Problem 80 - Pearson Physics

Question

Think \& Calculate 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 (see Figure 14.20). A typical ear canal has a length of about \(2.4 \mathrm{~cm}\). (a) What are the fundamental frequency of the ear canal and the wavelength of that standing wave? (b) 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.) (c) Suppose a person has an ear canal that is shorter than \(2.4 \mathrm{~cm}\). 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. [Notice that the frequencies found in parts (a) and (b) correspond closely to the frequencies of enhanced sensitivity in Figure 14.19.]

Step-by-Step Solution

Answer

(a) Fundamental frequency: ~3560 Hz, wavelength: ~0.096 m. (b) Third harmonic: frequency: ~10680 Hz, wavelength: ~0.032 m. (c) Shorter canal = higher frequency.

Step 1: Understand the Problem

The ear canal functions as an organ pipe, closed at one end. The problem asks us to find the fundamental frequency and the wavelength of the standing wave for a canal length of 2.4 cm. We also need to calculate the frequency and wavelength for the third harmonic and reason about how the fundamental frequency changes with canal length.

Step 2: Use the Formula for Fundamental Frequency

The fundamental frequency of a pipe closed at one end is given by the formula \[ f_1 = \frac{v}{4L} \]where \( v \) is the speed of sound in air (approximately 343 m/s) and \( L \) is the length of the pipe. Substitute \( L = 2.4 \text{ cm} = 0.024 \text{ m} \) into the formula.

Step 3: Calculate the Fundamental Frequency

Using the formula:\[ f_1 = \frac{343}{4 \times 0.024} \approx 3560 \text{ Hz} \]So, the fundamental frequency is approximately 3560 Hz.

Step 4: Calculate the Wavelength of the Fundamental Frequency

To find the wavelength \( \lambda_1 \), use the wave equation \[ \lambda_1 = \frac{v}{f_1} \]Plug in the values \( v = 343 \text{ m/s} \) and \( f_1 \approx 3560 \text{ Hz} \).

Step 5: Calculate the Wavelength

The wavelength is:\[ \lambda_1 = \frac{343}{3560} \approx 0.096 \text{ m} \]Thus, the wavelength of the fundamental frequency is approximately 0.096 meters.

Step 6: Identify the Harmonics

For a pipe closed at one end, the harmonics are odd multiples of the fundamental frequency. The fundamental frequency is the first harmonic, so the third harmonic is the second-lowest frequency.

Step 7: Calculate the Frequency of the Third Harmonic

The third harmonic (denoted as \( f_3 \)) is given by \[ f_3 = 3f_1 \]Substituting the fundamental frequency:\[ f_3 = 3 \times 3560 = 10680 \text{ Hz} \]

Step 8: Calculate the Wavelength of the Third Harmonic

Use the wave equation to find the wavelength:\[ \lambda_3 = \frac{v}{f_3} \]Substitute the values \( v = 343 \text{ m/s} \) and \( f_3 = 10680 \text{ Hz} \).

Step 9: Calculate the Third Harmonic Wavelength

The wavelength is:\[ \lambda_3 = \frac{343}{10680} \approx 0.032 \text{ m} \]So, the wavelength of the third harmonic is approximately 0.032 meters.

Step 10: Analyze Effect of Canal Length on Frequency

The fundamental frequency is inversely proportional to the length of the canal (\( f_1 = \frac{v}{4L} \)). If the canal length decreases, then the frequency increases. Therefore, if a person has a shorter ear canal, the fundamental frequency is greater than 3560 Hz.

Key Concepts

Standing Wave

When we think about sound waves in a closed pipe, like the human ear canal, it's important to understand the concept of a standing wave. Standing waves occur when waves reflect back and forth within a medium, such as in an organ pipe, creating specific patterns of nodes (points of no motion) and antinodes (points of maximum motion). In the case of the ear canal, which is closed at one end, we find a node at the closed end and an antinode at the open end. This setup is similar to a closed organ pipe, where waves bounce between the eardrum and the opening, leading to standing waves that fit the specific length of the canal. These waves help us identify different harmonics, or sound frequencies, present in the ear canal.

Harmonics

Harmonics in a closed pipe system refer to the different frequencies at which standing waves can form. For the human ear canal, which is modeled as a pipe closed at one end, the harmonics are odd multiples of the fundamental frequency. This means the sequence of harmonics follows a pattern: first harmonic (fundamental), third harmonic, fifth harmonic, and so on. The fundamental frequency is the lowest frequency that can create a standing wave in the canal. Each higher harmonic, like the third, represents a higher frequency with additional nodes and antinodes, creating more complex wave patterns. Harmonics play a crucial role in musical acoustics and the perception of sound, determining the richness and timbre of notes.

Wavelength

The wavelength is a measure of the distance between consecutive points of a wave in the same phase, such as from crest to crest or trough to trough. In the context of sound waves in the ear canal, the wavelength of a wave is determined by the speed of sound and the frequency of the wave. The relationship is given by the formula \( \lambda = \frac{v}{f} \), where \( \lambda \) is the wavelength, \( v \) is the speed of sound, and \( f \) is the frequency. In a closed pipe like the ear canal, shorter wavelengths correspond to higher frequencies. Understanding wavelength helps in visualizing how sound waves fit into the length of the ear canal to form standing waves.

Speed of Sound

The speed of sound is a crucial factor in determining wave behavior in different mediums. In air, sound travels at approximately 343 meters per second under normal conditions. This speed affects how quickly sound waves can travel through the air, influencing the frequency and wavelength of sounds in the ear canal. Since the speed of sound is relatively constant in given conditions, it allows us to calculate various properties of sound waves, like their frequency and wavelength, using the equations \( f = \frac{v}{\lambda} \) and \( \lambda = \frac{v}{f} \). Understanding the speed of sound helps explain how sounds are produced, transmitted, and perceived, especially in natural resonators like the human ear.