/*! 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 13 The point of the needle of a sew... [FREE SOLUTION] | 91Ó°ÊÓ

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The point of the needle of a sewing machine moves in SHM along the \(x\) -axis with a frequency of 2.5 \(\mathrm{Hz}\) . At \(t=0\) its position and velocity components are \(+1.1 \mathrm{cm}\) and \(-15 \mathrm{cm} / \mathrm{s}\) , respectively (a) Find the acceleration component of the needle at \(t=0\) . (b) Write equations giving the position, velocity, and accelerationcomponents of the point as a function of time.

Short Answer

Expert verified
(a) The acceleration at \(t=0\) is \(-118 \, \mathrm{cm/s^2}\). (b) Use SHM equations with found parameters for \(x(t)\), \(v(t)\), and \(a(t)\).

Step by step solution

01

Understand SHM properties

Simple Harmonic Motion (SHM) follows the equation \(x(t) = A \cos(\omega t + \phi)\), where \(x(t)\) is the position, \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. The velocity \(v(t)\) is given by the derivative:\(v(t) = -A\omega \sin(\omega t + \phi)\) and the acceleration \(a(t) = -A\omega^2 \cos(\omega t + \phi)\).
02

Calculate angular frequency

The angular frequency \(\omega\) is determined by \(\omega = 2\pi f\), where \(f = 2.5 \, \mathrm{Hz}\). Thus, \(\omega = 2\pi \times 2.5 = 5\pi \, \mathrm{rad/s}\).
03

Use initial conditions to find phase constant

Given initial position \(x(0)=1.1 \,\mathrm{cm}\) and velocity \(v(0)=-15 \, \mathrm{cm/s}\), we can use them to solve for \(\phi\). From \(x(0) = A \cos(\phi)\) and \(v(0) = -A\omega \sin(\phi)\), substitute \(\omega = 5\pi\). Write both equations and solve for \(\phi\).
04

Determine amplitude A

Use the equations \(x(0) = A \cos(\phi)\) and \(v(0) = -A\omega \sin(\phi)\) simultaneously to determine the amplitude \(A\). The solution of these equations will allow us to find \(A\) and \(\phi\).
05

Find acceleration at t=0

The acceleration at \(t=0\) is \(a(0) = -A\omega^2 \cos(\phi)\). With \(\omega = 5\pi\) and \(\phi\) determined, substitute to find \(a(0)\).
06

Write the function equations

Substitute the determined \(A\), \(\omega\), and \(\phi\) into the general SHM equations. Position: \(x(t) = A \cos(5\pi t + \phi)\), velocity: \(v(t) = -A\cdot5\pi \sin(5\pi t + \phi)\), acceleration: \(a(t) = -A\cdot (5\pi)^2 \cos(5\pi t + \phi)\).

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Key Concepts

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

Angular Frequency
Angular frequency, represented by \( \omega \), is a fundamental concept in Simple Harmonic Motion (SHM). It defines how fast an object oscillates through its cycle. Think of it as the rate at which the oscillation cycles are completed per unit of time. This is expressed in radians per second. For instance, in this exercise, the needle oscillates with a frequency of 2.5 Hz, which means it completes 2.5 cycles each second. To find the angular frequency, you use the formula \( \omega = 2\pi f \), with \( f \) being the frequency in Hz. This converts our frequency into the angular frequency, resulting in \( \omega = 5\pi \, \mathrm{rad/s} \). The angular frequency is crucial because it helps describe the oscillatory motion accurately in the equations for position, velocity, and acceleration.
Phase Constant
The phase constant, denoted as \( \phi \), is a pivotal factor in determining the starting point of the oscillation. In other words, it tells us where the object is in its cycle at \( t=0 \). In many problems, especially involving SHM, we'll have initial conditions that help us find this value. For example, when you know the initial position \( x(0) = 1.1 \, \mathrm{cm} \) and initial velocity \( v(0) = -15 \, \mathrm{cm/s} \), these conditions can be plugged into the SHM equations.By setting \( x(0) = A \cos(\phi) \) and \( v(0) = -A\omega \sin(\phi) \), and knowing \( \omega = 5\pi \), we solve these simultaneous equations to find \( \phi \). The phase constant aligns the mathematical model to the actual motion in space and time, highlighting how starting conditions impact the motion.
Amplitude
Amplitude \( A \) represents the extent of the oscillation from its equilibrium position. It measures the maximum displacement on either side of the equilibrium point. In the context of our sewing machine problem, amplitude will tell us how far the needle moves in its simple harmonic path.To determine \( A \), we use initial conditions provided in the problem: the initial position and velocity. By solving the equations \( x(0) = A \cos(\phi) \) and \( v(0) = -A\omega \sin(\phi) \), we can uncover both \( A \) and \( \phi \). Amplitude doesn't change over time in SHM, making it a fixed value that contributes to defining the motion's scale and energy.
Position-Velocity-Acceleration Relationship
In Simple Harmonic Motion, position, velocity, and acceleration are intricately linked. They describe how an object moves back and forth in a predictable pattern. The position function \( x(t) = A \cos(\omega t + \phi) \) provides the displacement from equilibrium at any time \( t \).The velocity is the time derivative of the position, calculated as \( v(t) = -A\omega \sin(\omega t + \phi) \). This shows how the speed and direction change as the needle passes through its cyclical path. Acceleration comes next as the derivative of velocity, \( a(t) = -A\omega^2 \cos(\omega t + \phi) \), revealing how the rate of velocity changes, influenced directly by both the amplitude and angular frequency squared.Together, these equations demonstrate the periodic nature of SHM: when the position is maximum, velocity is zero, and acceleration reaches its highest negative or positive value. Understanding this relationship is crucial for predicting the behavior of oscillatory systems.

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Most popular questions from this chapter

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

A thrill-seeking cat with mass 4.00 \(\mathrm{kg}\) is attached by a harness to an ideal spring of negligible mass and oscillates verticallyin SHM. The amplitude is \(0.050 \mathrm{m},\) and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; ( \((b)\) at its lowest point; (c) at its equilibrium position.

An object with mass 0.200 \(\mathrm{kg}\) is acted on by an elastic restoring force with force constant 10.0 \(\mathrm{N} / \mathrm{m}\) . (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-0.300 \mathrm{m}\) to \(+0.300 \mathrm{m} .\) On your graph, let \(1 \mathrm{cm}=0.05 \mathrm{J}\) vertically and \(1 \mathrm{cm}=0.05 \mathrm{m}\) borizontally. The object is set into oscillation withan initial potential energy of 0.140 \(\mathrm{J}\) and an initial kinetic energy of 0.060 \(\mathrm{J}\) . Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one-half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

A 0.500 \(\mathrm{kg}\) glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) , undergoes SHM with an amplitude of 0.040 \(\mathrm{m}\) . Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m} ;\) (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x=-0.015 \mathrm{m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

A \(0.150-\mathrm{kg}\) toy is undergoing SHM on the end of a horizontal spring with force constant \(k=300 \mathrm{N} / \mathrm{m}\) . When the object is 0.0120 \(\mathrm{m}\) from its equilibrium position, it is observed to have a speed of 0.300 \(\mathrm{m} / \mathrm{s}\) . What are (a) the total energy of the object at any point of its motion; (b) the amplitude of the motion; (c) the maximum speed attained by the object during its motion?

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