Question

A male alligator emits a subsonic mating call that has a frequency of 18.6 Hz by taking a large breath into his chest cavity and then releasing it.

If the hollow chest cavity of an alligator behaves approximately as a pipe open at only one end, estimate its length. (Express your answer to three significant figures.)

Answer 1

The length of the alligator's chest cavity is approximately 4.61 meters, calculated by treating it as a resonating tube closed at one end and using the given subsonic mating call frequency of 18.6 Hz.

Step-by-step explanation:

To estimate the length of the alligator's chest cavity, we can treat it as a resonating tube that is closed at one end and open at the other, which means it produces sound at its fundamental frequency or the first harmonic.

The formula for the fundamental frequency (f) for a tube closed at one end is given by:

f = v / (4L)

where:
v = speed of sound in air (assume v = 343 m/s at room temperature)
L = length of the tube
f = frequency of the sound

Given that the alligator's mating call has a frequency of 18.6 Hz, we can rearrange the formula to solve for L:

L = v / (4f)

Substituting the given values:

L = 343 m/s / (4 x 18.6 Hz)

L = 343 m/s / 74.4 s−¹

L = 4.61 m

Therefore, the approximated length of the alligator's chest cavity, to three significant figures, is 4.61 m.

Answer 2

The estimated length of the pipe is 74 m.

Step-by-step explanation:

The length of a pipe that is open at one end can be estimated using the formula L = (4/n) * λ, where L is the length of the pipe, n is the harmonic number (1 for the fundamental frequency), and λ is the wavelength of the sound wave.

To find the wavelength, we can use the formula λ = v/f, where λ is the wavelength, v is the speed of sound, and f is the frequency of the sound wave. In this case, the frequency is given as 18.6 Hz, and the speed of sound in air is 344 m/s.

Substituting these values into the formula, we get: λ = 344/18.6 = 18.5 m. Now we can substitute the wavelength into the formula for the length of the pipe: L = (4/1) * 18.5 = 74 m.