Problem 87 - Calculus 3

Question

A mass weighing 2 lb stretches a spring 2 ft. Find the equation of motion if the spring is released from 2 in. below the equilibrium position with an upward velocity of 8 ft/sec. What is the period and frequency of the motion?

Step-by-Step Solution

Answer

The equation of motion is \( x(t) = \frac{-1}{6}\cos(4t) + 2\sin(4t) \). The period is approximately 1.57 seconds, and the frequency is 0.6366 Hz.

Step 1: Determine the Spring Constant

To find the spring constant (k), use Hooke's Law in the static case: the weight stretches the spring by a certain amount. Since 2 lb stretches the spring by 2 ft, we use the formula:\[ F = kx \]where \( F = 2 \text{ lb} \), and \( x = 2 \text{ ft} \). Solving for \( k \):\[ k = \frac{F}{x} = \frac{2}{2} = 1 \text{ lb/ft} \]

Step 2: Convert Initial Conditions

The initial displacement of 2 inches must be converted to feet as the rest of our units are in feet. Thus, 2 inches is equivalent to \( 2 / 12 = 1/6 \text{ ft} \).

Step 3: Model as a Differential Equation

The motion of the spring-mass system can be modeled by the differential equation:\[ m\ddot{x} + kx = 0 \]where \( m \) is the mass in slugs (where 1 slug = 32.2 lb*ft/s²), and \( k = 1 \text{ lb/ft} \). First, convert the weight to mass in slugs:\[ m = \frac{2 \text{ lb}}{32.2 \text{ ft/s}^2} \approx 0.0621 \text{ slugs} \]Thus, the equation becomes:\[ 0.0621\ddot{x} + x = 0 \]

Step 4: Solve the Differential Equation

Assuming a solution of the form \( x(t) = A\cos(\omega t) + B\sin(\omega t) \), substitute and solve for \( \omega \):\[ \omega^2 = \frac{k}{m} = \frac{1}{0.0621} \approx 16.1 \]Thus, \( \omega = \sqrt{16.1} \approx 4 \).

Step 5: Apply Initial Conditions

The initial position is \( x(0) = -1/6 \text{ ft} \) and initial velocity is \( v(0) = 8 \text{ ft/s} \). Thus:1. Using \( x(0) = A = -1/6 \)2. Differentiate \( x(t) \), \( \dot{x}(t) = -4A\sin(4t) + 4B\cos(4t) \), apply \( v(0) = 8 \): \[ 4B = 8 \rightarrow B = 2 \]So the solution is:\[ x(t) = \frac{-1}{6}\cos(4t) + 2\sin(4t) \]

Step 6: Find Period and Frequency

The frequency \( f \) is given by \( f = \frac{\omega}{2\pi} \), and the period \( T = \frac{1}{f} \). With \( \omega = 4 \):\[ f = \frac{4}{2\pi} \approx 0.6366 \text{ Hz} \]\[ T = \frac{1}{0.6366} \approx 1.57 \text{ seconds} \]

Key Concepts

Spring-Mass System

A spring-mass system is a classic model used in physics and engineering to understand how objects move when subjected to a restoring force, like that of a spring. In this model, a mass is attached to a spring, which can stretch or compress. The motion of the system depends on several key factors:

• The stiffness of the spring (often measured by the spring constant, k), which determines how much force the spring exerts when it is stretched or compressed.
• The mass of the object attached to the spring, which affects how much the object will accelerate under the force of the spring's tension.
• External forces, like gravity, that might affect the system.

This system is governed by differential equations, specifically Hooke's Law for linear springs, as they describe the relationship between force, displacement, and the spring constant.

Hooke's Law

Hooke's Law is a principle of physics that describes how springs work. It states that the force exerted by a spring is directly proportional to the amount it is displaced from its equilibrium position. Mathematically, it is expressed as:\[ F = kx \] where:

• \( F \) is the force exerted by the spring,
• \( k \) is the spring constant, which measures the stiffness of the spring, and
• \( x \) is the displacement from the equilibrium position.

In our spring-mass system example, a 2 lb force stretches the spring 2 feet. Using Hooke's Law, we can determine that the spring constant \( k \) is 1 \( \text{lb/ft} \). This calculation is crucial as it forms the basis for modeling the motion of the mass attached to the spring using differential equations.

Initial Conditions

Initial conditions refer to the starting state of the system before it begins to move. In differential equations, these are necessary for determining the specific behavior of a system over time. For our spring-mass system:

• The initial position is the displacement of the mass from its equilibrium state, which is given as 2 inches below the equilibrium. Since computation is done in feet, this converts to \( \frac{1}{6} \) ft.
• The initial velocity is also crucial, here provided as an upward velocity of 8 ft/s.

These initial values allow us to solve the differential equation by substituting them into the solution for the highest specificity. They ensure that we can accurately sketch the path of the mass over time, giving us the complete motion equation.

Periodic Motion

Periodic motion refers to the repeated and continuous movement of an object along a path in a regular interval of time. In the context of spring-mass systems, this motion is characterized by oscillations back and forth around an equilibrium position.The described motion of our spring-mass system is: \[ x(t) = \frac{-1}{6}\cos(4t) + 2\sin(4t) \] This equation shows that the motion is regular and predictable. It repeats at a consistent frequency and period, both of which are essential to understanding the system's behavior.

• Frequency (\( f \)): This is how often the motion repeats itself in a second. In our case, the frequency is approximately 0.637 Hz.
• Period (\( T \)): This is the time it takes for one complete cycle of motion. For this system, the period is approximately 1.57 seconds.

Understanding periodic motion helps us predict the future position and velocity of the mass, making it a vital concept in the analysis of harmonic oscillators.