/*! 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 56 When the displacement of a mass ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 56

When the displacement of a mass on a spring is half of the amplitude of its oscillation, what fraction of the mass's energy is kinetic energy?

Short Answer

Expert verified
Answer: When the displacement is half the amplitude, the fraction of the mass's energy that is kinetic energy is 3/4 or 75%.

Step by step solution

01

Understand the relation between displacement and potential energy

The potential energy of a mass on a spring is given by the formula: \[PE = \frac{1}{2}kx^2,\] where \(PE\) is the potential energy, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position.
02

Find the potential energy at half amplitude

The problem states that the displacement is half the amplitude. Let \(A\) represent the amplitude, so the displacement is \(\frac{1}{2}A\). We can plug this into the potential energy formula to get: \[PE = \frac{1}{2}k\left(\frac{1}{2}A\right)^2 = \frac{1}{8}kA^2.\]
03

Understand the relation between total energy and amplitude

Total mechanical energy for a mass on a spring in simple harmonic motion is the sum of potential and kinetic energies. At the maximum displacement (amplitude), all energy is potential energy. Therefore, the total energy can be expressed as: \[E_{total} = \frac{1}{2}kA^2.\]
04

Find the kinetic energy at half-amplitude

At the given displacement, we know the total energy and potential energy. The kinetic energy, \(KE\), can be found by subtracting the potential energy from the total energy: \[KE = E_{total} - PE = \frac{1}{2}kA^2 - \frac{1}{8}kA^2 = \frac{3}{8}kA^2.\]
05

Calculate the fraction of kinetic energy in total energy

To find the fraction of kinetic energy in total energy, divide the kinetic energy by total energy: \[\frac{KE}{E_{total}} = \frac{\frac{3}{8}kA^2}{\frac{1}{2}kA^2} = \frac{3}{4}.\] Hence, when the displacement of the mass on a spring is half of the amplitude of its oscillation, the fraction of mass's energy that is kinetic energy is \(\frac{3}{4}\) or 75%.

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

When the amplitude of the oscillation of a mass on a stretched string is increased, why doesn't the period of oscil lation also increase?

Object \(A\) is four times heavier than object B. Fach object is attached to a spring, and the springs have equal spring constants. The two objects are then pulled from their equilibrium positions and released from rest. What is the ratio of the periods of the two oscillators if the amplitude of \(A\) is half that of \(B ?\) a) \(T_{A}: T_{\mathrm{g}}=1: 4\) c) \(T_{A}: T_{\mathrm{B}}=2\) : b) \(T_{A}: T_{B}=4: 1\) d) \(T_{A}: T_{B}=1: 2\)

A shock absorber that provides critical damping with \(\omega_{\gamma}=72.4 \mathrm{~Hz}\) is compressed by \(6.41 \mathrm{~cm} .\) How far from the equilibrium position is it after \(0.0247 \mathrm{~s} ?\)

An object in simple harmonic motion is isochronous, meaning that the period of its oscillations is independent of their amplitude. (Contrary to a common assertion, the operation of a pendulum clock is not based on this principle. A pendulum clock operates at fixed, finite amplitude. The gearing of the clock compensates for the anharmonicity of the pendulum.) Consider an oscillator of mass \(m\) in one-dimensional motion, with a restoring force \(F(x)=-c x^{3}\) where \(x\) is the displacement from equilibrium and \(c\) is a constant with appropriate units. The motion of this ascillator is periodic but not isochronous. a) Write an expression for the period of the undamped oscillations of this oscillator. If your expression involves an integral, it should be a definite integral. You do not need to evaluate the expression. b) Using the expression of part (a), determine the dependence of the period of oscillation on the amplitude. c) Generalize the results of parts (a) and (b) to an oscillator of mass \(m\) in one-dimensional motion with a restoring force corresponding to the potential energy \(U(x)=\gamma|x| / \alpha\), where \(\alpha\) is any positive value and \(\gamma\) is a constant

A mass, \(M=1.6 \mathrm{~kg}\), is attached to a wall by a spring with \(k=578 \mathrm{~N} / \mathrm{m}\). The mass slides on a frictionless floor. The spring and mass are immersed in a fluid with a damping constant of \(6.4 \mathrm{~kg} / \mathrm{s}\). A horizontal force, \(\mathrm{F}(\mathrm{t})=\mathrm{F}_{\mathrm{d}} \cos \left(\omega_{\mathrm{d}} \mathrm{t}\right)\) where \(F_{d}=52 \mathrm{~N},\) is applied to the mass through a knob, caus. ing the mass to oscillate back and forth. Neglect the mass of the spring and of the knob and rod. At approximately what frequency will the amplitude of the mass's oscillation be greatest, and what is the maximum amplitude? If the driving frequency is reduced slightly (but the driving amplitude remains the same), at what frecuency will the amplitude of the mass's ascillation be half of the maximum amplitude?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Wave Optics

Read Explanation

Energy Physics

Read Explanation

Electricity

Read Explanation

Force

Read Explanation

Translational Dynamics

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.