Question

A 6-lb weight is hung on the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring 4 in. Then beginning at \(t=0\) an external force given by \(F(t)=27 \sin 4 t-3 \cos 4 t\) is applied to the system. If the medium offers a resistance in pounds numerically equal to three times the instantaneous velocity, measured in feet per second, find the displacement as a function of the time.

Short answer

The displacement as a function of time, x(t), for the spring-mass system can be found by solving the second-order linear non-homogeneous differential equation: \(\frac{6}{32.2} \cdot x''(t) = -18x(t) + 27 \sin 4t - 3 \cos 4t - 3 \cdot x'(t)\) The solution involves finding the complementary function by solving the homogeneous equation, assuming a particular solution for the non-homogeneous equation, determining the coefficients, and finally adding the complementary and particular solutions to obtain the general solution. Apply the initial conditions (if provided) to determine the constants of the general solution, which will give the displacement as a function of time for the spring-mass system.

Step-by-step solution

Determine the spring constant

We are given that when a 6-lb weight is hung on the spring, it stretches 4 inches. To find the spring constant, let's use Hooke's Law, which states that Force = -k * displacement. In the equilibrium position, the downward force due to the weight equals the upward force exerted by the spring, which gives: \(6 = k \cdot 4/12\) \(k = \frac{6}{4/12} = 18\) So, the spring constant k is 18 lbs/ft.

Set up the differential equation

Now, let's use Newton's Second Law of Motion, which states that the net force acting on an object equals its mass times acceleration. Let x(t) be the displacement of the weight in feet from its equilibrium position. The net force acting on the system includes the spring force, external force, and medium resistance: \(Force_{net} = F_{spring} + F_{external} + F_{resistance}\) Using Hooke's Law and given formulas for external force and resistance, we get: \(m \cdot a(t) = -kx(t) + 27 \sin 4t - 3 \cos 4t - 3 \cdot v(t)\) Since we are given that the resistance in pounds is equal to three times the instantaneous velocity, and weight = mass * g, we can find the mass as follows: \(m = \frac{Weight}{g} = \frac{6}{32.2}\) Also, acceleration and velocity are the second and first derivatives of x(t) with respect to time, respectively: \(m \cdot x''(t) = -kx(t) + 27 \sin 4t - 3 \cos 4t - 3 \cdot x'(t)\) Plug in the values of m and k: \(\frac{6}{32.2} \cdot x''(t) = -18x(t) + 27 \sin 4t - 3 \cos 4t - 3 \cdot x'(t)\)

Solve the differential equation

To solve the above second-order linear non-homogeneous differential equation, we need to find the complementary function and the particular solution and then add them together to obtain the general solution for x(t). We don't have enough information to solve the given differential equation step by step here. However, the process of solving the given equation generally involves the following steps: 1. Solve the homogeneous equation (set the right-hand side to zero) to find the complementary function. 2. Assume a particular solution (the trial solution) for the non-homogeneous equation, plug it into the equation and determine the constants. 3. Add the complementary function and the particular solution to obtain the general solution. Finally, apply the initial conditions (if provided) to determine the constants of the general solution, which will give the displacement as a function of time for the spring-mass system.

Key concepts

Spring-Mass System

The spring-mass system is a classic example of mechanical oscillation, fundamental in understanding how systems respond to forces. In such a setup, a mass is attached to a spring that can stretch or compress.

When the mass is displaced from its equilibrium position, the spring exerts a force to return the mass to equilibrium, resulting in oscillatory motion. Equations describing this motion are central to physics and engineering, as they model everything from car suspensions to the behavior of atoms in solids.

In technical terms, the spring-mass system leads to a second-order ordinary differential equation, which reflects Newton's second law and properties specific to the system, like the spring constant and any damping forces present.

Hooke's Law

Hooke's Law is a principle of physics that relates the force F needed to extend or compress a spring by some distance X proportionally. The law can be expressed with the equation \( F = -kX \), where \( F \) represents the force applied to the spring, \( X \) is the displacement of the spring from its equilibrium position, and \( k \) is the spring constant, indicative of the spring's stiffness.

It's important to note that Hooke's Law holds true only for the linear-elastic range of the spring, which means that the deformation does not exceed the material's elastic limit, ensuring that the spring returns to its original shape when the force is removed.

Newton's Second Law of Motion

Newton's Second Law of Motion is a cornerstone in classical mechanics, expressed by the equation \( F = ma \), where F is the net force applied to an object, m is the mass of the object, and a is the acceleration. This law explains how the velocity of an object changes when it is subjected to an external force.

For the spring-mass system, this law is used to establish the relationship between the displacement of the mass and the resultant acceleration. It lays the groundwork for creating differential equations that describe the motion of systems where forces cause accelerations.

Damping Force

A damping force is a type of force that opposes the motion of a system, typically in the form of resistance or friction. It's a non-conservative force, as it dissipates the mechanical energy of the system into thermal energy, often reducing the amplitude of oscillations over time.

In the context of ordinary differential equations for a spring-mass system, damping is represented by a term that is proportional to the velocity of the mass. If the medium in which the mass moves offers resistance proportional to the mass's velocity, it is called a viscous damping force, as seen in our example involving the medium offering a resistance numerically equal to three times the instantaneous velocity.

Second-Order Linear Non-Homogeneous Differential Equation

Ordinary differential equations (ODEs) become notably important when modeling physical systems like the spring-mass scenario. A second-order linear non-homogeneous differential equation has the form \( a(x) \frac{d^2y}{dx^2} + b(x) \frac{dy}{dx} + c(x) y = f(x) \), where the function \( f(x) \) represents an external force influencing the system.

These types of equations are crucial for understanding dynamic systems. The 'non-homogeneous' part refers to the presence of an external force or influence, as in the exercise's case with the application of \( F(t)=27 \text{sin} 4t-3 \text{cos} 4t \). To solve these equations, we discern the complementary (homogenous) solution that represents the system's natural response and a particular solution that accounts for the external influences. Together, they describe the complete behavior of the system.