Problem 27 - Advanced Engineering Mathematics

Question

A mass weighing 10 pounds stretches a spring 2 feet. The mass is attached to a dashpot damping device that offers a damping force numerically equal to \(\beta(\beta>0)\) times the instantaneous velocity. Determine the values of the damping constant \(\beta\) so that the subsequent motion is (a) overdamped, (b) critically damped, and (c) underdamped.

Step-by-Step Solution

Answer

(a) \(\beta > 1.984\), (b) \(\beta = 1.984\), (c) \(\beta < 1.984\).

Step 1: Calculate the spring constant

Using Hooke's Law, we find the spring constant \(k\). The weight (force) of the mass is 10 pounds and it stretches the spring 2 feet. Hence, Hooke's Law \(F = kx\) becomes \(10 = k \times 2\). Solving for \(k\), we get \(k = 5\ \text{lb/ft}\).

Step 2: Find the mass in slugs

Convert the weight of the mass to slugs (the mass unit in the imperial system). \(\text{mass} = \frac{10}{32.2} \approx 0.3106\) slugs, since gravitational acceleration \(g = 32.2\ \text{ft/s}^2\).

Step 3: Set up the damping equation

The motion of the damped spring is governed by the second-order differential equation: \(m \frac{d^2x}{dt^2} + \beta \frac{dx}{dt} + kx = 0\). Here, \(m = 0.3106\) slugs and \(k = 5\ \text{lb/ft}\).

Step 4: Condition for Overdamping

For the system to be overdamped, the discriminant of the characteristic equation \(m r^2 + \beta r + k = 0\) must be positive. Thus, \(\beta^2 - 4mk > 0\). Substituting \(m = 0.3106\) and \(k = 5\), it becomes \(\beta > 2\sqrt{5 \times 0.3106} = 1.984\).

Step 5: Condition for Critical Damping

Critical damping occurs when the discriminant of the characteristic equation equals zero: \(\beta^2 = 4mk\). Using \(m = 0.3106\) and \(k = 5\), solve \(\beta = 2\sqrt{5 \times 0.3106} \approx 1.984\).

Step 6: Condition for Underdamping

For underdamping, the discriminant of the characteristic equation must be negative, meaning \(\beta^2 - 4mk < 0\). Using \(\beta < 1.984\).

Key Concepts

Overdamped

In a physical system, when we talk about something being **overdamped**, we refer to a scenario where the damping force is so strong that it causes the system to return to its equilibrium position very sluggishly. Imagine you are trying to push a heavy door that closes on its own; if the door takes a long time to close completely because of resistance, it can be considered overdamped.
In our exercise, to have an overdamped motion, the damping constant \( \beta \) must be large enough. To determine the condition for overdamping mathematically, we use the characteristic equation of the form \( m r^2 + \beta r + k = 0 \). For overdamping, the discriminant of this equation should be positive, i.e., \( \beta^2 - 4mk > 0 \). This ensures that the system has two distinct real roots, leading to a slow non-oscillatory return to equilibrium.
For the given mass and spring system, with \( m = 0.3106 \) slugs and \( k = 5 \) lb/ft, the value of \( \beta \) must be greater than approximately 1.984. When \( \beta > 1.984 \), the damping is sufficient to prevent oscillations, and the system returns slowly to its original state.

Critically Damped

A critically damped system represents a sweet spot in damping scenarios. This is the case where the system returns to equilibrium as quickly as possible without oscillating. It is the ideal situation in many mechanical and electronic systems where we want a fast stabilization without overshooting the target.
In our spring-mass-damper system, critical damping happens when the discriminant of the characteristic equation equals zero. Mathematically, this means \( \beta^2 = 4mk \). This results in a double real root for the characteristic equation, allowing the system to rapidly settle to its equilibrium without oscillations. In practical terms, picture a door that closes quickly but does not slam or bounce back upon closure.
For our specific calculations, with \( m = 0.3106 \) slugs and \( k = 5 \) lb/ft, the damping constant \( \beta \) needs to be exactly 1.984 to achieve this critically damped state. Balancing just right, this ensures swift damped motion without any unwanted vibrations.

Underdamped

When a system is **underdamped**, it means that the damping force is not strong enough to prevent the system from oscillating. This can be useful in systems where some level of oscillation is acceptable or desirable. A typical example might be a car's suspension, where a bit of bounce is needed for a smooth ride on uneven roads.
For underdamping to occur, the damping constant \( \beta \) should be less than the value that leads to critical damping. Specifically, we need the discriminant of the characteristic equation \( \beta^2 - 4mk \) to be negative, resulting in complex roots that signify oscillatory behavior.
In our modeled exercise, with a mass of 0.3106 slugs and spring constant of 5 lb/ft, \( \beta \) must be less than 1.984 to have underdamped motion. This underdamping allows the system to oscillate with diminishing amplitude over time, creating a smooth glide back to equilibrium as it slowly stops.