/*! 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 19 Objects of equal mass are oscill... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is \(174 \mathrm{~N} / \mathrm{m}\). The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring \(2 .\) The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2

Short Answer

Expert verified
The spring constant of spring 2 is 696 N/m.

Step by step solution

01

Recall the Formula for Maximum Velocity in Simple Harmonic Motion

The magnitude of maximum velocity for simple harmonic motion (SHM) is given by \( v_{max} = A \omega \), where \( A \) is the amplitude and \( \omega \) is the angular frequency. In SHM, angular frequency \( \omega \) is given by \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass.
02

Relate the Maximum Velocities of Both Springs

According to the problem, the maximum velocities are the same for both springs. Denoting the amplitude and spring constant for spring 1 as \( A_1 \) and \( k_1 \), and for spring 2 as \( A_2 \) and \( k_2 \) respectively, we have the equation \( A_1 \sqrt{\frac{k_1}{m}} = A_2 \sqrt{\frac{k_2}{m}} \).
03

Incorporate Amplitude Relationship

We know that \( A_1 = 2A_2 \). Substituting this into the equation from Step 2 gives us \( 2A_2 \sqrt{\frac{k_1}{m}} = A_2 \sqrt{\frac{k_2}{m}} \).
04

Simplify the Equation

Cancel out the terms \( A_2 \) and \( \sqrt{m} \) from both sides of the equation because they are non-zero. This leaves us with \( 2\sqrt{k_1} = \sqrt{k_2} \).
05

Solve for the Spring Constant of Spring 2

Square both sides to eliminate the square roots, resulting in \( 4k_1 = k_2 \). Substituting the given value of \( k_1 = 174 \, \mathrm{N/m} \), we find \( k_2 = 4 \times 174 \, \mathrm{N/m} = 696 \, \mathrm{N/m} \).

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.

Spring Constant
The spring constant, often denoted by the symbol \( k \), is a crucial parameter in the study of springs and simple harmonic motion (SHM). It represents the stiffness of a spring.
The higher the spring constant, the stiffer the spring. It is measured in newtons per meter (N/m).
The formula for spring force is given by Hooke's Law: \( F = kx \), where \( F \) is the force applied to the spring and \( x \) is the displacement from its resting position.
In the context of SHM, the spring constant helps determine how quickly or slowly an object will oscillate when attached to the spring. Higher spring constants result in faster oscillations, all else being equal.
For example, in our problem, Spring 1 has a spring constant \( k_1 = 174 \, \mathrm{N/m} \).
To find the spring constant of Spring 2, the relationship \( 4k_1 = k_2 \) is used, ultimately resulting in \( k_2 = 696 \, \mathrm{N/m} \). This means Spring 2 is stiffer compared to Spring 1.
Amplitude
Amplitude refers to the maximum extent of the oscillation from the equilibrium position.
In simple harmonic motion, amplitude is represented by the symbol \( A \).
It tells us how far the oscillating object moves on either side of its equilibrium position.In the given problem, it's noted that Spring 1 has twice the amplitude of Spring 2, i.e., \( A_1 = 2A_2 \).
So, if Spring 2 has an amplitude of, say, \( A_2 = x \), Spring 1 will have an amplitude of \( 2x \).
A larger amplitude means the object moves further from the equilibrium position and typically involves more energy in the system. However, amplitude does not affect the period or frequency of the oscillation in an ideal SHM system, where the only opposing force is the restoring force of the spring.
Oscillation
Oscillation is a repeated back and forth motion of an object around an equilibrium position.
It is the essence of simple harmonic motion. In an ideal oscillation with no external forces, the motion is sinusoidal. An important characteristic of oscillations in SHM is that they are uniform and symmetric about the equilibrium position.
Each cycle of motion takes the same time regardless of the amplitude. This property is derived from the Hooke’s Law and Newton’s second law. In the example of the springs, each object on its spring oscillates between two extreme points - the highest and the lowest point.
Different spring constants lead to different oscillation frequencies and behaviors, but the nature of oscillation remains consistent, being smooth and periodic.
Angular Frequency
Angular frequency, denoted by \( \omega \), describes how fast an object moves through an angle in SHM.
In terms of linear oscillation, it relates to how quickly the object oscillates back and forth.The angular frequency is linked to the spring constant and mass by the formula \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass.
It is measured in radians per second, reflecting how swiftly the phase of oscillation progresses.Angular frequency is a fundamental characteristic of SHM influencing the period \( T \) and frequency \( f \) of the oscillation:
  • Period \( T \) is the time it takes to complete one full cycle of oscillation, given by \( T = \frac{2\pi}{\omega} \).
  • Frequency \( f \) is the number of oscillations per second, given by \( f = \frac{\omega}{2\pi} \).
Changes in angular frequency directly affect how quickly the system returns to its equilibrium position, with higher frequencies resulting in faster oscillations.

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

The maximum applied force of compression that one man's thighbone can endure without breaking is \(7.0 \times 10^{4} \mathrm{~N}\). The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{~m}^{2}\), and its length is \(0.55 \mathrm{~m}\). The mass of the man is \(65 \mathrm{~kg}\). He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his left of gravity can fall without his breaking a bone.

An 86.0 -kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{~N} / \mathrm{m} .\) He accidentally slips and falls freely for \(0.750 \mathrm{~m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

Interactive LearningWare \(10.2\) at explores the approach taken in problems such as this one. A spring is mounted vertically on the floor. The mass of the spring is negligible. A certain object is placed on the spring to compress it. When the object is pushed further down by just a bit and then released, one up/down oscillation cycle occurs in \(0.250 \mathrm{~s}\). However, when the object is pushed down by \(5.00 \times 10^{-2} \mathrm{~m}\) to point \(P\) and then released, the object flies entirely off the spring. To what height above point \(P\) does the object rise in the absence of air resistance?

A rifle fires a \(2.10 \times 10^{-2}\) kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{~m}\) from its unstrained length. The pellet rises to a maximum height of \(6.10 \mathrm{~m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Concept Simulation 10.2 at allows you to explore the effect of the acceleration due to gravity on pendulum motion, which is the focus of this problem. Astronauts on a distant planet set up a simple pendulum of length \(1.2 \mathrm{~m}\). The pendulum executes simple harmonic motion and makes 100 complete vibrations in \(280 \mathrm{~s}\). What is the acceleration due to gravity?

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.