Question

One cue your hearing system uses to localize a sound (i.e., to

tell where a sound is coming from) is the slight difference in the

arrival times of the sound at your ears. Your ears are spaced

approximately 20 cm apart. Consider a sound source 5.0 m from

the center of your head along a line 45 to your right. What is the

difference in arrival times? Give your answer in microseconds.

Hint: You are looking for the difference between two numbers that are nearly the same. What does this near equality imply about the necessary precision during intermediate stages of the calculation?

Short answer

The difference in arrival times at your left and right ears is$412\mu s.$

Step-by-step solution

Given data

The sound source is $5cm$distant at an angle of ${45}^o$with the center of the head.

Two ears $E_{L,}{E}_R$(say) are separated by a distance of$20cm=0.2m$

Therefore, the distance of each ear from the center of head is,

$d=\frac{0.2m}{2}=0.1m.$

So, the set up should be as follows:

Determination of arrival time

The distance between the source and the left ear $E_L$is,

localid="1650010052598" role="math" $$\begin{gathered} d_L=\sqrt{x^2+{\left(y+0.1m\right)}^2} \\ {d}_L=\sqrt{{\left(5.0m\times \cos {45}^o\right)}^2+{\left[\left(5.0m\times \sin {45}^o\right)+0.1m\right]}^2} \\ {d}_L=\sqrt{{\left(\frac{5}{\sqrt{2}}\right)}^2+{\left(\frac{5}{\sqrt{2}}+0.1m\right)}^2} \\ {d}_L=5.0712m \end{gathered}$$

Similarly, the distance between the source and the right ear is,

$$\begin{gathered} d_R=\sqrt{x^2+{\left(y-0.1m\right)}^2} \\ {d}_R=\sqrt{{\left(5.0m\times \cos {45}^o\right)}^2+{\left[\left(5.0m\times \sin {45}^o\right)-0.1m\right]}^2} \\ {d}_R=\sqrt{{\left(\frac{5}{\sqrt{2}}\right)}^2+{\left(\frac{5}{\sqrt{2}}-0.1m\right)}^2} \\ {d}_R=4.9298m \end{gathered}$$

Thus, the difference between these two sound waves are $$\begin{gathered} ∆d=d_L-d_R \\ ∆d=\left(5.0712-4.9298\right)m \\ ∆d=0.1414m \end{gathered}$$

The difference between arrival time of the sound wave having a speed of $343m/s$is,

$$\begin{gathered} t=\frac{∆d}{v} \\ t=\frac{0.1414m}{343m/s} \\ t=4.122\times {10}^{-4}s \\ t=412\mu s \end{gathered}$$