Problem 39 - Physics

Question

A tube is open only at one end. A certain harmonic produced by the tube has a frequency of \(450 \mathrm{~Hz}\). The next higher harmonic has a frequency of \(750 \mathrm{~Hz}\). The speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\). (a) What is the integer \(n\) that describes the harmonic whose frequency is \(450 \mathrm{~Hz} ?\) (b) What is the length of the tube?

Step-by-Step Solution

Answer

(a) The integer \( n \) is 3. (b) The tube length is approximately 0.476 meters.

Step 1: Understanding Harmonics in Open Tube at One End

In a tube open only at one end, the harmonics are odd multiples of the fundamental frequency. The frequencies are given by the formula \( f_n = (2n - 1) \frac{v}{4L} \), where \( v \) is the speed of sound, \( L \) is the length of the tube, and \( n \) is an integer describing the harmonic.

Step 2: Determine the Harmonic Numbers

Given that the frequencies are 450 Hz and 750 Hz for consecutive harmonics, denote the lower frequency as \( f_1 = 450 \mathrm{~Hz} \) and the higher as \( f_2 = 750 \mathrm{~Hz} \). Since both are harmonics of an open tube at one end, \( f_2/f_1 = (2m-1)/(2n-1) = 750/450 = 5/3 \). Solving for \( m \) and \( n \), assume the smallest odd integers for consecutive harmonics are \( 3 \) and \( 5 \). This gives us \( n = 3 \) and \( n+2 = 5 \).

Step 3: Calculate the Length of the Tube

Use the formula \( f_n = (2n - 1) \frac{v}{4L} \) and substitute \( f_1 = 450 \mathrm{~Hz} \), \( n = 3 \), and \( v = 343 \mathrm{~m/s} \). This gives \( L = \frac{v}{4f_n}(2n-1) \). Substitute and solve: \[ L = \frac{343}{4 \times 450}(5) = \frac{343 \times 5}{1800} \]. Calculating this yields \( L \approx 0.476 \) meters.

Key Concepts

Open Tube Harmonics

An open tube, open at only one end, produces sound waves in a unique way. When it resonates, it creates standing waves, which are the result of the interference of the incident and reflected waves within the tube. This type of tube is special as it supports only odd-numbered harmonics. Why? Because at the closed end, there must be a node (a point of no displacement), while at the open end, there is an antinode (a point of maximum displacement). Thus, the possible harmonics in such a tube occur at frequencies which are odd multiples (1st, 3rd, 5th, etc.) of the fundamental frequency.
Understanding the behavior of open tube harmonics is crucial because it explains the specific frequency patterns we observe. Since only odd harmonics are present, you can determine the frequency of these harmonics using the formula:

• The frequency of the nth harmonic, \( f_n \), is given by \( f_n = (2n - 1) \frac{v}{4L} \).
• Where \( v \) is the speed of sound, \( L \) is the length of the tube, and \( n \) is the harmonic number (odd integers: 1, 3, 5, ...).

Frequency Calculation

Frequency calculation in tubes involves determining the specific sound frequencies that form standing waves inside them. In the case of a tube open at one end, each harmonic frequency depends on the length of the tube and its physical properties.
To calculate the harmonic frequencies within the tube:

• Identify the formula \( f_n = (2n - 1) \frac{v}{4L} \).
• "\( f_n \)" is the frequency of the nth harmonic, \( v \) is the speed of sound, and \( L \) is the tube's length.

For a given harmonic frequency, such as 450 Hz, the next possible harmonic frequency might be 750 Hz, if derived using the correct harmonic numbers, based on the odd number series. Divide the higher by the lower frequency to find the ratio and match it with the odd integers to find the corresponding harmonic numbers (3 and 5 in this case).

Speed of Sound

The speed of sound is an essential factor in harmonic resonance within tubes. It's the speed at which sound waves travel through a medium—in this case, air. The speed depends on the medium's temperature and pressure. In normal air conditions at room temperature (about 20°C), the speed of sound is approximately 343 m/s. This value is crucial when calculating frequencies because it directly influences how quickly the sound waves bounce back and forth in the tube to create standing waves.
The speed of sound helps determine the particular frequencies at which these standing waves occur, defining how the tube can naturally resonate. In frequency calculations for open tubes, knowing the speed of sound allows us to accurately apply the frequency formula \( f_n = (2n - 1) \frac{v}{4L} \) by replacing \( v \) with 343 m/s.

Acoustics

Acoustics is the science of sound, and it plays a vital role in understanding harmonics in tubes. It covers how sound waves travel, interact, and can create resonance in spaces and objects like a tube open at one end. Acoustics teaches us how different parameters, such as tube length, wave frequency, and the speed of sound, influence the creation and perception of sound.
When studying tubes, acoustics helps explain why certain frequencies amplify due to standing waves forming—this is resonance. In a tube, this happens when waves reflect back from the open or closed ends, interfering in such a way that they reinforce each other at specific frequencies. These frequencies are not random but are harmonics based on physical characteristics of the tube and the sound's speed, illustrating fundamental acoustics principles.