Problem 31 - College Physics

Question

A mass is oscillating with amplitude \(A\) at the end of a spring. (a) 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? (b) How far is the mass from the equilibrium position when the kinetic energy is \(\frac{1}{10}\) of the total energy?

Step-by-Step Solution

Answer

(a) \(\frac{A}{\sqrt{2}}\), (b) \(\frac{3A}{\sqrt{10}}\)."

Step 1: Understanding the Problem

This problem involves a mass oscillating on a spring, undergoing harmonic motion. We need to find distances from the equilibrium position where certain energy conditions hold: (a) when the elastic potential energy equals the kinetic energy and (b) when the kinetic energy is one-tenth of the total energy.

Step 2: Energy Relations in Harmonic Motion

For a mass-spring system, the total mechanical energy \(E\) is constant and can be expressed as the sum of kinetic energy \(K\) and potential energy \(U\): \[ E = K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \]where \(m\) is the mass, \(v\) is the velocity, \(k\) is the spring constant, and \(x\) is the displacement from equilibrium.

Step 3: Part (a): Solving for Equal Energy Condition

Here, the condition is \( K = U \). From the relation \( E = K + U \), we have:- If \( K = U \), then \( K = U = \frac{E}{2} \).Hence, \(\frac{1}{2}mv^2 = \frac{1}{2}kx^2\), which simplifies using the fact that total energy is only potential energy at amplitude (\( x = A \)), \( E = \frac{1}{2}kA^2 \).Setting: \[ \frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}kx^2 \implies kA^2 = kx^2 \] so, \[ A^2 = 2x^2 \Longrightarrow x = \frac{A}{\sqrt{2}} \].

Step 4: Part (b): Solving for Kinetic Energy Condition

In this case, the condition is \( K = \frac{1}{10}E \), leading to:\[ \frac{1}{2}mv^2 = \frac{1}{10}E \Rightarrow \frac{1}{2}kx^2 = \frac{9}{10}E \] (since potential energy \( U = E - K = \frac{9}{10}E \)).Using \( E = \frac{1}{2}kA^2 \), we substitute: \[ \frac{1}{2}kx^2 = \frac{9}{10}\left(\frac{1}{2}kA^2\right) \]This simplifies to \[ x^2 = \frac{9}{10}A^2 \Longrightarrow x = \frac{3}{\sqrt{10}}A \].

Step 5: Final Result Interpretation

For part (a), the mass is \(\frac{A}{\sqrt{2}}\) from the equilibrium when kinetic energy equals potential energy. For part (b), the mass is \(\frac{3A}{\sqrt{10}}\) from equilibrium when kinetic energy is \(\frac{1}{10}\) of total energy.

Key Concepts

Mass-Spring System

In a mass-spring system, we have a mass attached to a spring, which is capable of oscillating back and forth. This type of system is a classic example of simple harmonic motion. The oscillation occurs when the spring is either compressed or stretched, thereby storing potential energy. Once released, the mass moves towards its equilibrium position. As it moves, potential energy is converted into kinetic energy, making the mass speed up.

• Equilibrium Position: This is the position where the spring is neither stretched nor compressed.
• Amplitude: This is the maximum distance the mass moves from the equilibrium position.
• Restoring Force: The spring exerts a force to bring the mass back to equilibrium, described by Hooke's Law: \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement.

Energy Relations

In harmonic motion, energy continuously shifts between two forms: kinetic and potential. At any point, the sum of these two energies remains constant, known as the total mechanical energy, \( E \). This is essential because it defines energy conservation in a mass-spring system. The relationship between these energies is given by:\[ E = K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \] where \( K \) is kinetic energy and \( U \) is potential energy.

• Energy Conservation: At any displacement, the total energy \( E \) does not change, which means potential and kinetic energy interchange as the system oscillates.
• Maximum Values: At maximum displacement (amplitude), all energy is potential, while at the equilibrium position, it's all kinetic.

Kinetic Energy

Kinetic energy in a mass-spring system is linked to the velocity of the mass. It reaches its maximum value when the mass passes through the equilibrium position because here, the velocity is highest. Kinetic energy is defined as:\[ K = \frac{1}{2}mv^2 \]where \( m \) is the mass and \( v \) is the velocity.

• Relation to Velocity: Directly proportional to the square of the object's velocity, the kinetic energy determines how fast the mass moves.
• Energy Conditions: When kinetic energy equals potential energy, the system is at half of its amplitude. The formulas used help determine how far the mass is from equilibrium in such cases.

Potential Energy

Potential energy in a spring system relies on the displacement from its equilibrium position. The elastic potential energy is observed at its peak when the mass is at its maximum amplitude, as it translates into zero velocity and consequently zero kinetic energy. The potential energy is given by:\[ U = \frac{1}{2}kx^2 \]where \( k \) is the spring constant and \( x \) is the displacement.

• Displacement Impact: Directly proportional to the square of the displacement. The farther the spring is stretched or compressed, the more potential energy it stores.
• Energy Dependence: When potential energy equals kinetic energy, the mass has moved \( \frac{A}{\sqrt{2}} \) away from the equilibrium position. This balance helps in determining specific points of oscillation.