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You blow across the open mouth of an emply test tube and produce the fundamental standing wave of the air column inside the test tube. The speed of sound in air is 344 \(\mathrm{m} / \mathrm{s}\) and the test tube acts as a stopped pipe. (a) If the length of the air column in the test tube is \(14.0 \mathrm{cm},\) what is the frequency of this standing wave? (b) What is the frequency of the fundamental standing wave in the air column if the test tube is half filled with water?

Short Answer

Expert verified
(a) 614.29 Hz. (b) 1228.57 Hz.

Step by step solution

01

Understand the Fundamental Standing Wave in a Stopped Pipe

A stopped pipe is closed at one end. For such a pipe, the fundamental harmonic has a node at the closed end and an antinode at the open end. The length of the pipe is a quarter of the wavelength of the sound.
02

Calculate the Wavelength for the Fundamental Standing Wave

Given that the length of the air column is 14.0 cm, or 0.14 meters, the length represents \(\frac{\lambda}{4}\) for the fundamental frequency. Therefore, the wavelength \(\lambda\) is,\[\lambda = 4 \times 0.14 = 0.56 \text{ meters}\]
03

Calculate the Frequency of the Fundamental Standing Wave (Part a)

The formula to find the frequency \(f\) of the wave is given by \[f = \frac{v}{\lambda}\] where \(v\) is the speed of sound (344 m/s). Substituting the values, we get:\[f = \frac{344}{0.56} = 614.29 \text{ Hertz}\] Thus, the frequency of the fundamental standing wave is approximately 614.29 Hz.
04

Understand the Change When the Tube is Half-Filled with Water

When the test tube is half-filled with water, the length of the air column is halved. This effectively changes the air column length to 7.0 cm, or 0.07 meters.
05

Calculate the New Wavelength for the Reduced Air Column

Following a similar logic, since the length of the tube (0.07 m) now represents \(\frac{\lambda}{4}\), the new wavelength is,\[\lambda = 4 \times 0.07 = 0.28 \text{ meters}\]
06

Calculate the Frequency of the New Fundamental Standing Wave (Part b)

Using the same frequency formula \(f = \frac{v}{\lambda}\) with the new wavelength:\[f = \frac{344}{0.28} = 1228.57 \text{ Hertz}\] Thus, the frequency of the fundamental standing wave when the tube is half-filled with water is approximately 1228.57 Hz.

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Key Concepts

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

Sound Frequency
Sound frequency refers to the number of vibrations or cycles of a sound wave that occur in one second. It is measured in Hertz (Hz), where one Hertz equals one cycle per second. Higher frequencies result in higher pitched sounds, whereas lower frequencies produce lower pitched sounds. In musical terms, sound frequency determines the pitch of a note. The frequency of a sound wave can be calculated using the formula:
  • \( f = \frac{v}{\lambda} \)
where \( f \) is the frequency, \( v \) is the speed of sound, and \( \lambda \) represents the wavelength. This formula implies that if the speed of sound remains constant, the frequency varies inversely with the wavelength.
Understanding sound frequency is crucial in various fields, including acoustics and audio engineering. For students, grasping this concept can aid in experiments involving sound waves, such as measuring sound properties within different mediums.
Stopped Pipe
A stopped pipe is a tube that is closed at one end and open at the other. This configuration of a pipe influences the formation of standing waves within it. In a stopped pipe, the closed end acts as a node (point of minimal movement) and the open end serves as an antinode (point of maximum movement).
This arrangement means that the length of the stopped pipe constitutes a quarter of the wavelength of the sound that produces the fundamental frequency. This is because a complete standing wave in a stopped pipe must adjust its nodes and antinodes according to the pipe’s physical constraints.
  • For a pipe length \( L \), the wavelength \( \lambda \) is given by:\( L = \frac{\lambda}{4} \).
Understanding stopped pipes is essential when studying acoustics and the design of musical instruments, as it illustrates how physical boundaries affect sound wave behavior.
Wavelength Calculation
Wavelength calculation is a pivotal step in understanding sound waves within mediums like pipes. The wavelength \( \lambda \) of a sound wave in a stopped pipe is particularly significant as it determines the fundamental frequency of the sound. By knowing either the length of the air column or the frequency of the sound, one can calculate the remaining variables using the relationship between wavelength, speed, and frequency.
  • For a stopped pipe, the wavelength is calculated using:\( \lambda = 4 \times L \), where \( L \) is the length of the pipe.
When the length of the air column changes, such as when a pipe is partially filled with water, the wavelength is directly affected which in turn alters the frequency of the sound produced. By understanding and calculating wavelength, students can predict how alterations in physical dimensions affect acoustic properties, thus honing their practical problem-solving skills in physical acoustics.

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Most popular questions from this chapter

CP Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F .\) (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { beat }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 Hz. One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?

Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from A. What is the closest you can be to \(B\) and be at a point of destructive interference?

Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 860 Hz. Point \(P\) is 12.0 m from \(A\) and 13.4 \(\mathrm{m}\) from \(B\) . Is the interference at \(P\) constructive or destructive? Give the reasoning behind your answer.

A loud factory machine produces sound having a displace ment amplitude of \(1.00 \mu \mathrm{m},\) but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 \(\mathrm{Pa.}\) Under the conditions of this factory, the bulk modulus of air is \(1.42 \times 10^{5}\) Pa. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit? Is this frequency audible to the workers?

Example 16.1 (Section 16.1\()\) showed that for sound waves in air with frequency 1000 \(\mathrm{Hz}\) , a displacement amplitude of \(1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2}\) Pa. Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s} .\) For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain whyyour answer is much less than \(1.2 \times 10^{-8} \mathrm{m}\) .

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