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 1250 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.

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.

b. What is the ratio of this time difference to the period of the sound wave? Express your answer using two significant figures.

a) t_l - t_r = 12.54 us

b) (t_l - t_r) / T = 0.0157

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

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?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

d_l = sqrt (2.6^2 + 0.15^2)

d_l = sqrt (6.7825)

d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

t_l - t_r = (1 / v) * (d_l - d_r)

Where, v = 343 m/s speed of sound in air:

t_l - t_r = (1 / 343) * (2.6043 - 2.6)

t_l - t_r = (0.0043 / 343)

t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

T = 1 / f

T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

(t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

(t_l - t_r) / T = 0.0157