Problem 33 - College Physics

Question

\(\bullet\) \(\bullet\) 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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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 \( PE = \frac{1}{2} E_{total} \).

Step 4: 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:\[ x = \frac{A}{\sqrt{2}} \]

Key Concepts

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’s 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’s 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.