/*! 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 27 A 1.15-kg mass oscillates accord... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 27

A 1.15-kg mass oscillates according to the equation \(x=0.650 \cos 7.40 t\) where \(x\) is in meters and \(t\) in seconds. Determine \((a)\) the amplitude, \((b)\) the frequency, \((c)\) the total energy, and \((d)\) the kinetic energy and potential energy when \(x=0.260 \mathrm{~m}\).

Short Answer

Expert verified
Amplitude: 0.650 m, Frequency: 1.18 Hz, Total energy: 21.3 J, KE: 19.17 J, PE: 2.13 J.

Step by step solution

01

Identify the Amplitude

The amplitude of the oscillation can be directly obtained from the equation \( x = 0.650 \cos 7.40t \). This equation is of the form \( x = A\cos(\omega t) \), where \( A \) is the amplitude. Therefore, the amplitude is \( A = 0.650 \) meters.
02

Calculate the Frequency

The angular frequency \( \omega \) is given in the equation as \( 7.40 \) rad/s. To find the frequency \( f \), use the relation \( \omega = 2\pi f \). Solving for \( f \), we have \( f = \frac{\omega}{2\pi} = \frac{7.40}{2\pi} \approx 1.18 \) Hz.
03

Calculate the Total Energy

The total energy \( E \) in a simple harmonic oscillator is given by \( E = \frac{1}{2}mA^2\omega^2 \). Plugging in the values, we have:\[ E = \frac{1}{2} \times 1.15 \times (0.650)^2 \times (7.40)^2 \approx 21.3 \text{ J} \]
04

Calculate the Potential Energy at \( x=0.260 \) m

The potential energy \( U \) when the object is at a displacement \( x \) is given by \( U = \frac{1}{2}k x^2 \), where \( k = m\omega^2 \). Find \( k \):\[ k = 1.15 \times (7.40)^2 \approx 62.9 \text{ N/m} \]Then, calculate \( U \):\[ U = \frac{1}{2} \times 62.9 \times (0.260)^2 \approx 2.13 \text{ J} \]
05

Calculate the Kinetic Energy at \( x=0.260 \) m

The kinetic energy \( K \) is given by the difference between the total energy and the potential energy: \( K = E - U \). Using previously calculated values:\[ K = 21.3 - 2.13 \approx 19.17 \text{ J} \]

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!

Key Concepts

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

Amplitude
The amplitude of an oscillation represents the maximum distance the oscillating object can move from its equilibrium position. It is a measure of the maximum extent of the oscillation. This is directly represented by the constant in front of the cosine function in the equation of motion. In our exercise, the equation given is \( x = 0.650 \cos 7.40t \), where the amplitude \( A \) is \( 0.650 \) meters.
Amplitude is crucial because it tells us how far the object will travel on either side of its resting position.
Think of amplitude as the height of a wave from the center line to its peak or trough.
Frequency
Frequency refers to the number of oscillations an object makes in one second. It is measured in Hertz (Hz). In this exercise, the angular frequency \( \omega \) is provided as \( 7.40 \) rad/s. The relationship between angular frequency and regular frequency is given by \( \omega = 2\pi f \).
  • Solving for \( f \): \( f = \frac{\omega}{2\pi} = \frac{7.40}{2\pi} \approx 1.18 \) Hz.
Frequency is immensely important because it lets us know how fast the oscillations are happening.
Higher frequencies mean more rapid oscillations.
Total Energy
Total energy in a simple harmonic oscillator refers to the sum of its kinetic and potential energies. This energy remains constant throughout the motion, assuming there's no energy loss due to friction or air resistance.
The total energy \( E \) for our system is given by the formula:
  • \( E = \frac{1}{2}mA^2\omega^2 \).
Substituting the known values:
- Mass \( m = 1.15 \) kg
- Amplitude \( A = 0.650 \) m
- Angular frequency \( \omega = 7.40 \) rad/s,
we find \( E \approx 21.3 \) J.
Understanding the total energy helps in knowing the capacity of the system's motion and energy transformations it undergoes.
Kinetic Energy
Kinetic energy in oscillations represents the energy associated with the motion of the mass as it moves back and forth. The kinetic energy is maximum when the mass passes through the equilibrium position and is zero at the points of maximum displacement.
This is calculated by subtracting potential energy from total energy:
  • \( K = E - U \).
In the problem, \( U = 2.13 \) J (at \( x = 0.260 \) m), while \( E = 21.3 \) J,
so the kinetic energy \( K \approx 19.17 \) J.
Recognizing kinetic energy in oscillations helps us track the speed and momentum of the oscillating object.
Potential Energy
Potential energy in an oscillating system is the energy stored due to its position. It is maximum at the points of maximum displacement and zero at the equilibrium point.
To find potential energy when our mass is displaced by \( 0.260 \) meters, we use:
  • \( U = \frac{1}{2} k x^2 \).
Here, \( k = m \omega^2 = 62.9 \text{ N/m} \)
So, potential energy \( U \approx 2.13 \) J at \( x = 0.260 \) m.
Knowing potential energy allows us to understand how much stored energy can be converted back into motion.

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

(II) At \(t=0,\) a \(785-\mathrm{g}\) mass at rest on the end of a horizontal spring \((k=184 \mathrm{N} / \mathrm{m})\) is struck by a hammer which gives it an initial speed of 2.26 \(\mathrm{m} / \mathrm{s} .\) Determine \((a)\) the period and frequency of the motion, \((b)\) the amplitude, (c) the maximum acceleration, (d) the position as a function of time, \((e)\) the total energy, and \((f)\) the kinetic energy when \(x=0.40 A\) where \(A\) is the amplitude.

A \(950-\mathrm{kg}\) car strikes a huge spring at a speed of \(25 \mathrm{~m} / \mathrm{s}\) (Fig. \(14-41\) ), compressing the spring \(5.0 \mathrm{~m}\). ( \(a\) ) What is the spring stiffness constant of the spring? (b) How long is the car in contact with the spring before it bounces off in the opposite direction?

(II) A tuning fork oscillates at a frequency of 441 \(\mathrm{Hz}\) and the tip of each prong moves 1.5 \(\mathrm{mm}\) to either side of center. Calculate \((a)\) the maximum speed and \((b)\) the maximum acceleration of the tip of a prong.

(II) An 1150 kg automobile has springs with \(k=16,000 \mathrm{N} / \mathrm{m}\) . One of the tires is not properly balanced; it has a little extra mass on one side compared to the other, causing the car to shake at certain speeds. If the tire radius is \(42 \mathrm{cm},\) at what speed will the wheel shake most?

A 950 -kg car strikes a huge spring at a speed of 25 \(\mathrm{m} / \mathrm{s}\) (Fig. 41\()\) , compressing the spring 5.0 \(\mathrm{m}\) . (a) What is the spring stiffness constant of the spring? (b) How long is the car in contact with the spring before it bounces off in the opposite direction?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Wave Optics

Read Explanation

Waves Physics

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.