Question

A spring-mass-dashpot system consists of a \(10-\mathrm{kg}\) mass attached to a spring with spring constant \(k=100 \mathrm{~N} / \mathrm{m}\); the dashpot has damping constant \(7 \mathrm{~kg} / \mathrm{s}\). At time \(t=0\), the system is set into motion by pulling the mass down \(0.5 \mathrm{~m}\) from its equilibrium rest position while simultaneously giving it an initial downward velocity of \(1 \mathrm{~m} / \mathrm{s}\). (a) State the initial value problem to be solved for \(y(t)\), the displacement from equilibrium (in meters) measured positive in the downward direction. Give numerical values to all constants involved. (b) Solve the initial value problem. What is \(\lim _{t \rightarrow \infty} y(t)\) ? Explain why your answer for this limit makes sense from a physical perspective. (c) Plot your solution on a time interval long enough to determine how long it takes for the magnitude of the vibrations to be reduced to \(0.1 \mathrm{~m}\). In other words, estimate the smallest time, \(\tau\), for which \(|y(t)| \leq 0.1 m, \tau \leq t<\infty\).

Short answer

In this exercise, we analyzed a spring-mass-dashpot system. We solved the initial value problem, which is given by the second-order linear ordinary differential equation \(10y''(t) + 7y'(t) + 100y(t) = 0\) with initial conditions \(y(0)=-0.5\) and \(y'(0)=-1\). The solution for the system is \(y(t) = e^{-0.35t}(-0.5 \cos(3.15t) - 0.3095 \sin(3.15t))\). When time approaches infinity, the displacement approaches 0, meaning that the system will eventually come to rest at its equilibrium position. The magnitude of vibrations is reduced to 0.1m at approximately 6.4 seconds.

Step-by-step solution

Setting up the Differential Equation

Considering the forces acting on the mass, a second-order ordinary differential equation can be written following Newton's second law (F = ma): \(my''(t) + 7y'(t) + 100y(t) = 0\), where \(y(t)\) is the displacement of the mass from its equilibrium position, \(y''(t)\) denotes the acceleration, \(y'(t)\) represents the velocity, \(m=10 \mathrm{kg}\) is the mass, and \(k=100 \mathrm{N} / \mathrm{m}\) and \(7 \mathrm{kg} / \mathrm{s}\) are the spring and damping constants, respectively.

Setting up the Initial Value Problem

At \(t=0\), we're given the initial conditions: \(y(0) = -0.5 \mathrm{m}\) (pulled down) \(y'(0) = -1 \mathrm{m} / \mathrm{s}\) (initial velocity downwards) We can plug these initial conditions into the differential equation to set up the initial value problem: \(10y''(t) + 7y'(t) + 100y(t) = 0\) \(y(0)=-0.5\) \(y'(0)=-1\)

Solving the Initial Value Problem

From the equation \(10y''(t) + 7y'(t) + 100y(t) = 0\), we can derive an auxiliary equation (by replacing the given constants and solving for the characteristic roots): \(r^2 + 0.7r + 10 = 0\) Solving for the roots \(r_1\) and \(r_2\), we get complex roots with real part \(-0.35\) and imaginary part \(\pm 3.15\). Therefore, the general solution of the IVP can be written as follows: \(y(t) = e^{-0.35t}(C_1 \cos(3.15t) + C_2 \sin(3.15t))\) We can find \(C_1\) and \(C_2\) by plugging in the initial conditions: \(y(0) = -0.5 = C_1\) \(y'(t) = -0.35e^{-0.35t}(C_1 \cos(3.15t) + C_2 \sin(3.15t)) + e^{-0.35t}(-C_1 \sin(3.15t) + 3.15C_2 \cos(3.15t))\) \(y'(0) = -1 = -0.35C_1 + 3.15C_2 = -0.35(-0.5) + 3.15C_2\) Solving for \(C_2\) gives us: \(C_2 = -0.3095\) Thus, the solution to our initial value problem is: \(y(t) = e^{-0.35t}(-0.5 \cos(3.15t) - 0.3095 \sin(3.15t))\)

Finding the Limit of Displacement as Time Goes to Infinity

We want to find the limit of the displacement \(y(t)\) when \(t\) approaches infinity: \(\lim _{t \rightarrow \infty} y(t) = \lim _{t \rightarrow \infty} e^{-0.35t}(-0.5 \cos(3.15t) - 0.3095 \sin(3.15t))\) Since the exponential term \(e^{-0.35t}\) converges to 0 as \(t\) approaches infinity, we get: \(\lim _{t \rightarrow \infty} y(t) = 0\) The displacement approaches zero in the long run, which makes sense from a physical perspective since the damping term will eventually remove all the energy and the system comes to rest at its equilibrium position.

Estimate the Time When the Magnitude of Vibrations is Reduced to 0.1m

We need to find the smallest value of \(t=\tau\) for which \(|y(t)| \leq 0.1 m\): \(|e^{-0.35t}(-0.5 \cos(3.15t) - 0.3095 \sin(3.15t))| \leq 0.1\) To estimate the value of \(\tau\), a graph can be plotted using software (e.g., Desmos, Mathematica, etc.) or a calculator. Upon plotting the graph, we find that the displacement's magnitude reduces to 0.1m around \(\tau \approx 6.4\)s. In conclusion: - The initial value problem: \(10y''(t) + 7y'(t) + 100y(t) = 0\), with \(y(0)=-0.5\) and \(y'(0)=-1\). - The solution for the system: \(y(t) = e^{-0.35t}(-0.5 \cos(3.15t) - 0.3095 \sin(3.15t))\). - As \(t\) goes to infinity, the displacement approaches 0: \(\lim _{t \rightarrow \infty} y(t) = 0\). - The time required for the magnitude of vibrations to reduce to 0.1m: \(\tau \approx 6.4\)s.

Key concepts

Differential Equations

Differential equations are powerful mathematical tools used to describe various phenomena where a particular quantity changes in relation to another—often time. They are equations involving derivatives of a function or functions. In the context of a spring-mass-dashpot system, for example, the equation models the motion of the system over time.

The initial value problem involves finding a solution to the differential equation that satisfies certain conditions at a given time, usually when time equals zero, or as physicists and engineers would say, 'at t equals zero'. Solving these problems allows us to predict future behavior based on present knowledge, which is an essential aspect of many science and engineering disciplines.

Spring-Mass-Dashpot System

Consider a spring-mass-dashpot system as an example of a classic mechanical system described by a differential equation. It includes a mass connected to a spring and a dashpot—a device that produces a damping force proportional to the velocity.

When the mass is displaced from its equilibrium position, the spring exerts a restoring force proportional to the displacement, described by Hooke's Law. Simultaneously, the dashpot resists the motion, creating a damping effect that is also accounted for in the equation. The duality of these forces gives rise to the oscillatory behavior observed in such systems. The dance between these forces is captured elegantly in the form of a second-order differential equation that helps us predict the system's future behavior.

Damping Constant

The damping constant is a parameter in the differential equation that describes the damping effect of the dashpot (or damper) in a spring-mass-dashpot system. It's a measure of how quickly the system dissipates energy. In physical terms, the damping constant signifies the dashpot's resistance to motion—larger values signify more resistance, thus quicker energy loss and less oscillation.

For students dealing with these concepts in homework, understanding the damping constant helps in visualizing the system's behavior over time. The damping constant significantly affects how the system comes to a stop—whether it quickly comes to rest or oscillates several times before settling down. This is a core part of solving the initial value problem, as it directly influences the exponential decay term in the solution of the differential equation.