/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Explain why the frequency of a d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 8

Explain why the frequency of a damped system is lower than that of the equivalent undamped system.

Short Answer

Expert verified
The frequency of a damped system is lower than that of the equivalent undamped system because damping leads to energy dissipation in the system. As frequency is directly related to the energy of the system, a loss in system energy leads to a decrease in frequency.

Step by step solution

01

Understand System Damping

First, we need to understand what damping means in the context of a system. Damping is a phenomenon that causes a decrease in amplitude of the oscillations of an oscillatory system. In a physical system, damping is caused by friction that dissipates energy. In our case, we differentiate between damped and undamped systems.
02

Effect on Frequency

In an undamped system, the energy remains constant and no energy is lost to friction or other non-conservative forces, hence, the frequency of oscillation will remain the same. This is because frequency is directly linked to the energy of the system. In a damped system, the energy of the system decreases due to the work done against the damping forces, and this results in decreased frequency. The system loses energy faster than it acquires it.
03

Conclusion

So, the key point here is the energy dissipation in the damped system. This results in a lower frequency in the damped system as compared to the undamped system, because frequency is directly related to the energy of the system. As more energy is lost to damping in a damped system, the frequency decreases relative to an undamped system where no energy is lost.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(450-\mathrm{g}\) mass on a spring is oscillating at \(1.2 \mathrm{~Hz}\), with total energy \(0.51 \mathrm{~J}\). What's the oscillation amplitude?

The human eye and the muscles that hold it can be modeled as a mass-spring system with typical values \(m=7.5 \mathrm{~g}\) and \(k=2.5 \mathrm{kN} / \mathrm{m}\). What's the resonant frequency of this system? Shaking your head at this frequency blurs vision, as the eyeball undergoes resonant oscillations.

A \(280-\mathrm{g}\) volleyball with a \(21.0-\mathrm{cm}\) diameter is suspended by a wire and is undergoing torsional oscillations at \(1.57 \mathrm{~Hz}\). Find the torsional constant of the wire.

This problem explores what would happen if a hole were drilled through Earth's center and out the other side, and an object were dropped into the hole. Approximating Earth as a uniform solid sphere, the gravitational acceleration within the planet (including inside the hypothetical hole) would be \(g(r)=g_{0}\left(r / R_{\mathrm{E}}\right)\), where \(g_{0}\) is the value at Earth's surface, \(r\) is the distance from Earth's center, and \(R_{\mathrm{E}}\) is Earth's radius. This gravitational acceleration is directed toward Earth's center. (a) Write an expression for the force on a mass \(m\) at any point \(r\) in the hole, apply Newton's second law, and show that you get an equation analogous to Equation 13.3. Neglect air resistance. (b) Use your analogy to find an expression for the period of the simple harmonic motion that results when the mass is dropped into the hole. (c) Use appropriate values to find a numerical value for the period, and compare with the period for circular low- Earth orbit that we found in Chapter 8 .

A particle undergoes simple harmonic motion with amplitude \(27 \mathrm{~cm}\) and maximum speed \(4.6 \mathrm{~m} / \mathrm{s}\). Find the (a) angular frequency, (b) period, and (c) maximum acceleration.
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Circular Motion and Gravitation

Read Explanation

Nuclear Physics

Read Explanation

Classical Mechanics

Read Explanation

Dynamics

Read Explanation

Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.