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Chapter 11: Problem 33

\(\bullet\) \(\bullet\) A mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\) ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

Short Answer

Expert verified
The mass is \( \frac{A}{\sqrt{2}} \) from the equilibrium position.

Step by step solution

01

Understanding the Conservation of Energy

In a spring-mass system, the total mechanical energy is conserved if there is no damping. The total energy, which is the sum of kinetic energy (KE) and potential energy (PE), remains constant. So, when the elastic potential energy equals the kinetic energy, each is half of the total mechanical energy.
02

Express Total Energy in Terms of Amplitude

The total mechanical energy in the spring-mass system is given by the elastic potential energy when the mass is at amplitude position (maximum displacement), which is: \[ E_{total} = \frac{1}{2} k A^2 \] where \( k \) is the spring constant and \( A \) is the amplitude. At maximum displacement, the kinetic energy is zero.
03

Set Elastic Potential Energy Equal to Kinetic Energy

At a position \( x \) from equilibrium, the potential energy \( PE \) can be expressed as:\[ PE = \frac{1}{2} k x^2 \]If the potential energy equals the kinetic energy, then each is half of the total energy:\[ \frac{1}{2} k x^2 = \frac{1}{4} k A^2 \] This simplification results from setting \( PE = \frac{1}{2} E_{total} \).
04

Solve for the Displacement x

To find the displacement \( x \), solve the equation:\[ \frac{1}{2} k x^2 = \frac{1}{4} k A^2 \]Cancel out \( \frac{1}{2} k \) from both sides:\[ x^2 = \frac{1}{2} A^2 \]Take the square root of both sides:\[ x = \frac{A}{\sqrt{2}} \]

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Key Concepts

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

Energy Conservation in Harmonic Motion
In a spring-mass system experiencing harmonic motion, energy conservation is a fundamental principle. It dictates that the total mechanical energy remains constant over time, provided that there is no external friction or damping acting on the system. This energy is split between two forms:
  • Elastic Potential Energy (stored in the spring)
  • Kinetic Energy (associated with the mass's motion)
When the mass moves, energy transfers between these forms. At one moment, it might be mostly potential energy, and as it passes through the equilibrium point, it's largely kinetic energy. Understanding that the sum of these energies stays constant helps us predict the system鈥檚 behavior.
Spring-Mass System Dynamics
A spring-mass system consists of a mass attached to a spring, oscillating back and forth due to the restorative force of the spring. This force obeys Hooke's Law: \( F = -kx \), where:
  • \( F \) is the force exerted by the spring
  • \( k \) is the spring constant, representing the stiffness of the spring
  • \( x \) is the displacement from the equilibrium position
This system exhibits simple harmonic motion, characterized by a sinusoidal pattern of movement that repeats in cycles. The properties of this motion, including its frequency and period, depend on the mass and the spring constant, but not on the amplitude.
The maximum displacement from the equilibrium, known as amplitude \( A \), is where the elastic potential energy peaks, and the kinetic energy is minimal. Conversely, at the equilibrium point, energy is entirely kinetic.
Understanding Elastic Potential Energy
Elastic potential energy in a spring is the energy stored when the spring is compressed or stretched from its natural length. According to Hooke's Law, the formula for elastic potential energy (\( PE \)) is:\[ PE = \frac{1}{2} k x^2 \]Here:
  • \( k \) is the spring constant, indicating how stiff the spring is
  • \( x \) is the displacement from the equilibrium position
When the potential energy equals the kinetic energy, each is half of the total mechanical energy available in the system. At this point in the motion, the mass鈥檚 distance from the equilibrium position can be determined by the equation:\[ x = \frac{A}{\sqrt{2}} \]This reflects the balance between the transformed energies as the motion continues. Knowing this allows us to relate the position of the mass directly to its potential and kinetic energies during its cycle.

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Most popular questions from this chapter

\(\bullet\) Achilles tendon. The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is \(78.1 \mathrm{mm}^{2},\) its average length is \(25 \mathrm{cm},\) and its average Young's modulus is 1474 MPa. (a) How much tensile stress is required to stretch this muscle by 5.0\(\%\) of its length? (b) If we model the tendon as a spring, what is its force constant? (c) If a 75 kg sprinter exerts a force of 13 times his weight on his Achilles tendon, by how much will it stretch?

\(\bullet\) \(\bullet\) One end of a stretched ideal spring is attached to an airtrack and the other is attached to a glider with a mass of 0.355 \(\mathrm{kg}\) . The glider is released and allowed to oscillate in SHM. If the distance of the glider from the fixed end of the spring varies between 1.80 \(\mathrm{m}\) and \(1.06 \mathrm{m},\) and the period of the oscillation is \(2.15 \mathrm{s},\) find (a) the force constant of the spring, (b) the maximum speed of the glider, and (c) the magnitude of the maximum acceleration of the glider.

\(\bullet\) Shear forces are applied to a rectangular solid. The same forces are applied to another rectangular solid of the same material, but with three times each edge length. In each case the forces are small enough that Hooke's law is obeyed. What is the ratio of the shear strain for the larger object to that of the smaller object?

\(\bullet\) Effect of diving on blood. It is reasonable to assume that the bulk modulus of blood is about the same as that of water \((2.2 \mathrm{GPa}) .\) As one goes deeper and deeper in the ocean, the pressure increases by \(1.0 \times 10^{4}\) Pa for every meter below the surface. (a) If a diver goes down 33 \(\mathrm{m}\) (a bit over 100 \(\mathrm{ft} )\) in the ocean, by how much does each cubic centimeter of her blood change in volume? (b) How deep must a diver go so that each drop of blood compresses to half its volume at the surface? Is the ocean deep enough to have this effect on the diver?

\(\bullet\) A petite young woman distributes her 500 \(\mathrm{N}\) weight equally over the heels of her high-heeled shoes. Each heel has an area of 0.750 \(\mathrm{cm}^{2} .\) (a) What pressure is exerted on the floor by each heel? (b) With the same pressure, how much weight could be supported by two flat- bottomed sandals, each of area 200 \(\mathrm{cm}^{2} ?\)
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