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 1000 Hz . The bird is 2.5 m away, directly in front of the coyote’s right ear. The coyote’s ears are 15 cm apart.

Part A)

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?

Express your answer to two significant figures and include the appropriate units.

Part B)

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

Express your answer using two significant figures.

Answer 1

solution A > Assuming the speed of the sound = v= 340 m) 5 so BT = 0.15 (s) = 404X1024 seconds. 340 DT = 14,4x10-4 ] seconds. B> Period - T = = idoo = 100X10-3 seconds. 80 to a sus x 104 = = 4.4810-4 Joox 10-3 4048x10- 4,4810l