Problem 48 - Advanced Engineering Mathematics

Question

A32-pound weight stretchesa spring 6 inches. The weight moves through a medium offering a damping force numerically equal to \(\beta\) times the instantaneous velocity. Determine the values of \(\beta\) for which the system will exhibit oscillatory motion.

Step-by-Step Solution

Answer

The system will exhibit oscillatory motion if \( \beta < 16 \).

Step 1: Understand Hooke's Law

The spring constant, denoted by \( k \), is given by Hooke's Law which states that the force \( F \) is equal to \( k \times x \), where \( x \) is the displacement. Here, the weight is 32 pounds, and it stretches the spring by 6 inches, so \( k = \frac{32}{6} \times 12 = 64 \) pounds per foot (since we convert inches to feet).

Step 2: Write the Equation of Motion

The motion of the mass-spring-damping system is governed by the differential equation: \( m\ddot{x} + \beta\dot{x} + kx = 0 \). Here, \( m \) is the mass attached to the spring. We have \( m = \frac{32}{32.2} \) slugs since 1 pound-force is the weight of 1 slug at Earth's surface gravity.

Step 3: Specify the Damping Coefficient for Oscillation

For the system to exhibit oscillatory motion, the characteristic equation \( m\lambda^2 + \beta\lambda + k = 0 \) must have complex roots. This requires the discriminant \( \beta^2 - 4mk \) to be negative. Simplifying, we find \( \beta^2 < 4mk \).

Step 4: Calculate with Given Values

Replace \( m = 1 \), \( k = 64 \), then substitute into the inequality: \( \beta^2 < 4 \times 1 \times 64 \). This simplifies to \( \beta^2 < 256 \).

Step 5: Solve for \( \beta \)

Taking the square root of both sides, we get \( \beta < 16 \). Therefore, any value of \( \beta \) less than 16 will result in oscillatory motion.

Key Concepts

Hooke's Law

Hooke's Law lays the foundation for understanding how springs work in a mechanical system. It's all about the relationship between the force exerted on a spring and the displacement of the spring. According to Hooke's Law, the force, denoted as \( F \), is directly proportional to the displacement \( x \), and the formula is expressed as:

• \( F = k \times x \)

where \( k \) is the spring constant. This constant is unique to each spring, representing the stiffness - the higher \( k \), the stiffer the spring.
To put this into perspective, in the given exercise, a weight stretching the spring by 6 inches allows us to calculate \( k \). The weight (32 pounds) and the stretch (6 inches converted to 0.5 feet) give us:

• \( k = \frac{32}{0.5} = 64 \) pounds per foot.

This spring constant value is crucial for further calculations and understanding the system's behavior, particularly in terms of oscillations.

differential equation

In physics, especially when dealing with motion, differential equations help describe how quantities change over time. For a mass-spring-damping system, the equation of motion is a cornerstone.
The specific equation given in this exercise is:

• \( m\ddot{x} + \beta\dot{x} + kx = 0 \)

Here:

• \( m \) is the mass attached to the spring
• \( \ddot{x} \) is the acceleration, the second derivative of position with respect to time
• \( \beta \) is the damping coefficient
• \( \dot{x} \) is the velocity, the first derivative of position

and \( kx \) is the restoring force according to Hooke's Law.
This equation essentially captures the balance between the restoring spring force, resistance from damping, and the inertia of the mass. The challenge often lies in solving it to understand the system's motion under various damping scenarios.

spring constant

The spring constant \( k \) plays a vital role in predicting how a spring will behave under load. It quantifies the stiffness of the spring: a higher \( k \) means the spring is harder to stretch or compress.
To determine the spring constant from experimental data, we use Hooke's Law, where \( k \) is calculated as the force applied divided by the displacement:

• Given in the exercise: \( k = \frac{32}{0.5} = 64 \) pounds per foot.

This value tells us that for each foot of stretch, a force of 64 pounds is required. Understanding \( k \) is crucial, as it influences the angular frequency of oscillations when no damping occurs and even helps evaluate how the system will respond to damping effects.

damping coefficient

The damping coefficient, symbolized as \( \beta \), is central when examining systems like the mass-spring-damping system in the exercise. It signifies the amount of resistance the medium offers against the spring's motion.
In differential equations focused on motion, \( \beta \) directly impacts whether the system exhibits oscillations, no oscillations, or critical damping.
For oscillatory motion in a damped system, the requirement is that the characteristic equation must have complex roots. This condition is met when:

• \( \beta^2 < 4mk \)

Given the system's specifics in the exercise with \( m = 1 \) and \( k = 64 \), this translates to \( \beta < 16 \). Therefore, any damping coefficient value less than 16 yields an oscillatory behavior. Recognizing this threshold is vital for designing systems that harness controlled oscillations.