/*! 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 24 A cheerleader waves her pom-pom ... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 24

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; \((6)\) the acceleration and speed when the poin-poin's coordinate is \(x=+9.0 \mathrm{cm}\); (c) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asproach used parts \((\mathrm{a}),(\mathrm{b}),\) and \((\mathrm{c})\) can be found using the energy approach used in Section \(13.3,\) and which cannot? Explain.

Short Answer

Expert verified
(a) Maximum velocity: 0.961 m/s, maximum acceleration: 5.15 m/s虏. (b) At x = 9 cm, velocity: 0.831 m/s, acceleration: -2.57 m/s虏. (c) Time to x = 12 cm: 0.148 s. (d) Energy approach can be used for velocity but not acceleration.

Step by step solution

01

Understand Simple Harmonic Motion (SHM) Concepts

In SHM, a particle oscillates about an equilibrium position. The motion can be described by the equations: 1. Displacement: \( x(t) = A \cos(\omega t + \phi) \)2. Velocity: \( v(t) = -A\omega \sin(\omega t + \phi) \)3. Acceleration: \( a(t) = -A\omega^2 \cos(\omega t + \phi) \)where \( A \) is the amplitude, \( \omega \) is the angular frequency \( \omega = 2\pi f \), and \( \phi \) is the phase constant.
02

Calculate Maximum Velocity and Acceleration

First, find the angular frequency: \( \omega = 2\pi f = 2\pi \times 0.850 = 5.34 \, \mathrm{rad/s} \). The maximum velocity \( v_{\text{max}} \) is \( A\omega = 0.18 \times 5.34 = 0.961 \, \mathrm{m/s} \). The maximum acceleration \( a_{\text{max}} \) is \( A\omega^2 = 0.18 \times (5.34)^2 = 5.15 \, \mathrm{m/s^2} \).
03

Calculate Velocity and Acceleration at x=+9.0 cm

For displacement \( x = 9.0 \, \mathrm{cm} = 0.09 \, \mathrm{m} \), use energy conservation.Total energy \( E = \frac{1}{2}m(A\omega)^2 \).At position \( x \), potential energy \( PE = \frac{1}{2}m(\omega x)^2 \).Kinetic energy \( KE = E - PE = \frac{1}{2}mv^2 \).Solve for \( v \): \( v = \sqrt{(A^2 - x^2)\omega^2} = \sqrt{(0.18^2 - 0.09^2) \times 5.34^2} = 0.831 \, \mathrm{m/s} \).For acceleration, \( a = -\omega^2 x = -(5.34)^2 \times 0.09 = -2.57 \, \mathrm{m/s^2} \).
04

Calculate Time from Equilibrium to x = 12.0 cm

Using the equation for displacement, \( x(t) = A \cos(\omega t) \), determine the time \( t \) when \( x = 12.0 \, \mathrm{cm} = 0.12 \, \mathrm{m} \).\( \cos(\omega t) = \frac{x}{A} = \frac{0.12}{0.18} = \frac{2}{3} \).\( \omega t = \arccos(\frac{2}{3}) \).\( t = \frac{\arccos(\frac{2}{3})}{5.34} = 0.148 \, \mathrm{s} \).
05

Discuss Energy Approach

The energy approach can be used to find the maximum velocity and velocity at a point, since it involves kinetic and potential energy. However, acceleration, being directly a result of force (or second derivative of displacement), requires knowledge of kinematics and cannot be directly deduced from energy. Thus, parts (a) and (b)'s velocity calculations, as well as part (c), can use the energy approach. However, the acceleration calculations in (a) and (b) cannot be found using energy considerations alone.

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Key Concepts

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

Understanding Amplitude in Simple Harmonic Motion
In simple harmonic motion (SHM), amplitude is a crucial parameter. It is the maximum displacement of an object from its equilibrium position. Amplitude is denoted by the letter \( A \) and measured in meters (m). For this pom-pom waving exercise, the amplitude is given as 18.0 cm, or 0.18 m.

Amplitude tells us how far the oscillating object can move on either side of the center point. It isa positive constant, representing the extent of the oscillation.

Keep in mind that amplitude does not depend on other factors like frequency or the type of material. It solely represents the maximum range of motion of the oscillating particle.
Angular Frequency: The Heartbeat of SHM
Angular frequency is a concept that describes how quickly something is oscillating. It's often represented by the symbol \( \omega \) and is related to the physical parameters of the oscillation. Angular frequency is derived from the formula \( \omega = 2 \pi f \), where \( f \) is the frequency.

In our problem, with a frequency of 0.850 Hz, we calculated the angular frequency as \( \omega = 5.34 \) rad/s. This means our pom-pom completes approximately 5.34 full circular oscillations per second.
  • Higher angular frequency means faster oscillations.
  • It is measured in radians per second (rad/s).
Angular frequency is central to equations of motion in SHM, affecting both velocity and acceleration calculations.
Kinematics: Describing Motion in SHM
Kinematics focuses on describing motion without considering forces. In SHM, kinematics provides the equations for displacement, velocity, and acceleration.
  • Displacement: \( x(t) = A \cos(\omega t + \phi) \)
  • Velocity: \( v(t) = -A\omega \sin(\omega t + \phi) \)
  • Acceleration: \( a(t) = -A\omega^2 \cos(\omega t + \phi) \)
Kinematics allows us to predict the position and behavior of the pom-pom at any given time \( t \). It was used to calculate maximum velocity and acceleration in our solution. Additionally, it was crucial in finding how long it takes to move a certain distance.
Energy Approach in SHM: Balancing Act
The energy approach in simple harmonic motion involves analyzing potential and kinetic energy. In SHM, the total mechanical energy (\( E \)) is constant, and is the sum of kinetic energy (KE) and potential energy (PE).

Potential energy reaches its peak when the object is at its maximum displacement, while kinetic energy is maximized at the equilibrium position.
  • Potential Energy: \( PE = \frac{1}{2} m (\omega x)^2 \)
  • Kinetic Energy: \( KE = E - PE \)
Understanding how energy shifts between potential and kinetic forms helped solve questions about velocities at particular points without needing to perform detailed kinematic calculations, except for certain acceleration determinations which require kinematic analysis.

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Most popular questions from this chapter

When displaced from equilibrium, the two liydrogen atoms in an \(\mathrm{H}_{2}\) molecule are acted on by a restoring force \(F_{x}=-k x\) with \(k=580 \mathrm{N} / \mathrm{m}\) . Calculate the oscillation frequency of the \(\mathrm{H}_{2}\) molecule. (Hint: The mass of a hydrogen atom is 1.008 atomic mass units, or 1 u; see Appendix E. As in Example 13.7 (Section \(13.4 ),\) use \(m / 2\) instead of \(m\) in the expression for \(f\) )

An experimental package and its support structure, which are to be placed on board the International Space Station, act as an underdamped spring-mass system with a force constant of \(2.1 \times 10^{5} \mathrm{N} / \mathrm{m}\) and mass 108 \(\mathrm{kg}\) . A NASA requirement is that resonance for forced oscillations not occur for any frequency below 35 \(\mathrm{Hz}\) . Does this package meet the requirement?

Weighing Astronauts. This procedure has actually been used to "weigh" astronauts in space. A \(42.5-\mathrm{kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

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 spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) is hung vertically, and a \(0.200-\mathrm{kg}\) pan is suspended from its lower end. A butcher drops a 2.2\(\cdot \mathrm{kg}\) steak onto the pan from a height of 0.40 \(\mathrm{m}\) . The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak imto immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?
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