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Chapter 13: Problem 75

A 5.00 kg partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 \(\mathrm{m}\) below its equilibrium position and released, it vibrates with a period of 4.20 \(\mathrm{s}\) . (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is 0.050 \(\mathrm{m}\) above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point 0.050 \(\mathrm{m}\) below its equilibrium position to a point 0.050 \(\mathrm{m}\) above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

Short Answer

Expert verified
(a) 0.0472 m/s (b) -0.0112 m/s虏 (c) 1.05 s (d) spring shortens by 43.95 m.

Step by step solution

01

Calculate Spring Constant

To find the spring constant \( k \), we utilize the formula for the period of a mass-spring system: \[ T = 2\pi \sqrt{\frac{m}{k}} \]Given \( T = 4.20 \ \mathrm{s} \) and \( m = 5.00 \ \mathrm{kg} \), rearrange the formula to solve for \( k \):\[ k = \frac{4\pi^2 m}{T^2} \]Substitute the values:\[ k = \frac{4\pi^2 \times 5.00}{4.20^2} \approx 1.1150\, \mathrm{N/m} \]
02

Find Speed at Equilibrium Position

The speed at the equilibrium position is the maximum speed, and is given by:\[ v_{\text{max}} = \omega A \]where \( \omega = \sqrt{\frac{k}{m}} \) and \( A = 0.100 \ \mathrm{m} \).First calculate \( \omega \):\[ \omega = \sqrt{\frac{1.1150}{5.00}} \approx 0.4723 \ \mathrm{rad/s} \]Now calculate \( v_{\text{max}} \):\[ v_{\text{max}} = 0.4723 \times 0.100 = 0.0472 \ \mathrm{m/s} \]
03

Determine Acceleration 0.050 m Above Equilibrium

Acceleration at a point is given by:\[ a = \omega^2 x \]Where \( x = -0.050 \ \mathrm{m} \), since it's above the equilibrium, and \( \omega \approx 0.4723 \ \mathrm{rad/s} \). Substitute these into the formula:\[ a = (0.4723)^2 \times (-0.050) \approx -0.0112 \ \mathrm{m/s^2} \]
04

Time to Move 0.050 m Below to 0.050 m Above

For harmonic motion, the time to move from \( 0.050 \ \mathrm{m} \) below to \( 0.050 \ \mathrm{m} \) above the equilibrium is half the time of oscillation from equilibrium to maximum displacement, given by quarter period:\[ t = \frac{T}{4} \]Substitute \( T = 4.20 \ \mathrm{s} \):\[ t = \frac{4.20}{4} = 1.05 \ \mathrm{s} \]
05

Determine Spring Shortening After Partridge is Removed

The spring's displacement from equilibrium when the partridge is attached is:\[ x = \frac{mg}{k} \]Substitute \( m = 5.00 \ \mathrm{kg} \), \( g = 9.81 \ \mathrm{m/s^2} \), and \( k \approx 1.1150 \ \mathrm{N/m} \):\[ x = \frac{5.00 \times 9.81}{1.1150} \approx 43.95 \ \mathrm{m} \]This is the amount by which the spring is stretched when the partridge is attached, so it will shorten by this length when removed.

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Key Concepts

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

harmonic motion
Harmonic motion refers to a type of periodic motion where an object moves back and forth over the same path. Imagine a swing or pendulum; they repetitively move through their central position, reach extreme points, and then return. In physics, harmonic motion is often called simple harmonic motion (SHM) when the restoring force is directly proportional to the displacement from an equilibrium position.

Key characteristics of simple harmonic motion include:
  • A repeating path around the equilibrium position.
  • Acceleration always directed towards the equilibrium position.
  • Frequencies and periods that remain constant regardless of amplitude (assuming ideal conditions).

In the case of the 5.00 kg partridge, harmonic motion is observed as it oscillates around its equilibrium position. The maths of SHM provides a framework to predict positions, velocities, and accelerations at any point in the cycle, allowing us to solve various dynamics problems.
spring constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It shows how resistant a spring is to being compressed or stretched. A higher spring constant means the spring is stiffer and requires more force to change its length.

For a mass-spring system, the spring constant can be calculated if the period and mass are known. The formula used is:\[T = 2\pi \sqrt{\frac{m}{k}}\]Rearranging gives:\[k = \frac{4\pi^2 m}{T^2}\]

In our exercise, the spring constant of the system with a period of 4.20 seconds and mass of 5.00 kg was calculated to be approximately 1.1150 N/m. This value helps us understand how much the spring will stretch or compress based on the applied forces, like the weight of the partridge.
equilibrium position
The equilibrium position in a mechanical system, like a spring, is the point where forces are balanced, and the system experiences no net force. For a mass-spring system, it's the point where the spring neither compresses nor elongates when the mass is attached.

Whenever the partridge oscillates, it passes through this equilibrium position. It's significant because it's where the speed of the object is the highest during harmonic motion due to all potential energy being converted into kinetic energy. At this point, the spring force equals the gravitational force, keeping the system in balance.

In the given problem, the partridge is initially pulled below this position. Upon release, it moves up passing the equilibrium position at maximum speed, demonstrating the energy transformations inherent in SHM.
mass-spring system
A mass-spring system consists of a mass attached to a spring. It's a standard model for demonstrating oscillatory systems and studying simple harmonic motion.

Here's how it works:
  • When displaced from the equilibrium, the spring applies a force to return the mass to this point.
  • This restoring force follows Hooke's Law \( F = -kx \), where \( x \) is the displacement.
  • The system undergoes periodic motion with a characteristic frequency and period.

The 5.00 kg partridge attached to a spring is a perfect example of a mass-spring system. The dynamics of its motion reveal key parameters like amplitude, period, and frequency, all vital for predicting future positions and velocities. Moving from initial displacement, the system will continue to oscillate, showing the elegance and simplicity of such harmonic motion.

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Most popular questions from this chapter

Two identical atoms in a diatomic molecule vibrate as harmonic oscillators. However, their center of mass, midway between them, remains at rest. (a) Show that at any instant, the momenta of the atoms relative to the center of mass are \(\vec{p}\) and \(-\vec{p} .\) (b) Show that the total kinetic energy \(K\) of the two atoms at any instant is the same as that of a single object with mass \(m / 2\) with a momentum of magnitude \(p .\) (Use \(K=p^{2} / 2 m . )\) This result shows why \(m / 2\) should be used in the expression for \(f\) in have masses \(m_{1}\) and \(m_{2},\) show that the result of part (a) still holds and the single object's mass in part (b) is \(m_{1} m_{2} /\left(m_{1}+m_{2}\right) .\) The quantity \(m_{1} m_{2} J\left(m_{1}+m_{2}\right)\) is called the reducedmass of the system.

A block with mass \(M\) rests on a frictionless surface and is connected to a horizontal spring of force constant \(k\) . The other end of the spring is attached to a wall (Fig. 13.36\() .\) A second block with mass \(m\) rests on top of the first block. The coefficient of static friction between the blocks is \(\mu_{s}\) Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.

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.

Jerk. A guitar string vibrates at a frequency of 440 \(\mathrm{Hz}\) . A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

A 10.0 -kg mass is traveling to the right with a speed of 2.00 \(\mathrm{m} / \mathrm{s}\) on a smooth horizontal surface when it collides with and sticks to a second \(10.0-\mathrm{kg}\) mass that is initially at rest but is attached to a light spring with force constant 80.0 \(\mathrm{N} / \mathrm{m}\) . (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?
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