Problem 75 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Draw a potential energy diagram, \(U\) vs. \(x\), and analyze the motion of a mass \(m\) resting on a frictionless horizontal table and connected to a horizontal spring with stiffness constant \(k\). The mass is pulled a distance to the right so that the spring is stretched a distance \(x_{0}\) initially, and then the mass is released from rest.

Step-by-Step Solution

Answer

The energy diagram is a parabolic curve of \( U(x) = \frac{1}{2}kx^2 \), showing oscillatory motion, converting energy between potential and kinetic.

Step 1: Understand the System

We are considering a mass-spring system where the mass \( m \) is attached to a spring with spring constant \( k \). The mass is initially stretched to \( x = x_0 \) from its equilibrium position and then released. The table is frictionless, and we need to analyze the potential energy and the motion of the mass.

Step 2: Define the Potential Energy Function

The potential energy \( U(x) \) of a spring system is given by the formula \( U(x) = \frac{1}{2}kx^2 \), where \( x \) is the displacement from the equilibrium position. This equation will help us plot the energy diagram.

Step 3: Plot the Potential Energy Diagram

To draw the potential energy diagram, plot \( U(x) = \frac{1}{2}kx^2 \) against \( x \). This will be an upward-opening parabola with its vertex at \( x = 0 \), reflecting that potential energy is zero at the equilibrium position and increases as \( x \) increases in either direction from equilibrium.

Step 4: Analyze the Motion

Once released, the mass will oscillate back and forth about \( x = 0 \). Initially, at \( x = x_0 \), the total energy is potential (since it starts from rest), and the kinetic energy is zero. As the mass moves towards equilibrium, potential energy is converted to kinetic energy, becoming zero at the equilibrium position \( x = 0 \) where the speed is maximal. Then kinetic energy reverts back to potential energy as the mass continues beyond equilibrium.

Step 5: Identify Key Points on the Diagram

At maximum displacement \( x = x_0 \), potential energy is \( \frac{1}{2}kx_0^2 \), and kinetic energy is zero. As the mass moves through equilibria \( x = 0 \), potential energy is zero, and kinetic energy equals the initial potential energy. This conversion between potential and kinetic defines the oscillatory motion.

Key Concepts

Potential Energy

Potential energy in the context of harmonic motion is the stored energy in a system due to its position or configuration. For a mass-spring system, this potential energy comes from the deformation of the spring. When we have a spring with a spring constant \( k \) and it is displaced from its equilibrium position by a distance \( x \), it stores potential energy given by the equation:

• \( U(x) = \frac{1}{2}kx^2 \).

This equation tells us that the potential energy is proportional to the square of the displacement from the equilibrium. In a potential energy diagram, this relationship is represented as an upward-opening parabola. At \( x = 0 \), which is the equilibrium position, the potential energy is zero. As \( x \) increases either to the right or left (indicating stretching or compressing the spring), the potential energy increases. Thus, the equilibrium position is the point where the spring is neither compressed nor stretched, resulting in zero stored energy in the spring.

Spring Constant

The spring constant, denoted by \( k \), is a measure of a spring's stiffness. It quantifies how much force is needed to stretch or compress a spring by a certain amount. A higher \( k \) value indicates a stiffer spring that requires more force to deform. Understanding the spring constant is crucial because it directly influences how much potential energy can be stored in the system. According to Hooke's Law, the force \( F \) exerted by the spring is:

• \( F = -kx \).

This tells us that the force is proportional to the displacement \( x \) and the negative sign indicates that it acts in the opposite direction. When analyzing harmonic motion, a large spring constant means that for a given displacement, the force by the spring is greater, thus affecting the speed and acceleration of the mass. In potential energy terms, increasing \( k \) increases the steepness of the energy parabola, meaning that potential energy builds up more rapidly with displacement, affecting the system's dynamics.

Equilibrium Position

The equilibrium position in a mass-spring system is the point where the net force acting on the mass is zero, which occurs when the spring is at its natural length (neither stretched nor compressed). In this position, no potential energy is being stored, as seen when \( x = 0 \) in the potential energy formula \( U(x) = \frac{1}{2}kx^2 \). During harmonic motion, the equilibrium position is oscillation's central point where the kinetic energy of the mass is highest, while potential energy is at its minimum. The mass moves fastest at this position because all the potential energy has been converted into kinetic energy, and the system is in its most dynamically active state. As the system oscillates, the mass moves through the equilibrium position, gaining speed as potential energy transforms into kinetic, and slowing down as it converts back, either compressing or stretching the spring. Thus, understanding the equilibrium position is key to comprehending the fundamental behavior of harmonic motion, as it acts as a natural balance point where forces are momentarily in harmony, dictating the flow of energy in the system.