Question

A bugle can be represented by a cylindrical pipe of length \(L=1.35 \mathrm{~m} .\) The pipe is open at one end and closed at the other end (the end with the mouthpiece). 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

The three longest wavelengths of sound in the air around the bugle are 5.4 m, 1.8 m, and 1.08 m. Their corresponding frequencies are approximately 63.52 Hz, 190.56 Hz, and 317.59 Hz.

Step-by-step solution

Determine the longest three wavelengths of standing waves inside the closed pipe

Since we have a closed pipe, we have to consider the positions of nodes and anti-nodes within the pipe. For every oscillation mode, there is a node at the closed end and an anti-node at the open end. The longest wavelength corresponds to the fundamental (first harmonic) mode, which means there is only one node-antinode pair. From the node at the closed end, the length of the fundamental mode is 1/4 of the full wavelength. So, the longest wavelength inside the bugle can be found using the equation: \(\lambda_1 = \frac{4L}{1} = 4 \times 1.35 \mathrm{~m} = 5.4 \mathrm{~m}\) For the next longest wavelength, we have two node-antinode pairs (2nd harmonic). Therefore, \(\lambda_2 = \frac{4L}{3} = \frac{4 \times 1.35 \mathrm{~m}}{3} = 1.8 \mathrm{~m}\) For the third longest wavelength, we have three node-antinode pairs (3rd harmonic). Therefore, \(\lambda_3 = \frac{4L}{5} = \frac{4 \times 1.35 \mathrm{~m}}{5} = 1.08 \mathrm{~m}\)

Calculate the three lowest frequencies of the sound inside the bugle

To find the frequencies produced by these standing waves, we can use the wave equation: \(v = f\lambda\) where v is the speed of sound in the medium, f is the frequency, and \(\lambda\) is the wavelength. The speed of sound in air at room temperature is approximately \(v = 343 \mathrm{~m/s}\). For each wavelength, we can find the respective frequency: \(f_1 = \frac{v}{\lambda_1} = \frac{343 \mathrm{~m/s}}{5.4 \mathrm{~m}} \approx 63.52 \mathrm{~Hz}\) \(f_2 = \frac{v}{\lambda_2} = \frac{343 \mathrm{~m/s}}{1.8 \mathrm{~m}} \approx 190.56 \mathrm{~Hz}\) \(f_3 = \frac{v}{\lambda_3} = \frac{343 \mathrm{~m/s}}{1.08 \mathrm{~m}} \approx 317.59 \mathrm{~Hz}\)

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

For the sound produced in the air around the bugle, we consider it as an open-open pipe. Thus, there are no constraints on the nodes and antinodes, and we can use the same frequencies determined in Step 2. Using the wave equation, we can find the longest wavelengths of the sound produced in the air: \(\lambda_1' = \frac{v}{f_1} = \frac{343 \mathrm{~m/s}}{63.52 \mathrm{~Hz}} = 5.4 \mathrm{~m}\) \(\lambda_2' = \frac{v}{f_2} = \frac{343 \mathrm{~m/s}}{190.56 \mathrm{~Hz}} = 1.8 \mathrm{~m}\) \(\lambda_3' = \frac{v}{f_3} = \frac{343 \mathrm{~m/s}}{317.59 \mathrm{~Hz}} = 1.08 \mathrm{~m}\) The three longest wavelengths of sound in the air around the bugle are 5.4 m, 1.8 m, and 1.08 m, with corresponding frequencies of approximately 63.52 Hz, 190.56 Hz, and 317.59 Hz.