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You blow across the open mouth of an empty test tube and produce the fundamental standing wave of the air column inside the test tube. The speed of sound in air is 344 m/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 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

Understanding the Concept of a Stopped Pipe

A test tube acts as a stopped pipe (closed at one end), which results in the air column having a fundamental frequency that corresponds to a quarter of the wavelength. This is because the closed end must have a node and the open end an antinode. The fundamental frequency of a stopped pipe is given by \( f = \frac{v}{4L} \), where \( v \) is the speed of sound in air and \( L \) is the length of the air column.
02

Calculating the Frequency for a Full-Length Air Column

Using the formula for a stopped pipe, \( f = \frac{v}{4L} \), we substitute the given values: the speed of sound in air, \( v = 344 \text{ m/s} \), and the length of the air column, \( L = 14.0 \text{ cm} = 0.14 \text{ m} \). Thus, the frequency is calculated as:\[f = \frac{344 \, \text{m/s}}{4 \times 0.14 \, \text{m}} = 614.29 \, \text{Hz}\]}},{
03

Adjusting the Air Column Length for Partially Filled Test Tube

When the test tube is half-filled with water, the effective length of the air column is halved. Thus, the new length \( L' = \frac{0.14 \text{ m}}{2} = 0.07 \, \text{m}\).
04

Calculating the Frequency for the Shortened Air Column

Using the adjusted length in the same formula, \( f = \frac{v}{4L'} \), substitute the new length \( L' = 0.07 \, \text{m} \):\[f = \frac{344 \, \text{m/s}}{4 \times 0.07 \, \text{m}} = 1228.57 \, \text{Hz}\]

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Key Concepts

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

Stopped Pipe
A stopped pipe, also called a closed pipe, is a hollow tube with one end sealed and the other open. This configuration is common in various wind instruments, like clarinets and test tubes, when interacting with sound waves. The closed end of the pipe creates a boundary where the air can't move, forming a node here. Meanwhile, at the open end, air can move freely, forming an antinode. This setup means that the length of the stopped pipe supports a standing wave consisting of a quarter of a wavelength.

When air is blown across the open end, the vibrations inside the pipe produce a standing wave. This is the simplest form of vibration it can sustain and is known as the fundamental frequency. In a stopped pipe, the fundamental frequency corresponds to a standing wave with a single loop between the node and antinode, taking up a quarter wavelength.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system naturally vibrates. In this context, it refers to the simplest standing wave pattern in the air column of a stopped pipe. The formula to find the fundamental frequency of a standing wave in a stopped pipe is given by:
\[ f = \frac{v}{4L} \]Here:
  • \( f \) is the fundamental frequency.
  • \( v \) is the speed of sound in the air, measured in meters per second (m/s).
  • \( L \) is the length of the air column inside the stopped pipe, measured in meters.
For instance, in a test tube acting as a stopped pipe, if the air column's length is 14 cm, the fundamental frequency with sound speed at 344 m/s can be mathematically expressed as:

\[ f = \frac{344 \text{ m/s}}{4 \times 0.14 \text{ m}} = 614.29 \text{ Hz} \]
This indicates the air inside resonates primarily at approximately 614.29 hertz, creating the first harmonic.
Speed of Sound
The speed of sound refers to how fast sound waves travel through a medium. It is influenced by factors such as air temperature and pressure. For many calculations, a standardized speed of sound in air is used, which is typically around 344 meters per second at room temperature.

This value is crucial in determining wave properties such as frequency and wavelength in problems involving sound waves. It forms part of the fundamental frequency equation, linking the physical attributes of the medium (air in this case) and the resultant auditory properties We use this when calculating how sounds behave in different situations, like within a stopped pipe or through air columns of varying lengths.
Air Column Length
The air column length is the distance between the closed node and the open antinode in a stopped pipe. This length determines the wavelength of the standing wave formed. In a stopped pipe, the fundamental frequency occurs when the air column is one quarter of the sound's wavelength.

When half of the test tube is filled with water, the air column length effectively reduces by half. This change requires recomputation of the fundamental frequency, as the new air column length (\[ L' \]) affects the standing wave’s wavelength within the pipe.

For example, for a water-filled test tube, the air column becomes 0.07 meters, resulting in a recalculated fundamental frequency of:
\[ f = \frac{344 \text{ m/s}}{4 \times 0.07 \text{ m}} = 1228.57 \text{ Hz} \]This doubling of frequency as the air column length halves demonstrates how sensitive sound waves are to changes in the length of the column in closed-ended configurations.

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Most popular questions from this chapter

A loud factory machine produces sound having a displacement amplitude of 1.00 \(\mu\)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 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?

A 75.0-cm-long wire of mass 5.625 g is tied at both ends and adjusted to a tension of 35.0 N. When it is vibrating in its second overtone, find (a) the frequency and wavelength at which it is vibrating and (b) the frequency and wavelength of the sound waves it is producing.

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