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A \(50-\mathrm{cm}\) -long wire with a mass of \(1.0 \mathrm{g}\) and a tension of \(440 \mathrm{N}\) passes across the open end of an open-closed tube of air. The wire, which is fixed at both ends, is bowed at the center so as to vibrate at its fundamental frequency and generate a sound wave. Then the tube length is adjusted until the fundamental frequency of the tube is heard. What is the length of the tube? Assume \(v_{\text {sound }}=340 \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
The length of the tube can be calculated by determining the wire's fundamental frequency, using it to find the wavelength of the sound wave, then using this wavelength to find the tube's length for its fundamental frequency. The calculations involve formulas from wave physics, particularly those related to string vibration and open-closed tube resonance.

Step by step solution

01

Calculate the Frequency of the String

The fundamental frequency (\(f\)) of a string fixed at both ends can be calculated using the formula \(f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\), where \(L\) is the length of the string, \(T\) is the tension in the string and \(\mu\) is the mass per unit length. Given that the length \(L=0.50m\), the tension \(T=440N\) and the mass of wire \(m = 1.0g = 0.001kg\), the mass per unit length \(\mu = \frac{m}{L} = \frac{0.001kg}{0.50m} = 0.002kg/m\). Substituting these values in the formula we can calculate \(f\).
02

Calculate the Tube Length

In an open-closed tube, the fundamental frequency (first harmonic) occurs when the length of the tube is one-fourth of the wavelength (\(L=\lambda / 4\)). The speed of sound \(v_{sound}\) is related to the wavelength \(\lambda\) and the frequency \(f\) by the equation \(v_{sound} = f \times \lambda\), so \(\lambda = v_{sound} / f\). After finding the value of \(f\) from Step 1 and given that \(v_{sound} = 340m/s\), we can substitute these values to find \(\lambda\). Subsequently, we can calculate the length of the tube, \(L\), by using the relation \(L=\lambda / 4\).

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Key Concepts

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

Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system vibrates. In the context of a vibrating string, this frequency determines the first harmonic—the simplest form of vibration for the string. The formula to calculate this frequency for a string fixed at both ends is \(f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\). Here's how it works:
  • \(L\) is the length of the string.
  • \(T\) is the tension applied to the string.
  • \(\mu\) is the mass per unit length of the string, calculated by dividing the mass of the string by its length.
These factors contribute to the speed at which waves travel along the string, impacting the frequency. By tweaking parameters like tension or length, you can change the fundamental frequency. This principle is crucial in music, where string instruments depend on such adjustments to produce different notes.
Open-Closed Tube Resonance
An open-closed tube creates a unique type of resonance because of its boundary conditions. At one end, the tube is open, allowing air movement, while the other end is sealed, reflecting sound waves back. This setup supports specific resonant frequencies known as harmonics.

The fundamental frequency, or first harmonic, occurs when the length of the tube \(L\) equals one-fourth of the sound wave's wavelength in air (\(L = \lambda / 4\)). For each harmonic, there are particular patterns of vibration:
  • The tube's open end becomes an antinode (point of maximum amplitude).
  • The closed end is a node (point of zero amplitude).
This means that only odd harmonics are present, unlike in open-open or closed-closed systems where even harmonics can also exist. To find the tube's length for a specific harmonic, solving for \(\lambda\) using the speed of sound and fundamental frequency is key. These principles explain the operation of wind instruments, like organs or clarinets.
Wave Speed in Air
The speed of sound in air is crucial for calculating wavelengths and understanding wave behavior. At standard atmospheric conditions, this speed is approximately 340 meters per second (m/s). However, it can vary depending on factors like:
  • Temperature: Warmer air increases sound speed due to more rapid molecular movement.
  • Humidity: More humidity also boosts speed, as water vapor is less dense than dry air.
In this exercise, the sound speed is an integral factor for determining the wavelength and, subsequently, calculating the resonance length of the tube. The relationship between speed \(v\), frequency \(f\), and wavelength \(\lambda\) is given by the equation \(v = f \times \lambda\). Thus, knowing any two of these allows us to find the third. This interplay is foundational in acoustics, impacting how we perceive sound in different environments.

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Most popular questions from this chapter

An old mining tunnel disappears into a hillside. You would like to know how long the tunnel is, but it's too dangerous to go inside. Recalling your recent physics class, you decide to try setting up standing-wave resonances inside the tunnel. Using your subsonic amplifier and loudspeaker, you find resonances at \(4.5 \mathrm{Hz}\) and 6.3 Hz, and at no frequencies between these. It's rather chilly inside the tunnel, so you estimate the sound speed to be \(335 \mathrm{m} / \mathrm{s}\). Based on your measurements, how far is it to the end of the tunnel?

Two in-phase loudspeakers separated by distance \(d\) emit \(170 \mathrm{Hz}\) sound waves along the \(x\) -axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on. What are three possible values for d? Assume a sound speed of \(340 \mathrm{m} / \mathrm{s}\).

A carbon dioxide laser is an infrared laser. A \(\mathrm{CO}_{2}\) laser with a cavity length of 53.00 cm oscillates in the \(m=100,000\) mode. What are the wavelength and frequency of the laser beam?

What are the three longest wavelengths for standing sound waves in a 121 -cm- long tube that is (a) open at both ends and (b) open at one end, closed at the other?

The lowest note on a grand piano has a frequency of \(27.5 \mathrm{Hz}\) The entire string is \(2.00 \mathrm{m}\) long and has a mass of \(400 \mathrm{g}\). The vibrating section of the string is \(1.90 \mathrm{m}\) long. What tension is needed to tune this string properly?

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