IP Standing Waves in the Human Ear The human ear canal is much like an organ pip

Question

IP Standing Waves in the Human Ear The human ear canal is much like an organ pipe that is closed at one end (at the tympanic membrane or eardrum) and open at the other (Figure 1) . A typical ear canal has a length of about 2.4 cm.

Part A

What are the fundamental frequency and wavelength of the ear canal?

Express your answer using two significant figures.

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Part B

Express your answer using two significant figures.

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Part C

Find the frequency and wavelength of the ear canal's third harmonic. (Recall that the third harmonic in this case is the standing wave with the second-lowest frequency.)

Express your answer using two significant figures.

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Part D

Express your answer using two significant figures.

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Part E

Suppose a person has an ear canal that is shorter than 2.4 cm. Is the fundamental frequency of that person's ear canal greater than, less than, or the same as the value found in part A? Explain.

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Figure 1 of 1

IP Standing Waves in the Human Ear The human ear canal is much like an organ pipe that is closed at one end (at the tympanic membrane or eardrum) and open at the other (Figure 1) . A typical ear canal has a length of about 2.4 cm.

Part A

What are the fundamental frequency and wavelength of the ear canal?

Express your answer using two significant figures.

f1 =   Hz  

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Part B

Express your answer using two significant figures.

?1 =   cm  

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Part C

Find the frequency and wavelength of the ear canal's third harmonic. (Recall that the third harmonic in this case is the standing wave with the second-lowest frequency.)

Express your answer using two significant figures.

f3 =   Hz  

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Part D

Express your answer using two significant figures.

?3 =   cm  

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Part E

Suppose a person has an ear canal that is shorter than 2.4 cm. Is the fundamental frequency of that person's ear canal greater than, less than, or the same as the value found in part A? Explain.

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

3800 Character(s) remaining

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Figure 1 of 1

Explanation / Answer

A. In a closed air column, fundamental wavelength lambda0 = 4L = 9.6 cm
B. Wavelength lambda of harmonic n = lambda0/n; for n=3, lambda = 3.2 cm
C. Shorter L = shorter lambda0 = higher frequency.

To find the frequencies in A and B, divide sound speed c by wavelength. At 20 C, c = 343 m. Inside the ear canal the temperature could be higher.

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