/*! 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 60 A 50.0 -g hard-boiled egg moves ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 50.0 -g hard-boiled egg moves on the end of a spring with force constant \(k=25.0 \mathrm{N} / \mathrm{m} .\) Its initial displacement is 0.300 \(\mathrm{m} .\) A damping force \(F_{x}=-b v_{x}\) acts on the egg, and the amplitude of the motion decreases to 0.100 \(\mathrm{m}\) in 5.00 \(\mathrm{s}\) . Calculate the magnitude of the damping constant \(b\) .

Short Answer

Expert verified
The damping constant \(b\) is approximately \(4.39 \, \mathrm{N\, s/m}\).

Step by step solution

01

Understand the Problem

We need to find the damping constant \(b\) from the given information about the motion of a damped harmonic oscillator: an egg with some mass and spring attached. We are given the initial and final amplitude over a certain duration.
02

Write the Equation for Damped Harmonic Motion

The amplitude of a damped harmonic oscillator decreases over time according to the equation \( A(t) = A_0 e^{-bt/2m} \), where \(A_0\) is the initial amplitude, \(A(t)\) is the amplitude at time t, \(b\) is the damping constant, and \(m\) is the mass.
03

Substitute the Known Values

We are given \(A_0 = 0.300 \, \mathrm{m}\), \(A(5) = 0.100 \, \mathrm{m}\), \(t = 5.00 \, \mathrm{s}\), and \(m = 50.0 \, \mathrm{g} = 0.050 \, \mathrm{kg}\). Substitute these values into the amplitude equation to get: \[ 0.100 = 0.300 e^{-b(5.00)/2(0.050)}. \]
04

Solve for the Damping Constant b

Divide both sides by the initial amplitude to isolate the exponential term: \[ \frac{0.100}{0.300} = e^{-b(5.00)/2(0.050)}. \] This simplifies to \( e^{-b(5.00)/0.100} = \frac{1}{3} \). Next, take the natural logarithm of both sides to solve for \(b\): \[ b = -\ln\left(\frac{1}{3}\right) \cdot \frac{0.100}{5.00}. \] Evaluate this expression to find \(b\).
05

Calculate b

Calculate the values obtained in Step 4: \[ b = - \ln\left(\frac{1}{3}\right) \cdot 0.020. \] After computation, \(b\) approximately equals \(4.39 \, \mathrm{N\, s/m}\).

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.

Damping Constant
The damping constant, often denoted as \( b \), is a vital component in understanding damped harmonic motion. It represents the magnitude of the damping force that counteracts the movement of an oscillating system. In our scenario, the damping force is an opposing force proportional to the velocity of the egg attached to a spring. This force is mathematically expressed as \( F_x = -b v_x \), where \( v_x \) is the velocity.
If the damping constant is large, the oscillations will die out quickly, making systems seem more "resistant" to motion. Conversely, if the damping constant is small, the oscillations will persist longer. The damping constant is usually measured in \( ext{N} \, ext{s/m} \), highlighting its role in controlling how quickly the energy in the system dissipates. In our exercise, we've calculated \( b \) to be approximately 4.39 \( ext{N} \, ext{s/m} \), meaning the damping effect on the egg's motion is moderate.
Spring Force Constant
The spring force constant, represented as \( k \), is another key element in oscillatory systems. It defines the stiffness of the spring, indicating how much force is required to stretch or compress it by a unit length. The spring force itself can be described by Hooke's Law: \( F_s = -kx \), where \( x \) is the displacement.
A higher spring force constant signifies a stiffer spring that requires more force to stretch or compress, which results in a higher potential to return to its equilibrium position. Conversely, a lower \( k \) means the spring is more flexible. In our example, the spring force constant is given as 25.0 \( ext{N/m} \), indicating a relatively moderate stiffness that allows the egg to oscillate at a specific initial displacement.
Amplitude Decrease
Amplitude decrease in a damped harmonic oscillator refers to the reduction in peak displacement over time. Initially, the system may start with a certain amplitude, but due to resistance like air friction or internal spring damping, this amplitude fades.
The equation representing this is \( A(t) = A_0 e^{-bt/2m} \), where \( A_0 \) is the initial amplitude and \( A(t) \) is the amplitude at time \( t \). The damping constant \( b \) and mass \( m \) influence how fast this decrease happens.In the exercise, the egg's amplitude changes from 0.300 m to 0.100 m over 5 seconds, showing how damping affects the system in real-life scenarios. Understanding amplitude decrease is crucial for predicting system behavior under various damping conditions.
Exponential Decay
Exponential decay describes how a quantity diminishes at a rate proportional to its current value. In damped harmonic oscillators, this process captures how amplitude decreases over time due to damping forces.
The decay is expressed mathematically by an exponential function like \( A(t) = A_0 e^{-bt/2m} \), signifying a smooth, continuous reduction. The negative exponent emphasizes how the value reduces over intervals. In our problem, exponential decay is evident as the amplitude of the oscillating egg decreases according to the calculated damping constant. This decay helps us visualize how energy is sapped from the system, transitioning potential into kinetic energy and eventually into thermal energy, until the system halts. Understanding this concept is pivotal for applications such as signal processing and engineering, where precise predictions of motion over time are required.

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 a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 Hz. What will the frequency be if 0.220 kg are (a) added to the original mass and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

Inside a NASA test vehicle, a \(3.50-\) kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m}\) . The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached \(45.0 \mathrm{m} / \mathrm{s},\) its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

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. E14.42). The disk, when twisted and released, oscillates with a period of 1.00 s. Find the torsion constant of the fiber.

An object is undergoing SHM with period 0.300 s and amplitude 6.00 \(\mathrm{cm} .\) At \(t=0\) the object is instantaneously at rest at \(x=6.00 \mathrm{cm} .\) Calculate the time it takes the object to go from \(x=6.00 \mathrm{cm}\) to \(x=-1.50 \mathrm{cm} .\)

A 40.0 -N force stretches a vertical spring 0.250 \(\mathrm{m}\) . (a) What mass must be suspended from the spring so that the system will oscillare with a period of 1.00 \(\mathrm{s} ?\) (b) If the amplitude of the motion is 0.050 \(\mathrm{m}\) and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

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.