Question

A 10-kg object suspended from the end of a vertically hanging spring stretches the spring $9.8\mathrm{cm}$. At time $t=0$, the resulting spring-mass system is disturbed from its rest state by the given applied force, $F(t)$. The force $F(t)$ is expressed in newtons and is positive in the downward direction; time is measured in seconds. (a) Determine the spring constant, $k$. (b) Formulate and solve the initial value problem for $y(t)$, where $y(t)$ is the displacement of the object from its equilibrium rest state, measured positive in the downward direction. (c) Plot the solution and determine the maximum excursion from equilibrium made by the object on the $t$-interval $0\le t<\mathrm{\infty }$ or state that there is no such maximum. $$F(t)=20e^{-t}$$

Short answer

Answer: The displacement function y(t) for the spring-mass system is given by: $$y(t)=-\frac{2}{101}\cos (10t)-\frac{1}{505}\sin (10t)+\frac{2}{101}{e}^{-t}$$

Step-by-step solution

1. Calculate the weight of the object and equilibrium length of the spring

First, find the force exerted on the spring by the object due to gravity, using the formula $F_{gravity}=(mass)(gravity)$. The mass of the object is given as $10kg$ and the acceleration due to gravity $g$ is approximately $9.8m/s^2$. Calculate the weight: $$F_{gravity}=(10)(9.8)=98N$$ Using Hooke's law, $F_{spring}=kx$, where $x=9.8cm$, we can solve for the spring constant $k$.

2. Find the spring constant k using Hooke's Law

At equilibrium, the force exerted by the spring is equal to the weight of the object, so: $$kx=F_{gravity}$$ The spring stretches $9.8cm=0.098m$, so solving for $k$: $$k(0.098)=98$$ $$k=\frac{98}{0.098}$$ $$k=1000N/m$$ So the spring constant, $k$, is $1000N/m$. #b) Formulate and solve the initial value problem for y(t), where y(t) is the displacement of the object from its equilibrium rest state#

1. Write the equation of motion for the spring-mass system

The equation of motion for the spring-mass system is given by: $$m{y}^{″}(t)+ky(t)=F(t)$$ where $m$ is the mass of the object, $k$ is the spring constant, and $F(t)$ is the external force acting on the system. We are given: $m=10kg$, $k=1000N/m$, and $F(t)=20e^{-t}$ Plug in the values into the equation to get: $$10y^{″}(t)+1000y(t)=20e^{-t}$$

2. Solve the homogeneous equation and find complementary function

The homogeneous equation for this system is: $$10y^{″}(t)+1000y(t)=0$$ This is a second-order linear homogeneous equation with constant coefficients. The characteristic equation for this differential equation is: $$10r^2+1000=0$$ Solving for $r$: $$r^2=-100$$ $$r=\pm 10i$$ The complementary function is: $$y_c(t)=C_1\cos (10t)+C_2\sin (10t)$$ where $C_1$ and $C_2$ are constants to be determined by initial conditions.

3. Find the particular integral

To find the particular integral, let's use the guess method. The force function, $F(t)=20e^{-t}$, is an exponential function, so our guess for the particular integral will be: $$y_p(t)=A{e}^{-t}$$ Differentiate $y_p(t)$: $$y_p^{\prime }(t)=-A{e}^{-t}$$ $$y_p^{″}(t)=A{e}^{-t}$$ Substitute these into the equation for the non-homogeneous system: $$10(A{e}^{-t})+1000(A{e}^{-t})=20e^{-t}$$ Simplify and solve for A: $$1010A{e}^{-t}=20e^{-t}$$ $$A=\frac{20}{1010}=\frac{2}{101}$$ So the particular integral is: $$y_p(t)=\frac{2}{101}{e}^{-t}$$

4. Write the general solution and apply initial conditions

The general solution for the system is the sum of the complementary function and the particular integral: $$y(t)=y_c(t)+y_p(t)$$ $$y(t)=C_1\cos (10t)+C_2\sin (10t)+\frac{2}{101}{e}^{-t}$$ At $t=0$, the object is at rest, so $y(0)=0$ and $y^{\prime }(0)=0$. Apply these initial conditions to find $C_1$ and $C_2$: $$y(0)=C_1\cos (0)+C_2\sin (0)+\frac{2}{101}{e}^0=0$$ $$C_1=-\frac{2}{101}$$ Since the object is at rest, $y^{\prime }(t)=0$: $$y^{\prime }(0)=-10C_1\sin (0)+10C_2\cos (0)-\frac{2}{101}{e}^0=0$$ $$C_2=-\frac{2}{1010}=-\frac{1}{505}$$ The displacement function for the spring-mass system is: $$y(t)=-\frac{2}{101}\cos (10t)-\frac{1}{505}\sin (10t)+\frac{2}{101}{e}^{-t}$$ #c) Plot the solution and determine the maximum excursion from equilibrium#

1. Determine maximum excursion

Since we do not have a critical damping or oscillations decreasing with time, the object will oscillate indefinitely and there is no maximum excursion from equilibrium. In other words, we have to state that there is no such maximum for the given interval $0\le t<\mathrm{\infty }$.

Key concepts

Spring Constant Calculation

Understanding the spring constant is essential for analyzing the behavior of springs in various mechanical systems. It reflects the stiffness of a spring and the amount it resists deformation under load. In the context of a spring-mass system, as seen in the given exercise, the spring constant calculation is performed using Hooke's Law.

Hooke's Law states that the force exerted by a spring, denoted as $F_{spring}$, is directly proportional to the displacement $x$ from its equilibrium position, with the proportionality constant being the spring constant $k$. Mathematically, this is expressed as $F_{spring}=kx$. When a 10-kg mass stretches a spring by 9.8 cm, we use the weight of the object as the force $F_{gravity}=mg$, where $g$ is the acceleration due to gravity, to calculate $k$. Since at equilibrium, $F_{gravity}=F_{spring}$, we can solve for $k$ using the formula $k=\frac{F_{gravity}}{x}$. In this case, the spring constant is found to be 1000 N/m, indicating the amount of force needed to stretch or compress the spring by one meter.

Initial Value Problem

The initial value problem in differential equations is a specific scenario where the solution to a differential equation is sought under given conditions at a starting time $t=0$. In such problems, we know both the equation governing the system's behavior and the initial conditions.

To solve the initial value problem for a spring-mass system like our exercise, we establish the differential equation that describes the motion of the mass attached to the spring: $m{y}^{″}(t)+ky(t)=F(t)$, where $m$ is the mass, $y(t)$ is the displacement from equilibrium, $k$ is the spring constant, and $F(t)$ is the external force. After formulating this equation, the steps involve finding a complementary solution to the associated homogeneous equation and a particular integral for the non-homogeneous part. By imposing initial conditions, such as the starting position and velocity, we can determine any arbitrary constants to obtain the specific solution that describes the motion of the spring-mass system over time.

Equilibrium Solution

The equilibrium solution for a system, in general, represents a state where all forces balance, and there is no net change in the system's motion. For the spring-mass system, equilibrium is attained when the object is at rest, and the force of gravity is balanced by the spring's restoring force.

In our example, the mass creates a gravitational force $F_{gravity}=mg$ which is balanced by the spring's force $F_{spring}=kx$, leading to the equilibrium position $x$. The equilibrium solution is essential to understand since it serves as a baseline around which the system oscillates or returns to when disturbed. It also provides critical information for determining initial conditions, which are used to find the constants in our general solution of the system's differential equation. Recognizing that the equilibrium position is the point of zero displacement from rest informs our initial value problem by setting the starting point for the object's motion.

Damping and Oscillation

Damping and oscillation are two concepts crucial to the dynamics of a spring-mass system. Oscillation refers to the periodic motion around the equilibrium position, characterized by the mass moving back and forth due to the spring's restoring force. Damping, on the other hand, is an influence within or upon an oscillatory system that has the effect of reducing, restricting or preventing its oscillations.

In cases without damping, like the system presented in the exercise, the mass will oscillate indefinitely with a constant amplitude, as it has no mechanism to lose energy. The system's motion is described by a sinusoidal function for the complementary solution of the differential equation. However, in real-world applications, damping is often present due to friction or other resistive forces, causing the amplitude of oscillations to decrease over time until the object comes to rest. Understanding both concepts allows for a comprehensive analysis of how the system behaves over time, whether it will continue oscillating indefinitely, as with the undamped system in our example, or eventually come to a halt due to damping.