/*! 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 94 A \( 100-\mathrm{g}\) object is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \( 100-\mathrm{g}\) object is fixed to the end of a spring that has a spring constant of \(15 \mathrm{~N} / \mathrm{m}\). The object is displaced \(15 \mathrm{~cm}\) to the right and released from rest at \(t=0\) to slide on a horizontal, frictionless table. (a) Find the first three times when the object is at the equilibrium position. (b) Find the first three times when the object is \(10 \mathrm{~cm}\) to the left of equilibrium. (c) What is the first time that the object is \(5 \mathrm{~cm}\) to the right of equilibrium, moving toward the left?

Short Answer

Expert verified
The first three times when the object is at the equilibrium position are approximately 0.258 sec, 0.774 sec, 1.290 sec. The first three times when the object is \(10 cm\) to the left of equilibrium position are approximately 0.388 sec, 0.904 sec, 1.420 sec. The first time that the object is \(5 cm\) to the right of equilibrium, moving toward the left is approximately 0.454 sec.

Step by step solution

01

Identify the Known Variables

Here the mass of the object \(m\) is \(100 g = 0.1 kg\), the spring constant \(k\) is \(15 N/m\), initial displacement \(x_0\) is \(15 cm = 0.15 m\), hence the amplitude \(A\) of the motion is \(0.15 m\). The displacement for part (b) is \(10 cm = 0.1 m\) and for part (c) is \(5 cm = 0.05 m\).
02

Calculate the Period

We use the formula for the period \(T\) of a mass-spring system: \(T = 2\pi \sqrt{m/k}\), so \(T = 2\pi \sqrt{0.1/15} \approx 0.516 sec\). This is the time for one complete cycle of the motion.
03

Solve for Part (a)

The object completes half its cycle when it returns to the equilibrium position for the first time. Hence, the first time is \(T/2\), and the first three times are \(T/2, 3T/2, 5T/2\) which are approximately \(0.258 sec, 0.774 sec, 1.290 sec\).
04

Solve for Part (b)

We use the formula for displacement in SHM: \(x = A \cos(\omega t)\), where \(\omega = 2\pi/T\) is the angular frequency of the motion. Solving for \(t\) gives \(t = \cos^{-1}(x/A) / \omega\). Here \(x = -0.1 m\) because it is to the left of equilibrium position. So the first time is \(t = \cos^{-1}(-0.1/0.15) / (2\pi/0.516) \approx 0.388 sec\). This occurs once per cycle, so the first three times are \(0.388 sec, 0.388 sec + T, 0.388 sec + 2T\) which are approximately \(0.388 sec, 0.904 sec, 1.420 sec\).
05

Solve for Part (c)

We use the same formula for time as in step 4, but now \(x = 0.05 m\) because it is to the right of equilibrium position and moving toward the left so it is in the second half of the cycle. Because cosine is negative in the second half of a period, we need to subtract this time from a full period to get the actual time, hence \(t = T - \cos^{-1}(0.05/0.15) / \omega\). So the first time is approximately \(0.516 - \cos^{-1}(0.05/0.15) / (2\pi/0.516) = 0.454 sec\).

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.

Mass-Spring System
In a mass-spring system, we deal with a setup where a mass is attached to a spring that can be compressed or stretched. This system is fundamental to understanding simple harmonic motion (SHM) because it naturally exhibits oscillations. The spring exerts a force proportional to the displacement of the mass from its equilibrium position, a behavior described by Hooke's Law:
\[ F = -kx \] where \( F \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium. This force always acts to restore the mass to its equilibrium position.
In our example, the mass is \( 0.1 kg \) and the spring constant is \( 15 \text{ N/m} \). The system is set up so the object can slide frictionlessly, ensuring pure SHM. This arrangement allows us to study the periodic motion and how the mass moves back and forth in response to the spring's restoring force.
Equilibrium Position
The equilibrium position in a mass-spring system is the point where the spring exerts no force because it is neither compressed nor stretched. When the object is at equilibrium, the net force acting on it is zero, meaning there's no acceleration if friction or other forces are absent.
Understanding the equilibrium position is crucial because it acts as a reference point for oscillations. The object passes through this position twice during a complete cycle: once while moving in each direction. In this scenario, the equilibrium position helps us calculate when the object returns to center after being displaced. The first time the object reaches equilibrium after release is at half its period \( T/2 \), and it revisits there consistently every half period after that, predicting points of zero displacement within the oscillation cycle.
Angular Frequency
Angular frequency, denoted by \( \omega \), is a measure of how quickly an object rotates or oscillates in a circle. For SHM, this translates to how fast the oscillations occur. It is related to the mass and spring constant like this:
\[ \omega = \sqrt{\frac{k}{m}} \]where \( k \) is the spring constant and \( m \) is the mass. The angular frequency has units of radians per second. It allows us to calculate the period \( T \) and subsequently predict the timing of events like when the object will return to the equilibrium position or reach particular displacements.
In our setup, the period \( T \) is given by: \[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}} \approx 0.516 \text{ seconds} \] This calculation helps set the framework for understanding motion because angular frequency directly influences how fast the periodic oscillations occur.
Displacement in SHM
Displacement in simple harmonic motion describes how far the object is from the equilibrium position at any given time. It is often modeled by a sine or cosine function because the motion is periodic.
The general formula is: \[ x(t) = A \cos(\omega t + \phi) \] where:
  • \( x(t) \) is the displacement at time \( t \)
  • \( A \) is the amplitude, the maximum displacement from equilibrium
  • \( \omega \) is the angular frequency
  • \( \phi \) is the phase constant, which adjusts based on the initial conditions
In this exercise, when calculating for when the object is \(10 \text{ cm} \) to the left or \( 5 \text{ cm} \) to the right, we adjust our cosine function to solve for the specific times these displacements occur. Understanding this displacement formula allows you to predict positions over time, describing the object's location during its oscillatory 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

Starting with the force equation for a damped harmonic oscillator, show that a solution of the form \(x(t)=A e^{-(b / 2 m) t} \sin \omega_{1} t\) works. The differential equation and the lightly damped oscillation frequency are \(m \frac{d^{2} x}{d t^{2}}+b \frac{d x}{d t}+k x=0 \quad\) and \(\quad \omega_{1}=\sqrt{\omega_{0}^{2}-\frac{b^{2}}{4 m^{2}}}\)

A \(5.0-\mathrm{kg}\) object oscillates on a spring with a force constant of \(180 \mathrm{~N} / \mathrm{m}\). The damping coefficient is \(0.20 \mathrm{~kg} / \mathrm{s}\). The system is driven by a sinusoidal force of maximum value \(50 \mathrm{~N}\) and the angular frequency is \(20 \mathrm{rad} / \mathrm{s}\). (a) What is the amplitude of the oscillations? (b) If the driving frequency is varied, at what frequency will resonance occur?

In the case of the damped harmonic oscillator, what are the units of the damping constant, \(b\) ?

A spinning golf ball of radius \(R\) can be suspended in a stream of high velocity air (Figure 12-30). The ball is in equilibrium at the vertical center of the air stream \((y=0)\), and not moving except for its rotation at velocity \(v_{\text {Rot }}\). Show that the ball undergoes simple harmonic motion when it is released a small vertical distance \(y\) below the equilibrium position. Assume that for small \(y\) there is no net horizontal force, and that for small values of \(y\) the speed of the air drops off linearly from the center of the air stream. (For small \(y\), the speed varies as \(v(y)=v_{0}-b|y|\), where \(v_{0}\) is the speed of the air in the center of the stream and \(b\) is a constant.) SSM

An object on the end of a spring oscillates with a frequency of \(15 \mathrm{~Hz}\). Calculate (a) the period of the motion and (b) the number of oscillations that the object undergoes in \(2 \mathrm{~min}\). SSM

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.