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A tube \(1.20 \mathrm{~m}\) long is closed at one end. A stretched wire is placed near the open end. The wire is \(0.330 \mathrm{~m}\) long and has a mass of \(9.60 \mathrm{~g}\). It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

Short Answer

Expert verified
(a) Frequency: 71.5 Hz, (b) Tension: 392.4 N.

Step by step solution

01

Understand the tube's fundamental frequency

For a tube closed at one end, the fundamental frequency can be calculated using the formula \( f_1 = \frac{v}{4L} \), where \( v \) is the speed of sound (approximately \( 343 \text{ ms}^{-1} \) at room temperature), and \( L \) is the length of the tube. Substitute \( L = 1.20 \text{ m} \) into the equation to calculate \( f_1 \).
02

Calculate the Fundamental Frequency of the Tube

Substitute the length \( L = 1.20 \text{ m} \) and speed of sound \( v = 343 \text{ ms}^{-1} \) into the fundamental frequency formula: \[ f_1 = \frac{343}{4 \times 1.20} = 71.5 \text{ Hz} \]
03

Understand the wire's fundamental frequency

The fundamental mode of an oscillating wire fixed at both ends is given by \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, \( \mu \) is the linear mass density (mass per unit length), and \( L \) is the length of the wire. Our goal is to find \( T \).
04

Calculate the Linear Mass Density of the Wire

Convert the mass of the wire to kilograms: \( 9.60 \text{ g} = 0.00960 \text{ kg} \). The length of the wire is \( 0.330 \text{ m} \). The linear mass density is \( \mu = \frac{0.00960}{0.330} = 0.02909 \text{ kg/m} \).
05

Set Wire Frequency Equal to Tube Frequency

Since the wire causes resonance in the air column, both have the same fundamental frequency: \( f = 71.5 \text{ Hz} \). Set the wire's frequency formula equal to this value: \( 71.5 = \frac{1}{2 \times 0.330} \sqrt{\frac{T}{0.02909}} \).
06

Solve for the Tension in the Wire

Rearrange the equation from Step 5 to solve for \( T \): \[ T = \left( 2 \times 0.330 \times 71.5 \right)^2 \times 0.02909 \] Calculate \( T \) to get the tension: \[ T = 392.4 \text{ N} \]
07

Verify Results

Check to ensure calculations are based on correct physics equations and consistent units. The tube's fundamental frequency was calculated first and then used in finding tension. Cross-reference with typical values to ensure plausibility.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Frequency
Fundamental frequency is the lowest frequency at which a system naturally vibrates. It forms the basis from which other frequencies, known as harmonics, are derived. Understanding this concept is crucial in physics, especially in resonance phenomena where various systems can naturally oscillate. In our problem, the tube closed at one end has its fundamental frequency determined by a straightforward formula. For a closed tube like ours, the fundamental frequency (\( f_1 \) ) is calculated using:- \( f_1 = \frac{v}{4L} \)Here, \( v \) stands for the speed of sound, which is approximately \( 343 \text{ ms}^{-1} \) in air at room temperature, and \( L \) is the length of the tube. By substituting the actual length of the tube (1.20 m), we find out that the fundamental frequency is 71.5 Hz.This frequency signifies the lowest pitch the air column in the tube can produce when set into oscillation. It is a unique frequency that causes standing waves to form in the tube, resulting in resonance when similar frequencies from surrounding systems are encountered. Such fundamental frequencies lay the groundwork for understanding how harmonics behave.
Harmonics in Sound
Harmonics in sound are integral multiples of the fundamental frequency. These are the additional frequencies at which a system can resonate, producing overtones that combine with fundamental frequency to form complex sounds. Each harmonic represents an additional mode of vibration, contributing to the richness of sounds we hear. In our scenario, the wire and tube resonate together at their fundamental frequency due to their compatible harmonic structures. Harmonics help explain why specific systems like musical instruments produce distinctive sounds. For instance, a tube that's closed at one end supports only odd harmonics (like 3rd, 5th) because of its boundary conditions, while open tubes can support all harmonics. Harmonics in the wire oscillating at its fundamental frequency ensures it vibrates with precisely half a wavelength along its length. As this wire vibrates, its motion can bring surrounding air columns, like the one in our tube, into harmonics-affiliated resonance, influencing the sound's perceived qualities.
Tension in Oscillating Systems
Tension is a significant factor influencing the frequency at which a system oscillates. For musical strings or wires, tension determines how tight or loose the string is, directly impacting its vibrational frequency.The tension in the wire is vital for ensuring it oscillates at a frequency that causes resonance in nearby air columns. In solving for tension, we use the fundamental frequency equation for a wire:- \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \)Here, \( T \) represents tension and \( \mu \) is the linear mass density, calculated by dividing the mass of the wire by its length. By equating the wire's frequency with the tube's fundamental frequency (which is 71.5 Hz in our case), we can rearrange the formula to solve for \( T \).Through our calculation, the tension required to achieve this resonance is about 392.4 N. This tension determines how prominently the wire influences the air column's oscillations. Understanding and calculating tension allows us to precisely tune systems, be it for scientific applications or musical performances, to achieve desired acoustical characteristics.

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Most popular questions from this chapter

A certain loudspeaker system emits sound isotropically with a frequency of \(2000 \mathrm{~Hz}\) and an intensity of \(0.960 \mathrm{~mW} / \mathrm{m}^{2}\) at a distance of \(6.10 \mathrm{~m}\). Assume that there are no reflections. (a) What is the intensity at \(30.0 \mathrm{~m}\) ? At \(6.10 \mathrm{~m}\), what are (b) the displacement amplitude and (c) the pressure amplitude?

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