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Chapter 13: Problem 8

\(\bullet\bullet\) Atoms in a solid are in continuous vibrational motion due to thermal energy. At room temperature, the amplitude of these atomic vibrations is typically about \(10^{-9} \mathrm{~cm},\) and their frequency is on the order of \(10^{12} \mathrm{~Hz}\). (a) What is the approximate period of oscillation of a typical atom? (b) What is the maximum speed of such an atom?

Short Answer

Expert verified
(a) Period of oscillation is \(10^{-12}\) seconds. (b) Maximum speed is approximately \(6283.2\) cm/s.

Step by step solution

01

Understanding the Relationship Between Frequency and Period

The period of oscillation, denoted as \( T \), is the reciprocal of the frequency \( f \). Mathematically, this relationship is given by \( T = \frac{1}{f} \). The frequency of the atomic vibrations is given as \( 10^{12} \) Hz. We will use this to calculate \( T \).
02

Calculate the Period of Oscillation

Substitute the given frequency \( f = 10^{12} \) Hz into the formula \( T = \frac{1}{f} \). This gives \( T = \frac{1}{10^{12}} \) seconds, which simplifies to \( 10^{-12} \) seconds. Therefore, the period of oscillation of a typical atom is \( 10^{-12} \) seconds.
03

Understanding Maximum Speed During Oscillation

The maximum speed \( v_{ ext{max}} \) of an oscillating object is given by the formula \( v_{ ext{max}} = A imes 2\pi f \), where \( A \) is the amplitude of oscillation and \( f \) is the frequency. We will use the given amplitude \( A = 10^{-9} \) cm and frequency \( f = 10^{12} \) Hz to calculate the maximum speed.
04

Calculate the Maximum Speed of the Atom

Substitute \( A = 10^{-9} \) cm and \( f = 10^{12} \) Hz into the formula \( v_{ ext{max}} = A imes 2\pi f \). This results in \( v_{ ext{max}} = 10^{-9} \times 2\pi \times 10^{12} \). Simplifying this, \( v_{ ext{max}} = 2\pi \times 10^{3} \) cm/s. Evaluating \( 2\pi \) approximately as \( 6.2832 \), we find \( v_{ ext{max}} \approx 6.2832 \times 10^{3} \) cm/s, which is approximately \( 6283.2 \) cm/s.

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Key Concepts

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

Vibrational Motion
Atoms are not stationary; instead, they are constantly moving in a vibrational manner within a solid. This continuous **vibrational motion** is why atoms are never truly at rest, even when a substance seems not to be moving. When considering a solid, each atom vibrates around its equilibrium position.

This motion is essential because it contributes to various properties of materials, such as thermal conductivity and elasticity. **Key characteristics** of vibrational motion include:
  • **Frequency**, or how often the vibration occurs per unit time.
  • **Amplitude**, which refers to the maximum extent of the vibration from its resting position.
  • **Phase**, representing the position of the vibrating particle at a given time.
Understanding atomic vibrations is crucial in the study of solid-state physics as it helps explain thermal behavior and material properties.
Thermal Energy
Thermal energy is the energy that arises from the motion of atoms or molecules in a substance. At the atomic level, this energy causes atoms to vibrate. Even in seemingly cold objects, atoms still possess some degree of vibrational motion due to the presence of thermal energy.

**Thermal energy** has several key characteristics:
  • It's a form of kinetic energy, relating to the motion of particles.
  • Increases with temperature 鈥� as temperature rises, atoms and molecules vibrate faster.
In the context of solids, this energy is responsible for the continuous oscillation of atoms. Hence, even at room temperature, atoms have vibrational motion stemming from their inherent thermal energy. This energy transfer plays an important role in processes like conduction and changes in state.
Frequency of Oscillation
The **frequency of oscillation** refers to how often an atom completes a cycle of its vibrational motion per second. It is measured in hertz (Hz). For example, in the exercise, the frequency of atomic oscillations is given as \(10^{12}\) Hz. This means each atom vibrates one trillion times per second.

There are a few fundamental aspects of frequency:
  • **It's the inverse of the period**: Mathematically, frequency \( f \) is the reciprocal of the period \( T \) (\( f = \frac{1}{T} \)).
  • Higher frequencies mean more rapid oscillations, while lower frequencies mean slower oscillations.
Understanding the frequency of oscillation is vital for calculating the period of vibrations and determining the speed of vibrating atoms in solids.
Amplitude of Oscillation
The **amplitude of oscillation** is the maximum distance an atom moves from its equilibrium position during vibration. In the context of atomic vibrations, this measure provides insight into the extent of atomic displacement.

Several important points about amplitude include:
  • Amplitude typically decreases with stronger atomic bonds, meaning atoms return to equilibrium more strongly.
  • It is affected by temperature鈥攈igher thermal energy can increase amplitude.
  • Given in the exercise as \(10^{-9}\) cm, indicating atomic vibrations are extremely small.
Amplitude plays a key role in determining the maximum speed of atoms, as it's directly related to the energy involved in oscillation. Understanding both amplitude and frequency is essential for exploring deeper vibrational behaviors in materials.

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Most popular questions from this chapter

\(\bullet\bullet\) (a) If a pendulum clock were taken to the Moon, where the acceleration due to gravity is only one-sixth (assume the figure to be exact) that on the Earth, will the period of vibration (1) increase, (2) remain the same, or (3) decrease? Why? (b) If the period on the Earth is \(2.0 \mathrm{~s}\), what is the period on the Moon?

\(\bullet\bullet\) Show that the total energy of a mass-spring system in simple harmonic motion is given by \(\frac{1}{2} m \omega^{2} A^{2}\).

A student uses a 2.00 -m-long steel string with a diameter of \(0.90 \mathrm{~mm}\) for a standing wave experiment. The tension on the string is tweaked so that the second harmonic of this string vibrates at \(25.0 \mathrm{~Hz}\). (a) Calculate the tension the string is under. (b) Calculate the first harmonic frequency for this string. (c) If you wanted to increase the first harmonic frequency by \(50 \%,\) what would be the tension in the string? [Hint: See Table 9.2\(]\)

\(\bullet\) The displacement of an object is given by \(y=(5.0 \mathrm{~cm}) \cos [(20 \pi \mathrm{rad} / \mathrm{s}) t] .\) What are the object's (a) amplitude, (b) frequency, and (c) period of oscillation?

\(\bullet\bullet\) The AM frequencies on a radio dial range from \(550 \mathrm{kHz}\) to \(1600 \mathrm{kHz},\) and the FM frequencies range from \(88.0 \mathrm{MHz}\) to \(108 \mathrm{MHz}\). All of these radio waves travel at a speed of \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\) (speed of light). (a) Compared with the FM frequencies, the AM frequencies have (1) longer, (2) the same, or (3) shorter wavelengths. Why? (b) What are the wavelength ranges of the AM band and the FM band?
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