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What must be the length of an iron rod that has the fundamental frequency \(320 \mathrm{~Hz}\) when clamped at its center? Assume longitudinal vibration at a speed of \(5.00 \mathrm{~km} / \mathrm{s}\).

Short Answer

Expert verified
The length of the rod must be approximately 7.81 meters.

Step by step solution

01

Understanding the Problem

This problem involves longitudinal vibrations of an iron rod. The fundamental frequency of the rod is given as 320 Hz, and we have the speed of sound in iron as 5.00 km/s (which is 5000 m/s). We need to find the length of the rod when it is clamped at its center.
02

Recall Formula for Fundamental Frequency

For a rod clamped at the center and vibrating longitudinally, the fundamental frequency \( f \) can be calculated using the formula:\[ f = \frac{v}{2L} \]where \( f \) is the frequency, \( v \) is the speed of sound in the material, and \( L \) is the length of the rod.
03

Substitute Known Values

We are given \( f = 320 \, \text{Hz} \) and \( v = 5000 \, \text{m/s} \). Substituting these values into the formula:\[ 320 = \frac{5000}{2L} \]
04

Solve for the Length L

Rearrange the formula to solve for \( L \):\[ 2L = \frac{5000}{320} \]\[ L = \frac{5000}{640} \]\[ L = 7.8125 \, \text{meters} \]
05

Conclude the Calculation

Therefore, the length of the rod must be approximately 7.81 meters for the rod to have a fundamental frequency of 320 Hz.

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Key Concepts

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

Fundamental Frequency
When a material vibrates, it creates sound waves. The fundamental frequency refers to the lowest frequency at which an object can naturally vibrate. This frequency is important because it determines the resonance of the object. In the case of the iron rod clamped at its center, the fundamental frequency is the lowest note or pitch it can produce when it vibrates along its length. Understanding this helps in calculating the precise length of the rod to achieve a specific frequency of vibration.

The formula used to find the fundamental frequency in this problem is:
  • \( f = \frac{v}{2L} \)
where:
  • \( f \) is the fundamental frequency
  • \( v \) is the speed of sound in the rod's material
  • \( L \) is the length of the rod
This equation shows how frequency is linked to both the wave speed and the size of the object that is vibrating.
Acoustics
Acoustics is the study of sound. It involves understanding how sound is produced, controlled, transmitted, and received. In this scenario, acoustics focuses on how the iron rod transmits sound waves when it vibrates.

When dealing with sound, it's important to comprehend how different frequencies affect the sound that reaches our ears. High frequencies produce high-pitched sounds, while low frequencies produce lower-pitched sounds. The fundamental frequency is therefore a crucial concept in acoustics because it represents the primary pitch of a sound.

In practice, acoustics can be used to design musical instruments, improve architectural soundscapes, and enhance sound quality in audio devices. Understanding how objects like the iron rod vibrate gives us insight into creating and controlling sound in various environments.
Wave Speed
Wave speed is the velocity at which a wave travels through a medium. For longitudinal vibrations in a rod, it refers specifically to how fast sound waves move through the material of the rod, in this case, iron.

The wave speed (\( v \)) is a key determinant in the formula for fundamental frequency \( f \):
  • \( f = \frac{v}{2L} \)
In this problem, the wave speed through iron is given as 5000 m/s. This speed indicates how quickly vibrations travel from one end of the rod to the other.

This information is necessary not only for calculating frequency but also for applications in engineering and physics. It helps in understanding the material's ability to conduct sound, which is essential for applications such as music production, building construction, and even in designing equipment that is affected by vibrations.
Material Properties
Material properties dictate how well an object can conduct sound and vibrations. When examining an iron rod, several material properties come into play, such as density, elasticity, and tensile strength.

For the purpose of understanding its fundamental frequency, we focus on how these properties affect wave speed. Since iron has a high density and a specific modulus of elasticity, it is capable of allowing sound waves to propagate quickly, which is why it has a wave speed of 5000 m/s.

Material properties are not only crucial for determining characteristics like wave speed but also for practical applications. For instance:
  • Choosing the right materials for construction to minimize sound vibrations.
  • Designing musical instruments that require specific material attributes to produce desired notes or tones.
  • Engineering solutions that require soundproofing or sound enhancement.
Therefore, understanding material properties allows engineers and scientists to select or modify materials to achieve the desired acoustic or physical result.

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Most popular questions from this chapter

A glass tube that is \(70.0 \mathrm{~cm}\) long is open at both ends. Find the frequencies at which it will resonate when driven by sound waves that have a speed of \(340 \mathrm{~m} / \mathrm{s}\). A pipe that is open at both ends must have an antinode at each end. It will therefore resonate as in Fig. \(22-8\). From the diagram it can be seen that the resonance wavelengths \(\lambda_{n}\) are given by $$ L=n\left(\frac{\lambda_{n}}{2}\right) \quad \text { or } \quad \lambda_{n}=\frac{2 L}{n} $$ where \(n\) is an integer. But \(\lambda_{n}=v / f_{n}\), therefore $$ f_{n}=\left(\frac{n}{2 L}\right)(v)=(n)\left(\frac{340 \mathrm{~m} / \mathrm{s}}{2 \times 0.700 \mathrm{~m}}\right)=243 n \mathrm{~Hz} $$

When driven by a 120-Hz vibrator, a string has transverse waves of \(31 \mathrm{~cm}\) wavelength traveling along it. (a) What is the speed of the waves on the string? \((b)\) If the tension in the string is \(1.20 \mathrm{~N}\), what is the mass of \(50 \mathrm{~cm}\) of the string?

Steel and silver wires of the same diameter and same length are stretched with equal tension. Their densities are \(7.80 \mathrm{~g} / \mathrm{cm}^{3}\) and \(10.6 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. What is the fundamental frequency of the silver wire if that of the steel is \(200 \mathrm{~Hz}\) ?

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.

A horizontal cord \(5.00 \mathrm{~m}\) long has a mass of \(1.45 \mathrm{~g}\). What must be the tension in the cord if the wavelength of a 120 - \(\mathrm{Hz}\) wave on it is to be \(60.0 \mathrm{~cm}\) ? How large a mass must be hung from its end (say, over a pulley) to give it this tension? We know that the speed of a wave on a rope depends on both the tension and the mass per unit length. Moreover, Further, since $$ \begin{array}{l} v=\lambda f=(0.600 \mathrm{~m})\left(120 \mathrm{~s}^{-1}\right)=72.0 \mathrm{~m} / \mathrm{s} \\ v=\sqrt{(\text { Tension }) /(\text { Mass per unit length })} \end{array} $$ $$ \text { Tension }=(\text { Mass per unit length })\left(v^{2}\right)=\left(\frac{1.45 \times 10^{-3} \mathrm{~kg}}{5.00 \mathrm{~m}}\right)(72.0 \mathrm{~m} / \mathrm{s})^{2}=1.50 \mathrm{~N} $$ The tension in the cord balances the weight of the mass hung at its end. Therefore, $$ F_{T}=m g \quad \text { or } \quad m=\frac{F_{T}}{g}=\frac{1.50 \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=0.153 \mathrm{~kg} $$

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