Question

A mass weighing 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in the positive direction from its equilibrium position and released with no initial velocity. Assuming that there is no damping and that the mass is acted on by an external force of \(2 \cos 3 t\) lb, formulate the initial value problem describing the motion of the mass.

Short answer

Solution: The IVP representing the motion of the mass is x''(t) + 256x(t) = (1/2)cos(3t) with the initial conditions x(0) = 2/12 and x'(0) = 0.

Step-by-step solution

Determine the spring constant, k

We'll use Hooke's law: F = -kx where F is the force, k is the spring constant, and x is the displacement. We know that the mass (m) weighs 4 lb and stretches the spring by 1.5/12 ft (positive sign because it's a stretch). Therefore, F = mg = 4(32) lb = 128 lb. Solving for k using Hooke's law: 128 = -k(1.5/12) => k = -1024 lb/ft

Use Newton's second law

Using Newton's second law of motion, F = ma. In this case, F = -kx + F_ext, where F_ext is the external force acting on the mass. Thus, we can write the equation as: -kx + F_ext = m * x'' (double prime denotes the second derivative of x). Now substitute the given values and external force into the equation: -((-1024) x) + 2cos(3t) = 4 * x''

Write the initial value problem

We have the following information and conditions: - The mass is initially displaced by x(0)=2/12 ft - The mass has no initial velocity, which means x'(0)=0 Now, we can write the initial value problem as: IVP: x''(t) + 256x(t) = (1/2)cos(3t) Initial conditions: x(0) = 2/12, x'(0)=0

Key concepts

Hooke's Law

Understanding Hooke's Law is fundamental when studying mechanical systems involving springs. Simply put, Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position; mathematically, this is expressed as \( F = -kx \). The negative sign indicates that the spring force is restoring, meaning it acts in the opposite direction to the displacement.

Consider a spring with one end attached to a fixed point and the other to a mass. If we pull the mass and stretch the spring, Hooke’s Law becomes evident as the spring exerts a force to return to its original length. This is the principle behind simple harmonic motion in undamped systems without external forces. Hooke’s Law is the stepping stone to understanding more complex dynamic situations that involve springs, such as the exercise given where external forces are also considered.

Newton's Second Law of Motion

Newton's Second Law of Motion is a cornerstone of classical mechanics. This law tells us how an object will change its motion in response to forces acting upon it. The law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration \( F=ma \).

In context of the given exercise, we use this law to describe the behavior of a mass-spring system subject to both spring force and an external force. Understanding this relationship is crucial to formulating the differential equations that model the system's motion. Newton's second law helps us to marry the concept of force derived from Hooke's law to the acceleration acting on the mass, setting the stage for writing the complete equation of motion.

Differential Equations

Differential equations are mathematical tools used to model relationships involving rates of change. In physics and engineering, they are vital for describing how physical systems evolve over time. The given problem uses a second-order differential equation, meaning it includes a term for the second derivative \(x''(t)\), which represents acceleration.

Setting up a differential equation requires translating physical laws—like Hooke's Law and Newton's Second Law—into mathematical language. This equation then details how the system progresses from any given state when considering initial values, such as initial displacement and velocity, leading to what we call an 'initial value problem'. Solutions to these problems give us precise insights into the behavior of dynamic systems in motion.

Spring Constant

The spring constant, denoted as \(k\), is a measure of a spring's stiffness. It appears in both Hooke's Law \( F = -kx \) and the system's differential equation. A larger \(k\) means a stiffer spring that requires more force to stretch or compress a given distance. Conversely, a smaller \(k\) indicates a more flexible spring.

The value of \(k\) is critical in determining the motion of the mass attached to the spring. In our exercise, by knowing the force exerted by the weight of a mass and the resultant displacement, we can calculate the spring constant. This, in turn, tells us how 'responsive' the spring is to forces and influences the amplitude and period of oscillation in the motion of the mass.

Undamped Motion

Undamped motion refers to oscillatory motion that does not lose energy over time. In reality, systems often experience some form of resistance, like friction or air resistance, which dissipates energy and is known as damping. However, in an ideal undamped system, such as the one described in the exercise, the energy loss is assumed to be nonexistent.

The motion of a mass connected to an undamped spring is a perfect example of undamped harmonic motion, where the mass will continue to oscillate indefinitely. When an external force is applied, as in the problem at hand, the system is subject to 'forced' motion, but the lack of damping still means the oscillations do not diminish in amplitude over time, leading to a perpetual motion that is uniquely described by the initial value problem derived from the system's parameters.