Problem 92 - Calculus 3

Question

A 1-kg mass is attached to a vertical spring with a spring constant of 21 N/m. The resistance in the spring-mass system is equal to 10 times the instantaneous velocity of the mass. a. Find the equation of motion if the mass is released from a position 2 m below its equilibrium position with a downward velocity of 2 m/sec. b. Graph the solution and determine whether the motion is over damped, critically damped, or under damped.

Step-by-Step Solution

Answer

The motion is overdamped. The equation of motion is \( x(t) = e^{-7t} - 3e^{-3t} \).

Step 1: Understanding the Problem

This exercise involves finding the equation of motion for a damped spring-mass system. Given the mass, spring constant, and damping coefficient, we need to determine the type of damping based on the initial conditions provided. The initial conditions include the initial position and initial velocity.

Step 2: Write the Differential Equation

The spring constant is 21 N/m, and the damping coefficient is 10. The equation of motion is given by\[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \]where \( m = 1 \) kg, \( c = 10 \), \( k = 21 \). Simplifying gives:\[ \frac{d^2x}{dt^2} + 10 \frac{dx}{dt} + 21x = 0 \]

Step 3: Identify Type of Damping

To identify the type of damping, calculate the discriminant in the characteristic equation. The characteristic equation for \( \frac{d^2x}{dt^2} + 10 \frac{dx}{dt} + 21x = 0 \) is:\[ r^2 + 10r + 21 = 0 \]The discriminant \( \Delta = b^2 - 4ac = 10^2 - 4 \cdot 1 \cdot 21 = 100 - 84 = 16 \). Since \( \Delta > 0 \), the system is overdamped.

Step 4: Solve the Characteristic Equation

To find the roots of the characteristic equation:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{16}}{2} \]This gives roots \( r_1 = -7 \) and \( r_2 = -3 \). Thus, the general solution is:\[ x(t) = C_1 e^{-7t} + C_2 e^{-3t} \]

Step 5: Apply Initial Conditions

Using the initial conditions \( x(0) = -2 \) and \( \frac{dx}{dt}(0) = 2 \):1. Substitute \( t=0 \) into the general solution: \( -2 = C_1 + C_2 \)2. Differentiate the general solution, \( \frac{dx}{dt} = -7C_1 e^{-7t} - 3C_2 e^{-3t} \), and substitute \( t=0 \): \( 2 = -7C_1 - 3C_2 \)Solving the system of equations:- From \( C_1 + C_2 = -2 \)- Substitute to get \( C_1 = 1 \), \( C_2 = -3 \) gives:Final solution: \( x(t) = e^{-7t} - 3e^{-3t} \)

Step 6: Analyze Graph of Solution

Graph \( x(t) = e^{-7t} - 3e^{-3t} \) to observe the motion, confirming the overdamped characteristic as solution approaches equilibrium without oscillating.

Step 7: Determine Damping Type

With the discriminant \( \Delta > 0 \), the system is overdamped, which aligns with the graph of the solution showing no oscillations.

Key Concepts

Equation of Motion

In the realm of physics, an equation of motion is crucial for understanding how objects behave over time. Specifically, in a damped harmonic oscillator, the equation describes the position of a mass as a function of time, considering external forces like damping or resistance.

In the context of our spring-mass system, the equation of motion comes from Newton's second law, defined through a differential equation. We have a mass (\( m \)), a damping coefficient (\( c \)), and a spring constant (\( k \)). They relate through the differential equation: \[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \]

This equation embodies the essence of a damped oscillator. Each term represents a different factor affecting movement:

• Acceleration (\( \frac{d^2x}{dt^2} \)): Caused by the mass.
• Velocity (\( \frac{dx}{dt} \)): Damped by resistance.
• Displacement (\( x \)): Restoring force due to the spring.

Through this equation, the complex motion of the system is captured, allowing engineers and physicists to predict how the system reacts to various forces.

Spring-Mass System

A spring-mass system is a classic model used to study the principles of motion and force in physics. It entails a mass attached to a spring, like a weight bobbing up and down due to its elasticity.

Key elements characterize this system:

• Mass (\( m \)): The object that moves, whose weight is central in calculating forces.
• Spring Constant (\( k \)): Measures the spring's stiffness. The larger the constant, the more force required to stretch the spring. In our example, \( k = 21 \) N/m.
• Damping (\( c \)): Resistance, often proportional to velocity, that slows down motion. Here, the damping coefficient is \( c = 10 \).

Understanding these components allows us to model the dynamic behavior of the system. When analyzing a spring-mass system, attention is paid to initial conditions, such as displacement and velocity, which significantly affect the system's motion over time. This simplistic yet powerful model explains natural phenomena like seismic activities and is foundational in engineering for designing various mechanical systems.

Differential Equations

Differential equations are mathematical equations that involve derivatives, encapsulating how a quantity changes over time. In physics, they are fundamental for modeling dynamic systems, such as our spring-mass system.

The equation governing the spring-mass system is a second-order linear differential equation: \[ \frac{d^2x}{dt^2} + 10 \frac{dx}{dt} + 21x = 0 \]

Here, the derivatives \( \frac{d^2x}{dt^2} \) and \( \frac{dx}{dt} \) signify the acceleration and velocity, respectively. These differential equations are crucial for predicting how physical systems will evolve over time. In our problem:

• Solutions to this equation describe how the position of the mass changes.
• The characteristic equation derived helps determine the motion type, whether it is over damped, under damped, or critically damped.
• Solving it requires initial conditions for a specific solution, here given as initial displacement and velocity.

By solving these equations, one can model the behavior of various real-world systems, from simple mechanical devices to complex biological processes.