Question

A bugle can be represented by a cylindrical pipe of length \(L=1.35 \mathrm{~m} .\) Since the ends are open, the standing waves produced in the bugle have antinodes at the open ends, where the air molecules move back and forth the most. Calculate the longest three wavelengths of standing waves inside the bugle. Also calculate the three lowest frequencies and the three longest wavelengths of the sound that is produced in the air around the bugle.

Short answer

Answer: The three longest wavelengths inside the bugle are 2.70 meters, 1.35 meters, and 0.90 meters. The three lowest frequencies are approximately 127.04 Hz, 254.07 Hz, and 381.11 Hz. The three longest wavelengths of the sound produced in the air around the bugle are also 2.70 meters, 1.35 meters, and 0.90 meters.

Step-by-step solution

Understand the characteristics of open pipes

In an open pipe, both ends have antinodes. The simplest standing wave can be created, when we have only one antinode more than the endpoints, which is half a wavelength (λ/2). In the case of open pipes, the fundamental frequency or first harmonic has a length of L = λ/2. For higher harmonics, the length is given by L = n(λ/2), where n is the harmonic number.

Calculate the three longest wavelengths inside the bugle

Let's find the longest three wavelengths inside the bugle by using the formula L = n(λ/2), and solve for λ. We will do this for n = 1, 2, and 3, as they represent the three lowest harmonics. λ1 = 2 * L / 1 λ2 = 2 * L / 2 λ3 = 2 * L / 3 Plug in the given length, L = 1.35 meters: λ1 = 2 * 1.35 / 1 λ1 = 2.70 meters λ2 = 2 * 1.35 / 2 λ2 = 1.35 meters λ3 = 2 * 1.35 / 3 λ3 = 0.90 meters The three longest wavelengths inside the bugle are 2.70 meters, 1.35 meters, and 0.90 meters.

Calculate the three lowest frequencies

Now, let's find the three lowest frequencies of standing waves inside the bugle. We will use the equation f = v / λ, where f is the frequency, v is the speed of sound (approximately 343 meters/second for air), and λ is the wavelength. f1 = v / λ1 f2 = v / λ2 f3 = v / λ3 Use speed of sound in air, v = 343 meters/second: f1 = 343 / 2.70 f1 ≈ 127.04 Hz f2 = 343 / 1.35 f2 ≈ 254.07 Hz f3 = 343 / 0.90 f3 ≈ 381.11 Hz The three lowest frequencies are approximately 127.04 Hz, 254.07 Hz, and 381.11 Hz.

Calculate the wavelengths of the sound produced in the air around the bugle

The wavelengths of the sound produced in the air around the bugle are the same as the wavelengths inside the bugle. Therefore, the three longest wavelengths of the sound produced in the air around the bugle are 2.70 meters, 1.35 meters, and 0.90 meters.

Key concepts

Overtones and Harmonics

When we talk about overtones and harmonics in musical instruments like bugles, we're diving into the exciting world of sound waves that repeat or "harmonize." In an open pipe bugle, both ends have antinodes, which are the points where air moves the most.
This unique feature allows the bugle to produce a series of sound frequencies or harmonics that are integral multiples of the fundamental frequency.

• The **fundamental frequency** is the simplest sound produced, where the length of the bugle is half of a wavelength \(L = \frac{\lambda}{2}\).
• **Overtones** are the sounds above that basic note, and they're what we call multiple waves fitting into the pipe's length. Essentially, these are just higher harmonics.
• Each harmonic order (first, second, third) corresponds to a specific overtone, contributing to the bugle's rich sound.

This results in a complex mix of sounds, leading to the melodious tunes we hear. Understanding harmonics helps us predict overtones and create musical instruments with distinct sounds.

Wavelength Calculations

Calculating the wavelengths of sound waves in an open pipe like a bugle involves a bit of math, but it’s interesting once you get the hang of it. The length of the bugle is key: with both ends open, it allows for standing waves with antinodes at both ends.
The general formula for the wavelengths in an open pipe is:\[ L = n\left(\frac{\lambda}{2}\right) \]Where:\- **\(L\)** is the length of the pipe (1.35 meters in this case),
- **\(n\)** is the harmonic number (1, 2, 3…) representing each subsequent wave.
To find the wavelengths, rearrange the formula to solve for \(\lambda\):

• For the first harmonic (n=1): \(\lambda_1 = 2 \cdot L = 2.70\) meters.
• For the second harmonic (n=2): \(\lambda_2 = L = 1.35\) meters.
• And for the third harmonic (n=3): \(\lambda_3 = \frac{2}{3} \cdot L = 0.90\) meters.

Each successive harmonic or overtone will have a shorter wavelength than the one before. These calculations reveal the three longest wavelengths of the standing waves inside the bugle.

Sound Frequency

Frequency in sound waves tells us how many times a wave passes a certain point in one second—measured in Hertz (Hz). It is the pitch we hear: higher frequencies mean a higher pitch.
In the exercise, each wavelength we calculated corresponds to a specific sound frequency.To find frequency **\(f\)**, we use the formula:\[ f = \frac{v}{\lambda} \] - **\(v\)** is the speed of sound in air, approximately 343 meters per second.
- **\(\lambda\)** is the wavelength we calculated earlier.So for our three longest wavelengths inside the bugle:

• The frequency for \(\lambda_1 = 2.70\) meters is \(f_1 \approx 127.04\) Hz.


• The frequency for \(\lambda_2 = 1.35\) meters is \(f_2 \approx 254.07\) Hz.


• The frequency for \(\lambda_3 = 0.90\) meters is \(f_3 \approx 381.11\) Hz.

These calculated frequencies help us understand how sound is perceived by our ears as distinct notes, each with its particular tone and pitch, allowing musicians to create harmonious tunes with their instruments.