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Chapter 22: Problem 44

Determine the length of the shortest air column in a cylindrical jar that will strongly reinforce the sound of a tuning fork having a vibration rate of \(512 \mathrm{~Hz}\). Use \(v=340 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air.

Short Answer

Expert verified
The shortest air column reinforcing 512 Hz is 0.166 meters.

Step by step solution

01

Understanding the Problem

We need to find the length of an air column in a cylindrical jar where the frequency of sound produced by a tuning fork is reinforced. The frequency of the tuning fork is given as 512 Hz, and the speed of sound is 340 m/s. This situation represents a resonance condition.
02

Identifying the Relationship

In an open-closed cylindrical jar (one end open, one end closed), the resonance condition occurs at odd multiples of a quarter wavelength \(\left(\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \ldots\right)\). We are looking for the shortest length, which occurs at the first quarter wavelength (\(\frac{1}{4}\)).
03

Using the Resonance Formula

For the fundamental frequency, the length \(L\) of the air column is \(L = \frac{\lambda}{4}\), where \(\lambda\) is the wavelength. The wavelength \(\lambda\) can be calculated using \(\lambda = \frac{v}{f}\), where \(v\) is the speed of sound and \(f\) is the frequency of the tuning fork.
04

Calculating the Wavelength

Plug in the speed of sound and the frequency into the wavelength formula: \[\lambda = \frac{340 \, \text{m/s}}{512 \, \text{Hz}}.\] This gives us the wavelength \(\lambda\).
05

Determining the Length of the Air Column

Substitute the value of \(\lambda\) into the formula for \(L\): \[L = \frac{\lambda}{4} = \frac{\frac{340}{512}}{4}.\] Simplifying this gives us the shortest length of the air column.

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Key Concepts

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

Cylindrical Air Column
A cylindrical air column is simply an air-filled tube, where sound waves can travel. Imagine sound waves bouncing back and forth inside a cylinder, similar to air getting trapped in a flute or an organ pipe.

The length of the column is a crucial factor when it comes to resonance of sound waves in the cylinder. In our scenario, one end of the cylindrical jar is open, and the other is closed. This setup provides a specific scenario for understanding how such columns can amplify sound.

Understanding the shape and boundaries of this column helps us see how sound waves behave, creating areas of high and low pressure called nodes and antinodes, respectively. As a result, the right length of this column resonates with the sound from a tuning fork, amplifying it effectively.
Tuning Fork Frequency
The tuning fork is a device that produces sound at a specific frequency when struck. In this exercise, the tuning fork has a frequency of 512 Hz.

Frequency refers to how many cycles of sound waves pass a certain point in one second. So, in this case, 512 sound wave cycles pass each second. This very specific frequency corresponds to a particular musical note, aiding musicians and scientists when precise pitch is needed.

The tuning fork鈥檚 frequency plays a vital role in understanding resonance. To reinforce the sound from a tuning fork, we need the air column's length to be in harmony with the frequency of the tuning fork. Thus, sound waves are amplified through resonance when both tuning fork and air column frequencies align.
Speed of Sound in Air
The speed of sound is the rate at which sound waves travel through a medium鈥攊n our case, air. For this exercise, the speed of sound is provided as 340 m/s. This speed can vary based on factors such as air temperature and pressure, but 340 m/s is a common classroom approximation.

Understanding the speed of sound is essential as it helps calculate the wavelength of sound for a given frequency. The wavelength, being the distance between two consecutive crests or compressions of a wave, depends inversely on frequency.

With a known speed of sound and the frequency of the tuning fork, we can determine how long a single cycle of sound is in meters using the formula: \[\lambda = \frac{v}{f}\]where \(\lambda\) represents the wavelength, \(v\) is the speed of sound, and \(f\) is the frequency of the tuning fork.
Open-Closed Resonator
An open-closed resonator describes a system where one end of a column is open to the air, while the other end is closed. This type of resonator is significant in physics and music as it forms standing waves at odd multiples of quarter wavelengths. This characteristic allows it to effectively amplify certain frequencies.

In our cylindrical jar setup, the open-closed configuration creates a fundamental or first resonance condition, which is the shortest possible resonant length, at one quarter of the wavelength鈥攎eaning the air column resonates most strongly at this length.

To find this length, one uses the formula for the fundamental frequency:\[L = \frac{\lambda}{4}\]The longest wave that fits in this setup shows where the reinforcement or resonance effectively makes the sound louder, thereby helping understand how musical instruments or sound systems leverage air columns for powerful sound production.

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Most popular questions from this chapter

(a) At what point should a stretched string be plucked to make its fundamental tone most prominent? At what point should it be plucked and then at what point touched \((b)\) to make its first overtone most prominent and \((c)\) to make its second overtone most nrominent?

A string fastened at both ends resonates at \(420 \mathrm{~Hz}\) and \(490 \mathrm{~Hz}\) with no resonant frequencies in between. Find its fundamental resonant frequency. In general, \(f_{n}=n f_{1}\). We are told that \(f_{n}=420 \mathrm{~Hz}\) and \(f_{n+1}=490 \mathrm{~Hz}\). Therefore, $$ 420 \mathrm{~Hz}=n f_{1} \text { and } 490 \mathrm{~Hz}=(n+1) f_{1} $$ Subtract the first equation from the second to obtain \(f_{1}=70.0 \mathrm{~Hz}\).

A metal rod \(40 \mathrm{~cm}\) long is dropped, end first, onto a wooden floor and rebounds into the air. Compression waves of many frequencies are thereby set up in the bar. If the speed of compression waves in the bar is \(5500 \mathrm{~m} / \mathrm{s}\), to what lowest- frequency compression wave will the bar resonate as it rebounds? Both ends of the bar will be free, and so antinodes will exist there. In the lowest resonance form (i.e., the one with the longest segments), only one node will exist on the bar, at its center, as illustrated in.We will then have $$ L=2\left(\frac{\lambda}{4}\right) \quad \text { or } \quad \lambda=2 L=2(0.40 \mathrm{~m})=0.80 \mathrm{~m} $$ Then, from \(\lambda=v T=\mathrm{v} / \mathrm{f}\), $$ f=\frac{v}{\lambda}=\frac{5500 \mathrm{~m} / \mathrm{s}}{0.80 \mathrm{~m}}=6875 \mathrm{~Hz}=6.9 \mathrm{kHz} $$

A banjo string \(30 \mathrm{~cm}\) long oscillates in a standing-wave pattern. It resonates in its fundamental mode at a frequency of \(256 \mathrm{~Hz}\). What is the tension in the string if \(80 \mathrm{~cm}\) of the string have a mass of \(0.75 \mathrm{~g}\) ? First we'll find \(u\) and then the tension. The string vibrates in one segment when \(f=256 \mathrm{~Hz}\). Therefore, from (\mathrm{a}) \text { : }}\( and $$ $$ \frac{0.75 \times 10^{-3} \mathrm{~kg}}{0.80 \mathrm{~m}}=9.4 \times 10^{-4} \mathrm{~kg} / \mathrm{m} $$ Then, from, \)v=\sqrt{(\text { Tension }) /(\text { Mass per unit length })}$, $$ F_{T}=(154 \mathrm{~m} / \mathrm{s})^{2}\left(9.4 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\right)=22 \mathrm{~N} $$ \begin{array}{l} \frac{\lambda}{2}=L \quad \text { or } \quad \lambda=(0.30 \mathrm{~m})(2)=0.60 \mathrm{~m} \\ v=f \lambda=\left(256 \mathrm{~s}^{-1}\right)(0.60 \mathrm{~m})=154 \mathrm{~m} / \mathrm{s} \end{array} $$ The mass per unit length of the string is

A 60-cm-long bar, clamped at its middle, is vibrated lengthwise by an alternating force at its end. Its fundamental resonance frequency is found to be \(3.0 \mathrm{kHz}\). What is the speed of longitudinal waves in the bar? Because its ends are free, the bar must have antinodes there. The clamp point at its center must be a node. Therefore, the fundamental resonance is as shown in Because the distance from node to antinode is always \(\frac{1}{4} \lambda\), we see that \(L=2\left(\frac{1}{4} \lambda\right)\). Since \(L=0.60 \mathrm{~m}\), we find \(\lambda=1.20 \mathrm{~m}\). Then, from the basic relation (p. 274) \(\lambda=\mathrm{v} / f\), we have $$ u=\lambda f=(1.20 \mathrm{~m})(3.0 \mathrm{kHz})=3.6 \mathrm{~km} / \mathrm{s} $$
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