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A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a coyote is listening to a bird whistling at 1000 Hz. The bird is 3.0 m away, directly in front of the coyote’s right ear. The coyote’s ears are 15 cm apart.
a. What is the difference in the arrival time of the sound at the left ear and the right ear?
b. What is the ratio of this time difference to the period of the sound wave?

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?

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Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

[tex]d_2[/tex] = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

[tex]d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m[/tex]

Time

[tex]t=\dfrac{Distance}{Speed}[/tex]

Time difference would be

[tex]\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s[/tex]

The time difference is 0.0000109261200583 s

Time period is given by

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s[/tex]

The ratio is

[tex]\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583[/tex]

The ratio is 0.0109261200583

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