Chapter 17, Problem 11PE

To determine

(a)

The echo times for given temperatures.

Answer to Problem 11PE

The echo time at $5°\text{C}$ is $18\text{ ms}$ and the echo time at $35°\text{C}$ is $17.1\text{ ms}$.

Explanation of Solution

Given:

The distance between bat and insect is $3.00\text{ m}$.

The temperatures at which echo times is to be calculated are $5.00°\text{ C}$ and $35.0°\text{ C}$

Formula used:

The speed of sound at any given temperature $T$ is given by,

  $v_w=v_0\sqrt{\frac{T}{273}}$

The temperature in Kelvin is given by,

  $T_{\text{K}}=273+T_{\text{C}}$

The time taken to cover a certain distance is given by,

  $t=\frac{d}{v}$

Calculation:

The temperature $5°\text{C}$ in Kelvin is given as,

  $\begin{array}{c}{T}_{\text{K}}=273+T_{\text{C}}\\ =273+5\text{ K}\\ =\text{278 K}\end{array}$

The speed of sound at $278\text{ K}$ is calculated as,

  $\begin{array}{c}{v}_w=v_0\sqrt{\frac{T}{273}}\\ =\left(331\text{m}/\text{s}\right)\sqrt{\frac{\text{278}}{273}}\\ =\left(331\text{m}/\text{s}\right)\left(1.0091\right)\\ =334\text{m}/\text{s}\end{array}$

Echo distance is twice the prey distance, so the echo distance becomes $6\text{ m}$.

The time for travelling echo distance at $\text{278 K}$ is calculated as,

  $\begin{array}{c}t=\frac{d}{v}\\ =\frac{6\text{ m }}{334\text{m}/\text{s}}\\ =\left(0.01796\text{ s}\times \frac{{10}^3\text{ ms}}{1\text{ s}}\right)\\ \approx 18\text{ s}\end{array}$

The temperature $5°\text{C}$ in Kelvin is given as,

  $\begin{array}{c}{T}_{\text{K}}=273+T_{\text{C}}\\ =273+35\text{ K}\\ =30\text{8 K}\end{array}$

The speed of sound at $\text{308 K}$ is calculated as,

  $\begin{array}{c}{v}_w=v_0\sqrt{\frac{T}{273}}\\ =\left(331\text{m}/\text{s}\right)\sqrt{\frac{\text{308 K}}{273\text{ K}}}\\ =\left(331\text{m}/\text{s}\right)\left(1.0621\right)\\ =351.58\text{m}/\text{s}\end{array}$

The time for travelling echo distance at $\text{308 K}$ is calculated as,

  $\begin{array}{c}t=\frac{d}{v}\\ =\frac{6\text{ m}}{351.58\text{m}/\text{s}}\\ =\left(0.017066\text{ s}\times \frac{{10}^3\text{ ms}}{1\text{ s}}\right)\\ \approx 17.1\text{ ms}\end{array}$

Conclusion:

The echo time at $5°\text{C}$ is $18\text{ ms}$ and the echo time at $35°\text{C}$ is $17.1\text{ ms}$.

To determine

(b)

The uncertainty caused due to the change in temperature.

Answer to Problem 11PE

The uncertainty caused by temperature difference is $5\%$.

Explanation of Solution

Formula Used:

Percentage uncertainty in locating the prey at different temperatures is given by,

  $\%\text{uncertainty}=\frac{t_1-t_2}{t_1}\times 100\%$

Calculation:

Uncertainty is calculated as,

  $\begin{array}{c}\%\text{uncertainty}=\frac{t_1-t_2}{t_1}\times 100\%\\ =\frac{18\text{ ms}-17.1\text{ ms}}{18\text{ ms}}\times 100\%\\ =\frac{0.9}{18}\times 100\%\\ =5\%\end{array}$

Conclusion:

The uncertainty caused by temperature difference is $5\%$.

To determine

(c)

The significance of uncertainty and whether it could cause difficulties for the bat.

Explanation of Solution

Introduction:

A bat uses sound echoes to find its way further and to catch the prey. The time taken by the echo to return gives an idea of the distance of object or prey in front of it.

The bat uses the echo time to calculate distance of an object or the prey. The uncertainty in this echo location occurs due to difference in the temperature of air because the speed of sound is dependent on the temperature of the air. Thus, a prey at same distance can give two echo times depending on the temperature of the air. Due to this, a bat faces difficulties to adjust and locate the exact position of the prey which could lead to missing the prey.

Conclusion:

Thus, the uncertainty causes difficulties for bats in locating the objects or the prey.