Chapter 2, Problem 80PQ

(a)

To determine

The depth of the chasm.

Answer to Problem 80PQ

The depth of the chasm is $46.1\text{ m}$ .

Explanation of Solution

Assume that the pebble falls a distance $d$ into the chasm in a time interval $\Delta t_1$ and the sound of impact travels upward the same distance in a time interval $\Delta t_2$ before the rock climber hears it.

Write the equation for the total time interval.

$\Delta t=\Delta t_1+\Delta t_2$

Here, $\Delta t$ is the total time interval, $\Delta t_1$ is the time taken for the pebble to fall into the bottom of the chasm and $\Delta t_2$ is the time taken for the sound impact of the pebble to reach the rock climber.

Rewrite the above equation for $\Delta t_1$ .

$\Delta t_1=\Delta t-\Delta t_2$ (I)

Write the Newton’s equation of motion for the fall of the pebble into the chasm.

  $d=v_i\left(\Delta t_1\right)+\frac{1}{2}a{\left(\Delta t_1\right)}^2$

Here, $d$ is the distance travelled by the pebble before reaching the bottom of the chasm, $v_i$ is the initial speed of the pebble and $a$ is the acceleration of the pebble.

The initial speed of the pebble is zero. The pebble is under free fall during the time interval $\Delta t_1$ so that its acceleration is equal to the acceleration due to gravity.

Replace $v_i$ by $0$ and $a$ by $g$ in the above equation to get the final expression for $d$ .

$\begin{array}{c}d=0+\frac{1}{2}g{\left(\Delta t_1\right)}^2\\ =\frac{1}{2}g{\left(\Delta t_1\right)}^2\end{array}$ (II)

The sound from the impact travels at constant speed during the time interval $\Delta t_2$ and covers the distance $d$ .

Write the expression for distance covered during the time interval $\Delta t_2$ .

$d=v_s\Delta t_2$ (III)

Here, $v_s$ is the speed of sound.

Equate equations (II) and (III).

  $\frac{1}{2}g{\left(\Delta t_1\right)}^2=v_s\Delta t_2$

Put equation (I) in the above equation.

  $\begin{array}{c}\frac{1}{2}g{\left(\Delta t-\Delta t_2\right)}^2=v_s\Delta t_2\\ {\left(\Delta t\right)}^2-2\left(\Delta t\right)\left(\Delta t_2\right)+{\left(\Delta t_2\right)}^2=\frac{2v_s\Delta t_2}{g}\\ {\left(\Delta t_2\right)}^2-2\left(\Delta t+\frac{v_s}{g}\right)\Delta t_2+{\left(\Delta t\right)}^2=0\end{array}$

The above equation is quadratic in $\Delta t_2$ .

Write the expression for the solution of the above equation.

$\Delta t_2=\frac{2\left(\Delta t+\frac{v_s}{g}\right)\pm \sqrt{{\left(-2\left(\Delta t+\frac{v_s}{g}\right)\right)}^2-4{\left(\Delta t\right)}^2}}{2}$ (IV)

Conclusion:

It is given that the total time interval is $3.20\text{ s}$ , the speed of sound is $343\text{ m/s}$ and the value of acceleration due to gravity is $9.81{\text{ m/s}}^2$ .

Substitute $3.20\text{ s}$ for $\Delta t$, $343\text{ m/s}$ for $v_s$ and $9.81{\text{ m/s}}^2$. for $g$ in equation (IV) to find $\Delta t_2$ .

  $\begin{array}{c}\Delta t_2=\frac{2\left(3.20\text{ s}+\frac{343\text{ m/s}}{9.81{\text{ m/s}}^2}\right)\pm \sqrt{{\left(-2\left(3.20\text{ s}+\frac{343\text{ m/s}}{9.81{\text{ m/s}}^2}\right)\right)}^2-4{\left(3.20\text{ s}\right)}^2}}{2}\\ =0.134\text{ s}\end{array}$

Substitute $343\text{ m/s}$ for $v_s$ and $0.134\text{ s}$ for $\Delta t_2$ in equation (III) to find $d$ .

  $\begin{array}{c}d=\left(343\text{ m/s}\right)\left(0.134\text{ s}\right)\\ =46.1\text{ m}\end{array}$

Therefore, the depth of the chasm is $46.1\text{ m}$ .

(b)

To determine

The percentage of error that would result from assuming the speed of sound to be infinite.

Answer to Problem 80PQ

The percentage of error that would result from assuming the speed of sound to be infinite is $8.89\text{ \%}$ .

Explanation of Solution

If the travel time for the sound of impact is ignored, it would mean that equation (II) has to be used to determine the depth.

Write the equation for the percentage of error.

$\%\text{ of error}=\left|\frac{\text{true value}-\text{obtained value}}{\text{true value}}\right|\times 100\%$ (V)

Conclusion:

Substitute $9.81{\text{ m/s}}^2$. for $g$ and $3.20\text{ s}$ for $\Delta t$ in equation (II) to find $d$ .

  $\begin{array}{c}d=\frac{1}{2}\left(9.81{\text{ m/s}}^2\right){\left(3.20\text{ s}\right)}^2\\ =50.2\text{ m}\end{array}$

Substitute $46.1\text{ m}$ for true value and $50.2\text{ m}$ for obtained value in equation (V) to find the percentage of error.

  $\begin{array}{c}\%\text{ of error}=\left|\frac{46.1\text{ m}-50.2\text{ m}}{46.1\text{ m}}\right|\times 100\%\\ =8.89\text{ \%}\end{array}$

Therefore, the percentage of error that would result from assuming the speed of sound to be infinite is $8.89\text{ \%}$ .