Question

A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency \(512 \mathrm{~Hz}\). If the tube is opened at both ends what is the fundamental frequency that can be excited (in \(\mathrm{Hz}\) )? (A) 256 (B) 1024 (C) 128 (D) 512

Short answer

The fundamental frequency of the tube when opened at both ends is (B) 1024 Hz.

Step-by-step solution

Understand the fundamental frequency of a tube

The fundamental frequency of a tube depends on its length and whether it is open or closed at one or both ends. For a tube closed at one end, the fundamental frequency is determined by the following formula: \(f_1 = \frac{c}{4L}\), where \(f_1\) is the fundamental frequency, \(L\) is the length of the tube, and \(c\) is the speed of sound in air, which is approximately \(343 \frac{m}{s}\). For a tube opened at both ends, the fundamental frequency is given by: \(f_2 = \frac{c}{2L}\), where \(f_2\) is the new fundamental frequency.

Find the length of the tube using the closed tube formula

We are given the fundamental frequency of the tube when closed at one end (\(f_1\)) as \(512 Hz\). Therefore, we can find the length of the tube using the closed tube formula: \(L = \frac{c}{4f_1}\), Plugging in the given frequency and the speed of sound in air, we get: \(L = \frac{343}{4 \times 512} = \frac{343}{2048} m\)

Calculate the new fundamental frequency using the open tube formula

Now that we have the length of the tube, we can use the formula for the fundamental frequency of an open tube: \(f_2 = \frac{c}{2L}\), Plugging in the values of the speed of sound in air and the length of the tube, we get: \(f_2 = \frac{343}{2 \times \frac{343}{2048}} = 1024 Hz\)

Determine the correct answer

Comparing the calculated fundamental frequency (\(f_2 = 1024 Hz\)) to the given options: (A) 256 (B) 1024 (C) 128 (D) 512 The correct answer is (B) 1024 Hz.

Key concepts

Fundamental Frequency

Understanding the fundamental frequency of a tube is key to determining the sounds it can produce. It refers to the lowest frequency at which the tube naturally resonates. The formula for calculating this frequency depends on the tube's configuration. For a tube closed at one end, the fundamental frequency is given by the formula: \[ f_1 = \frac{c}{4L} \]Where:

• \( f_1 \) is the fundamental frequency.
• \( c \) is the speed of sound in air, typically around 343 m/s.
• \( L \) is the length of the tube.

With this knowledge, we can determine how modifications to the tube, such as opening both ends, affect its resonant frequencies.

Closed Tube Resonance

A tube closed at one end creates a unique resonance profile. This setup, often used in musical instruments such as clarinets, allows it to support specific sound waves. In closed tubes, only odd harmonics are present. The closed end forces a node, while the open end forces an antinode, configuring the sound wave into quarters of the wavelength. This restriction influences the fundamental frequency and leads to a lower pitch compared to an open tube of the same length. The fundamental frequency in a closed tube can be represented mathematically with the formula: \[ f_1 = \frac{c}{4L} \]This equation underscores how the tube's length and the speed of sound combine to define the resonance characteristics.

Open Tube Resonance

When a tube is open at both ends, the resonance pattern changes significantly. In this configuration, each end of the tube forms an antinode, allowing a more straightforward wave pattern. This setup is common in instruments like flutes. For open tubes, both even and odd harmonics are possible, leading to richer and more varied sound possibilities compared to closed tubes. The fundamental frequency for open tubes is expressed by the formula:\[ f_2 = \frac{c}{2L} \]In this case, the absence of a node at either end makes a wave complete a full half of its wavelength, effectively doubling the fundamental frequency of a closed tube of the same length.

Speed of Sound

The speed of sound in air is a crucial factor in determining the frequencies that can resonate within a tube. Typically, this speed is around 343 m/s at room temperature. It represents how fast sound waves travel through a medium. The speed of sound varies with temperature and medium. In tubes, higher temperatures can increase the speed of sound; consequently, the frequency of sound waves that resonate within the tube might also increase. Understanding this concept is vital when calculating the fundamental frequencies for different tube configurations. It directly influences the wavelength and adjusts the resonating frequencies when applied in formulas like:

• \( f_1 = \frac{c}{4L} \) for closed tubes.
• \( f_2 = \frac{c}{2L} \) for open tubes.

Acoustics in Tubes

Acoustics in tubes is an area that explores how sound behaves within hollow, cylindrical structures. The shape of the tube, whether closed or open, directly affects sound wave propagation and resonance. Tubes play a critical role in musical acoustics as they are commonly found in instruments like organs, trumpets, and woodwinds.

• Closed tubes produce a different sound spectrum due to constraints at one end leading to the dominance of odd harmonics.
• Open tubes, allowing for antinodes at both ends, produce a fuller range of harmonics.

Exploring tube acoustics allows us to understand sound wave behavior, influencing musical instrument design and architectural acoustics.