Question

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 1300 Hz . The bird is 2.6 m away, directly in front of the coyote’s right ear. The coyote’s ears are 15 cm apart.

What is the difference in the arrival time of the sound at the left ear and the right ear? 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?

What is the ratio of this time difference to the period of the sound wave?

Answer 1

Assuming I have the visual correct, the distance from the bird to the left ear is the hypotenuse of a right triangle with legs 2.60 meters and 0.15 meters. s = sqrt(2.6^2 + 0.15^2) = 2.60432 meters.

Part A: The extra distance traveled to get to the left ear is 0.00432 meters. If we take the sound to be traveling at 343 m/s, the time is given by  
t = d / v
t = 0.00432 / 343 m/s = 1.26E-5 seconds!

Part B: If the frequency is 1300 Hz, the period is just the inverse, or 1/1300 seconds. The ratio is:
1.26E-5 / 7.696E-4 = 0.0163, or roughly 1/61