Answered: A stone is dropped from rest into a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter14: Sound

Section: Chapter Questions

Problem 8P: A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.00 s later....

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Question

A stone is dropped from rest into a well. The sound of the splash is heard exactly 1.10 s later. Find the depth of the well if the air temperature is 11.0°C.
m

Expert Solution

Step 1

The formula for speed of sound in air is given by

$\begin{array}{l} v = (331 \frac{m}{s}) \sqrt{1 + (\frac{T_{C}}{273 ° C})} \\ = (331 \frac{m}{s}) \sqrt{1 + (\frac{11 ° C}{273 ° C})} \\ = 337.6 \frac{m}{s} \end{array}$

The depth of the well is given by

$h = \frac{g t^{2}}{2} [1]$

Here, $t$ is the time for the stone to fall in to the well.

The time for the sound to travel back up the well is given by

$t' = \frac{h}{v}$

Here, $t'$ is the time taken for the sound to travel back up the well.

The time taken for the sound to travel up the well is difference between

$t' = 1.10 - t$

Use $t' = \frac{h}{v}$ in the above equation.

$\begin{array}{l} \frac{h}{v} = 1.10 - t \\ \frac{g t^{2}}{2 v} = 1.10 - t \\ \frac{g t^{2}}{2 v} + t - 1.10 = 0 \end{array}$

Substitute $9.8 \frac{m}{s^{2}}$ for $g$ , and $337.6 \frac{m}{s}$ for $v$ in the above equation for $t$ .

$\begin{array}{l} \frac{(9.8 \frac{m}{s^{2}}) t^{2}}{2 (337.6 \frac{m}{s})} + t - 1.10 = 0 \\ (0.0145) t^{2} + t - 1.10 = 0 \end{array}$

Solve the above quadratic equation.

$t = 1.08 m or - 70.04 m$

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