Log out Go to App
Go to App

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

14.10 A pendulum is suspended from the ceiling of an elevator. When the elevator is at rest, the period of the pendulum is \(T\). The elevator accelerates upward, and the period of the pendulum is then a) still T. b) less than \(T_{-}\) c) greater than \(T\).

In a lab, a student measures the unstretched length of a spring as \(11.2 \mathrm{~cm}\). When a 100.0 - g mass is hung from the spring, its length is \(20.7 \mathrm{~cm}\). The mass-spring system is set into oscillatory motion, and the student obscrves that the amplitude of the oscillation decreases by about a factor of 2 after five complete cycles. a) Calculate the period of oscillation for this system, assuming no damping. b) If the student can measure the period to the nearest \(0.05 \mathrm{~s}\). will she be able to detect the difference between the period with no damping and the period with damping?

The period of oscillation of an object in a frictionless tunnel running through the center of the Moon is \(T=2 \pi / \omega_{0}\) \(=6485 \mathrm{~s}\), as shown in Fxample 142 . What is the period of oscillation of an object in a similar tunnel through the Earth \(\left(R_{\mathrm{I}}=6.37 \cdot 10^{6} \mathrm{~m} ; R_{\mathrm{M}}=1.74 \cdot 10^{6} \mathrm{~m} ; M_{\mathrm{E}}=5.98 \cdot 10^{24} \mathrm{~kg}\right.\) \(\left.M_{u}=7.36 \cdot 10^{22} \mathbf{k g}\right) ?\)

A \(3.00-\mathrm{kg}\) mass attached to a spring with \(k=140 . \mathrm{N} / \mathrm{m}\) is oscillating in a vat of oil, which damps the oscillations. a) If the damping constant of the oil is \(b=10.0 \mathrm{~kg} / \mathrm{s}\), how long will it take the amplitude of the oscillations to decrease to \(1.00 \%\) of its original value? b) What should the damping constant be to reduce the amplitude of the oscillations by \(99.0 \%\) in 1.00 s?

A 2.0 -kg mass attached to a spring is displaced \(8.0 \mathrm{~cm}\) from the equilibrium position. It is released and then oscillates with a frequency of \(4.0 \mathrm{~Hz}\) a) What is the energy of the motion when the mass passes through the equilibrium position? b) What is the speed of the mass when it is \(20 \mathrm{~cm}\) from the equilibrium position?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Linear Momentum

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Electromagnetism

Read Explanation

Famous Physicists

Read Explanation

Mechanics and Materials

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.

Sign-up for free