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Chapter 13: Problem 2

The vibration frequencies of molecules are much higher than those of macroscopic mechanical systems. Why?

Short Answer

Expert verified
Molecules vibrate at higher frequencies than macroscopic systems due to their lower mass and stronger force constants associated with atomic bonds.

Step by step solution

01

Understanding Frequency

Frequency refers to the number of vibrations, or cycles of motion, completed in a given time frame. It is measured in Hertz (Hz), which is equivalent to 1 cycle per second.
02

Comparing Sizes

Molecules are much smaller than macroscopic mechanical systems. Their reduced size corresponds to lower mass, and according to Hooke's Law, systems with less mass can vibrate at higher frequencies.
03

Considering Bond Strength

In addition to having lower mass, molecules have specific types of bonds holding their atoms together (ionic, covalent, metallic, etc.). These bonds are stronger when compared with the forces holding together macroscopic systems. Hence, the force constant is much larger for molecular systems, leading to higher frequencies as per Hooke's Law (f = 1/2蟺 鈭�(k/m), where 'k' represents the force constant and 'm' is the mass).

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Most popular questions from this chapter

An automobile suspension has an effective spring constant of \(26 \mathrm{kN} / \mathrm{m},\) and the car's suspended mass is \(1900 \mathrm{kg} .\) In the absence of damping, with what frequency and period will the car undergo simple harmonic motion?

A pendulum consists of a 320 -g solid ball \(15.0 \mathrm{cm}\) in diameter, suspended by an essentially mass-less string \(80.0 \mathrm{cm}\) long. Calculate the period of this pendulum, treating it first as a simple pendulum and then as a physical pendulum. What's the error in the simple-pendulum approximation? (Hint: Remember the parallel-axis theorem.)

How can a system have more than one resonant frequency?

The total energy of a mass-spring system is the sum of its kinetic and potential energy: \(E=\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2} .\) Assuming \(E\) remains constant, differentiate both sides of this expression with respect to time and show that Equation 13.3 results. (Hint: Remember that \(v=d x / d t .)\)

Derive the period of a simple pendulum by considering the horizontal displacement \(x\) and the force acting on the bob, rather than the angular displacement and torque.
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