Question

A spring fixed at its upper end is stretched 6 inches by a 10-pound weight attached at its lower end. The spring-mass system is suspended in a viscous medium so that the system is subjected to a damping force of \(5 \frac{d x}{d t}\) lbs. Describe the motion of the system if the weight is drawn down an additional 4 inches and released. What would happen if you changed the coefficient " 5 " to "4"? [You may need to consult your introductory physics text.]

Short answer

Given that the spring constant, damping ratio and the additional displacement have been determined, the motion of the system can be described by solving the formed ordinary differential equation. This will unveil the nature of the oscillations: whether the system is over-damped, under-damped or critically-damped. Changing the damping coefficient from '5' to '4' affects the damping ratio, which would in turn alter the system's response - potentially changing its damping status (becoming over, under or critically damped). The exact results depend on the numerical values of the parameters involved and the actual calculation made.

Step-by-step solution

Determine the spring constant

Find \(k\) by applying Hooke's Law, \(F=kx\), where \(F\) is the force applied (weight 10 pounds), and \(x\) is the displacement (6 inches or 0.5 feet). Solve to get the spring constant \(k = \frac{F}{x}\).

Calculate the damping ratio

The damping force is \(5 \frac{dx}{dt}\). The damping ratio, \(\zeta\), can be found by dividing the damping force by the force obtained through multiplying the mass \(m\) (which in this case is the weight divided by gravitational acceleration) and 2 times the square root of the spring constant, \( \zeta = \frac{5}{2m\sqrt{k}} \).

Formulate the ODE

Apply Newton's 2nd Law to formulate the equation of motion for the damper-spring system: \(m\frac{d^2x}{dt^2} + 5\frac{dx}{dt} + kx = 0\). Here, we have an additional displacement of 4 inches or (1/3) feet when the weight is released.

Solve the ODE

Solve the second order ordinary differential equation (ODE) to describe the motion. The solution depends on the value of the damping ratio, \(\zeta\). There are three cases to consider: If \(\zeta<1\), the system is under-damped and oscillates. If \(\zeta>1\), the system is over-damped and will not oscillate but returns to equilibrium without oscillating. If \(\zeta=1\), the system is critically-damped and also does not oscillate.

Analyze the effect of the changed damping coefficient

If the damping coefficient was '4' instead of '5', we would have to recompute the damping ratio and reanalyze the system. Observe the effect this change has on the system's behavior, especially with regards to oscillation.

Key concepts

Spring Constant Calculation

In a spring-mass system, the spring constant, denoted by \( k \), is a crucial parameter that determines how stiff a spring is. It's calculated using Hooke's Law: \( F = kx \). This law states that the force \( F \) required to stretch or compress a spring by a distance \( x \) is proportional to that distance. For this exercise, we are given that a 10-pound weight stretches the spring 6 inches (or 0.5 feet). Therefore, to find \( k \), we rearrange the equation to \( k = \frac{F}{x} \). With \( F = 10 \) pounds and \( x = 0.5 \) feet, we compute the spring constant as \( k = \frac{10}{0.5} = 20 \, \text{lb/ft} \). This constant tells us how much force is needed per unit length of stretch in the spring. Understanding spring constants is vital to predicting the behavior of the spring in dynamic scenarios, such as oscillations.

Damping Ratio Analysis

The damping ratio, represented by \( \zeta \), is an essential concept when analyzing damped harmonic oscillators. It helps in understanding the system's response to disturbances, particularly whether the system will oscillate and how quickly it will return to equilibrium.

• A damping force of \( 5 \frac{dx}{dt} \) is applied, where \( 5 \) represents the damping coefficient.
• The weight attached can be viewed as a mass \( m \), calculated by dividing the weight (10 pounds) by the gravitational force \( 32.2 \, \text{ft/s}^2 \).
• Then, the damping ratio is determined using the formula \( \zeta = \frac{5}{2m\sqrt{k}} \). This formula considers both the damping force and the spring constant, finding their ratio against mass-related factors.

If \( \zeta < 1 \), the system is underdamped, leading to oscillations. If \( \zeta = 1 \), it is critically damped, returning to equilibrium without oscillating. Finally, if \( \zeta > 1 \), the system is overdamped, settling without oscillations.

Ordinary Differential Equations in Physics

Ordinary differential equations (ODEs) play a fundamental role in modeling physical systems. For a damped harmonic oscillator, the motion is governed by a second-order ODE derived from Newton's 2nd Law: \[ m\frac{d^2x}{dt^2} + 5\frac{dx}{dt} + kx = 0 \]Here, each term has a specific interpretation that represents physical actions within the system:

• The first term \( m\frac{d^2x}{dt^2} \) corresponds to mass times acceleration, reflecting the inertia of the system.
• The second term \( 5\frac{dx}{dt} \) symbolizes the damping force.
• The last term \( kx \) is the restoring force due to the spring.

These ODEs model how the spring-mass-damper system responds over time. By solving the ODE, we can predict how the system behaves, given an initial disturbance such as a 4-inch displacement from equilibrium. The nature of the solution depends on the damping ratio \( \zeta \), which influences whether the system returns to equilibrium oscillating or not. Understanding how to formulate and solve these ODEs is crucial for analyzing and predicting physical phenomena in mechanics.