}$

Here,

$PQ$ is the distance travelled by the first echo.

$v$ is the speed of the sound wave in air.

Substitute $y$ for $PQ$ and $340\text{m}/\text{s}$ for $v$ in the above formula to find $t_1$.

$\begin{array}{l}{t}_1=\frac{2\left(PQ\right)}{v}\\ =\frac{2y}{340\text{m}/\text{s}}\end{array}$

Thus, the time interval of the first echo is $\frac{2y}{340\text{m}/\text{s}}$.

The formula to calculate the time interval of the second echo is,

$t_2=\frac{2\left(QR\right)}{v}$

Here,

$QR$ is the distance travelled by the second echo.

$v$ is the speed of the sound wave in air.

Substitute $x-y$ for $QR$ and $340\text{m}/\text{s}$ for $v$ in the above formula to find $t_2$.

$\begin{array}{l}{t}_1=\frac{2\left(QR\right)}{v}\\ =\frac{2\left(x-y\right)}{340\text{m}/\text{s}}\end{array}$

Thus, the time interval of the second echo is $\frac{2\left(x-y\right)}{340\text{m}/\text{s}}$.

The formula to calculate the time interval of the third echo is,

$t_3=\frac{2\left(PQ+QR\right)}{v}$

Here,

$\left(PQ+QR\right)$ is the distance travelled by the third echo.

$v$ is the speed of the sound wave in air.

Substitute $x$ for $\left(PQ+QR\right)$, $340\text{m}/\text{s}$ for $v$ in the above formula to find $t_3$.

$\begin{array}{l}{t}_3=\frac{2\left(PQ+QR\right)}{v}\\ =\frac{2x}{340\text{m}/\text{s}}\end{array}$

Thus, the time interval of the third echo is $\frac{2x}{340\text{m}/\text{s}}$.

The formula to calculate the difference in time between the first and second echo is,

$1.92\text{s}=t_2-t_1$

Here,

$t_1$ is the time interval of the first echo.

$t_2$ is the time interval of the second echo.

Substitute $\frac{2y}{340\text{m}/\text{s}}$ for $t_1$ and $\frac{2\left(x-y\right)}{340\text{m}/\text{s}}$ for $t_2$ in the above formula.

$\begin{array}{c}1.92\text{s}=\frac{2\left(x-y\right)}{340\text{m}/\text{s}}-\frac{2y}{340\text{m}/\text{s}}\\ 2x-2y-2y=1.92\text{s}\times 340\text{m}/\text{s}\\ 2x=652.8\text{m}+\text{4}y\\ x=\frac{652.8\text{m}+\text{4}y}{2}\end{array}$ (1)

The formula to calculate the difference in time between the second and third echo is,

$1.47\text{s}=t_3-t_2$

Here,

$t_3$ is the time interval of the third echo.

$t_2$ is the time interval of the second echo.

Substitute $\frac{2x}{340\text{m}/\text{s}}$ for $t_3$ and $\frac{2\left(x-y\right)}{340\text{m}/\text{s}}$ for $t_2$ in the above formula to find $y$.

$\begin{array}{c}1.47\text{s}=\frac{2x}{340\text{m}/\text{s}}-\frac{2\left(x-y\right)}{340\text{m}/\text{s}}\\ 2x-2x+2y=1.47\text{s}\times 340\text{m}/\text{s}\\ y=\frac{499.8\text{m}}{2}\\ =249.9\text{m}\end{array}$

Thus, the value of time difference between the first and second echo is $249.9\text{m}$.

Substitute $249.9\text{m}$ for $y$ in equation (1) to find $x$.

$\begin{array}{c}\\ x=\frac{652.8\text{m}/\text{s}+4\left(249.9\text{m}\right)}{2}\\ =826\text{m}\end{array}$

Conclusion:

Therefore, the distance between the cliff’s is $826\text{}\text{m}$.

To determine

The time interval after which the third echo arrive.

Answer to Problem 17.18P

The time interval after which the third echo arrive is

Explanation of Solution

Given info: The time of the second echo is $1.92\text{s}$ after the first and $1.47\text{s}$ after the third echo.

The formula to calculate the time interval of the fourth echo is,

$t_4=\frac{2\left(QR\right)}{v}$

Here,

$QR$ is the distance travelled by the fourth echo.

$v$ is the speed of the sound wave in air.

Substitute $x-y$ for $QR$ and $340\text{m}/\text{s}$ for $v$ in the above formula to find $t_4$.

$\begin{array}{l}{t}_4=\frac{2\left(QR\right)}{v}\\ =\frac{2\left(x-y\right)}{340\text{m}/\text{s}}\end{array}$

Thus, the time interval of the fourth echo is $\frac{2\left(x-y\right)}{340\text{m}/\text{s}}$.

The formula to calculate the time interval of the third echo is,

$t_3=\frac{2\left(PQ+QR\right)}{v}$

Here,

$\left(PQ+QR\right)$ is the distance travelled by the third echo.

$v$ is the speed of the sound wave in air.

Substitute $x$ for $\left(PQ+QR\right)$ and $340\text{m}/\text{s}$ for $v$ in the above formula to find $t_3$.

$\begin{array}{l}{t}_3=\frac{2\left(PQ+QR\right)}{v}\\ =\frac{2x}{340\text{m}/\text{s}}\end{array}$

Thus, the time interval of the third echo is $\frac{2x}{340\text{m}/\text{s}}$.

The formula to calculate the difference in time between the third and fourth echo is,

$t_3-t_4=t^{\prime }$

Here,

$t_3$ is the time interval of the first echo.

$t_4$ is the time interval of the second echo.

Substitute $\frac{2x}{340\text{m}/\text{s}}$ for $t_3$ and $\frac{2\left(x-y\right)}{340\text{m}/\text{s}}$ for $t_2$ in the above formula to find $t^{\prime }$.

$t^{\prime }=\frac{2x}{340\text{m}/\text{s}}-\frac{2\left(x-y\right)}{340\text{m}/\text{s}}$

Substitute $249.9\text{m}$ for $x$ and $826\text{m}$ for $x$ find $y$.

$\begin{array}{c}{t}^{\prime }=\frac{2x}{340\text{m}/\text{s}}-\frac{2\left(x-y\right)}{340\text{m}/\text{s}}\\ =\frac{2x-2y+2y}{340\text{m}/\text{s}}\\ =\frac{2\left(249.9\text{m}\right)}{340\text{m}/\text{s}}\\ =1.47\text{s}\end{array}$

Conclusion:

Therefore, the time interval after which the third echo arrive is $1.47\text{s}$.