/*! 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 34 \(\mathrm{A}\) mass is oscillati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\mathrm{A}\) mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\) ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

Short Answer

Expert verified
The mass is \(A / \sqrt{2}\) units from the equilibrium when the elastic potential energy equals the kinetic energy.

Step by step solution

01

Setup the Energy Equality

We know that at some point in the oscillation, the elastic potential energy equals the kinetic energy. This gives the equality: \(\frac{1}{2}kx^2 = \frac{1}{2}mv^2\).
02

Express the Velocity

We can also express the velocity at a given point in the oscillation using the conservation of energy which gives: \(v = \sqrt{{k/m} (A^2 - x^2)}\). This represents the velocity at any position \(x\) during the oscillation.
03

Substitute the Expression of Velocity in the Energy Equality

Substitute the value of \(v\) from step 2 into the energy equality from step 1. This gives us: \(\frac{1}{2}kx^2 = \frac{1}{2}m \cdot k/m (A^2 - x^2)\). Simplify this equation to: \(x^2 = A^2 - x^2\).
04

Solving for x

Rearranging the equation from step 3 leads to: \(2x^2 = A^2\). Then, \(x = A / \sqrt{2}\).

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 91Ó°ÊÓ!

Key Concepts

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

Elastic Potential Energy
Imagine you're playing with a slinky, stretching it out and then letting it go. When you stretch it, you're storing energy, much like pulling back a slingshot or compressing a spring. This is what we call elastic potential energy. It's the energy stored in elastic materials as the result of their stretching or compressing. Elastic potential energy is a form of potential energy and can be calculated using the formula \( \frac{1}{2}kx^2 \), where \( k \) is the spring constant (measuring the stiffness of the spring) and \( x \) is the displacement from the equilibrium position.

When a mass attached to a spring is set into motion, at certain points during its oscillation, the elastic potential energy is converted into kinetic energy, and vice versa. This transformation allows the system to oscillate, and understanding the relationship between these types of energy is crucial in oscillatory motion physics.

Kinetic Energy
As the mass attached to the spring moves, not only is elastic potential energy at play, but also kinetic energy. Kinetic energy is the energy of motion, and it depends on two variables: the mass of the object (\( m \)) and its velocity (\( v \)). We calculate kinetic energy using the formula \( \frac{1}{2}mv^2 \).

Kinetic energy is highest when the oscillating mass moves fastest, which is at the equilibrium position of the spring where it has no elastic potential energy – all the stored energy has been converted into motion. Understanding kinetic energy is essential when analyzing movements, from the simplest systems like playground swings to complex mechanical devices. It ties into Newton's laws of motion and provides a foundation for understanding dynamic systems in physics.

Conservation of Energy
A fundamental principle in physics is the conservation of energy, which states that energy cannot be created or destroyed, only transformed from one form to another or transferred from one object to another. In the context of oscillatory motion, like that of a mass on a spring, energy constantly switches between elastic potential energy and kinetic energy, but the total energy of the system remains constant, assuming no external forces, like friction, are present.

In our exercise, the moment when the elastic potential energy equals kinetic energy signifies a special position in the motion of the mass. It is a point where exactly half of the total energy is stored as elastic potential energy, and the other half is present as kinetic energy. This concept allows us to determine specific positions of the oscillating body and analyze the energy transformations occurring within the system.

Simple Harmonic Motion
The back and forth movement of a mass on a spring is an example of simple harmonic motion (SHM), a type of periodic motion where the restoring force is directly proportional to the displacement. One hallmark of SHM is that it is sinusoidal in time and demonstrates a single resonant frequency. In SHM, the motion repeats after a characteristic period, and the system's energy oscillates between kinetic and potential forms in a predictable manner.

Simple harmonic motion is not only found in springs but also in pendulums, vibrating strings, and even molecules. It's foundational to understanding the behavior of wave phenomena, such as sound and light waves. In our exercise, by equating the elastic potential energy to the kinetic energy, we delve into a crucial instant of this harmonic motion, revealing a deep connection between motion, forces, and energy.

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

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced \(0.120 \mathrm{~m}\) from its equilibrium position and released with zero initial speed, then after \(0.800 \mathrm{~s}\) its displacement is found to be \(0.120 \mathrm{~m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

A thin metal disk with mass \(2.00 \times 10^{-3} \mathrm{~kg}\) and radius \(2.20 \mathrm{~cm}\) is attached at its center to a long fiber (Fig. \(\mathbf{E 1 4 . 4 0}\) ). The disk, when twisted and released, oscillates with a period of \(1.00 \mathrm{~s}\). Find the torsion constant of the fiber.

Object \(A\) has mass \(m_{A}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{A} .\) Object \(B\) has mass \(m_{B}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{B}\). The amplitude \(A_{A}\) for object \(A\) is twice the amplitude \(A_{B}\) for the motion of object \(B\). Also, \(m_{B}=4 m_{A}\) and \(k_{A}=9 k_{B}\). (a) What is the ratio of the maximum speeds of the two objects, \(v_{\max , A} / v_{\max , B} ?\) (b) What is the ratio of their maximum accelerations, \(a_{\max , A} / a_{\max , B} ?\)

BIO Weighing Astronauts. This procedure has been used to "weigh" astronauts in space: A \(42.5 \mathrm{~kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes \(1.30 \mathrm{~s}\) to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes \(2.54 \mathrm{~s}\) for one cycle. What is the mass of the astronaut?

In a physics lab, you attach a \(0.200 \mathrm{~kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is \(2.60 \mathrm{~s}\). Find the spring's force constant.

See all solutions

Recommended explanations on Physics Textbooks

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.