Question

A spring and dashpot system is to be designed for a \(32-1 b\) weight so that the overall system is critically damped. (a) How must the damping constant \(\gamma\) and spring constant \(k\) be related? (b) Assume the system is to be designed so that the mass, when given an initial velocity of \(4 \mathrm{ft} / \mathrm{sec}\) from its rest position, will have a maximum displacement of 6 in. What values of damping constant \(\gamma\) and spring constant \(k\) are required?

Short answer

The relationship between damping constant γ and spring constant k for a critically damped spring and dashpot system is given by: $$k = \frac{\gamma^2}{4m}$$ For the given initial conditions (32 lb weight, 4 ft/s velocity, and maximum displacement of 6 inches), the damping constant γ and spring constant k are calculated to be: $$\gamma = 4e~\frac{ft}{s}$$ $$k = 4e^2~\frac{ft}{lb}$$

Step-by-step solution

Write down the equation of motion for a mass-spring-damper system (mass \(m\), damping constant \(\gamma\), and spring constant \(k\))

The equation of motion for a mass-spring-damper system can be written as: $$m\frac{d^2x}{dt^2} + \gamma\frac{dx}{dt} + kx = 0$$ where \(x(t)\) is the displacement of the mass as a function of time \(t\).

Determine the relationship between \(\gamma\) and \(k\) for the critically damped system.

A generally damped system has the characteristic equation: $$ms^2 + \gamma s + k = 0$$ For a critically damped system, the discriminant of this quadratic equation must be equal to zero: $$\Delta = \gamma^2 - 4mk = 0$$ Solving this equation for the relationship between the damping constant \(\gamma\) and the spring constant \(k\), we get: $$k = \frac{\gamma^2}{4m}$$

Convert weight to mass

The given weight is \(32~lb\). We need to convert it to mass for our calculations. The weight \(W\) is related to mass \(m\) by the equation: $$W = mg$$ Where \(g\) is the acceleration due to gravity which is approximately \(32.2~\frac{ft}{s^2}\) for our case. Hence, the mass \(m\) of the object can be calculated as: $$m = \frac{W}{g} = \frac{32}{32.2} = 1~slugs$$

Write down the initial conditions

We are given that the object has an initial velocity of \(4~\frac{ft}{s}\) and has a maximum displacement of \(6~in\) or \(0.5~ft\). We can express these initial conditions mathematically as follows: Initial velocity (\(v_0\)): \(v(0) = \frac{dx}{dt}(0) = 4~\frac{ft}{s}\) Maximum displacement (\(x_{max}\)): \(x_{max} = 0.5~ft\)

Use the given initial conditions to find the values of \(\gamma\) and \(k\)

For a critically damped system, the displacement function can be written as: $$x(t) = (A+Bt)e^{-\frac{\gamma t}{2m}}$$ Where \(A\) and \(B\) are constants. Using the initial conditions above, we have: 1. At \(t=0\), \(x_{max} = (A+B(0))e^{-\frac{\gamma(0)}{2m}} \Rightarrow A = 0.5~ft\) 2. At \(t=0\), \(\frac{dx}{dt} = 4 = (A+Bt)e^{-\frac{\gamma t}{2m}} \Rightarrow B = 8~\frac{ft}{s}\) Now we have the displacement function for the critically damped system: $$x(t) = (0.5+8t)e^{-\frac{\gamma t}{2m}}$$ To find \(\gamma\) and \(k\), we first find the value of \(\gamma\) by taking the second derivative of \(x(t)\) and equating it to zero to get the maximum displacement. Taking the second derivative, we get: $$\frac{d^2x}{dt^2} = 8e^{-\frac{\gamma t}{2m}}(1-\frac{\gamma^2 t^2}{4m^2})$$ Setting this equal to zero, we can find the value of \(t\) (let's call it \(t_{max}\)) when the maximum displacement occurs: $$1-\frac{\gamma^2 t_{max}^2}{4m^2} = 0 \Rightarrow t_{max} = \frac{2m}{\gamma}$$ Now we can find the value of \(\gamma\) by substituting the values of \(m\) and \(x_{max}\) in the above equation: $$x_{max} = x(t_{max}) = \frac{2m^2}{\gamma}e^{-1}$$ $$0.5 = \frac{2\cdot 1^2}{\gamma}e^{-1} \Rightarrow \gamma = 4e~\frac{ft}{s}$$ Finally, we can find the spring constant \(k\) by substituting the value of \(\gamma\) into the relationship from Step 2: $$k = \frac{\gamma^2}{4m} = \frac{(4e)^2}{4(1)} = 4e^2~\frac{ft}{lb}$$

Summary of results

In summary, we found the relationship between damping constant \(\gamma\) and spring constant \(k\) for the critically damped system as follows: $$k = \frac{\gamma^2}{4m}$$ Using the given initial conditions, the damping constant \(\gamma\) and the spring constant \(k\) for the spring and dashpot system are calculated as: $$\gamma = 4e~\frac{ft}{s}$$ $$k = 4e^2~\frac{ft}{lb}$$

Key concepts

Mass-Spring-Damper System

A mass-spring-damper system is a common model used in mechanical systems to describe oscillatory motion. This type of system involves a mass, which represents the object in motion, connected to a spring and a damper that impact its movement.

In this system:

• The mass (\(m\)) is the object's resistance to changes in motion. It is usually measured in slugs or kilograms.
• The spring is characterized by its spring constant (\(k\)), which determines how stiff the spring is.
• The damper, associated with the damping constant (\(\gamma\)), affects how quickly the system's oscillations diminish.

The equation of motion for a mass-spring-damper system is crucial for understanding how these elements interact. This mathematical expression relates the forces acting on the mass to its acceleration, velocity, and position over time.

Damping Constant

The damping constant (\(\gamma\)) is a measure of how much resistance the damper provides to motion. In a mass-spring-damper system, this constant influences how quickly vibrations are reduced.

For critical damping, \(\gamma\) is chosen such that it just eliminates any oscillation, allowing the mass to return to equilibrium as quickly as possible without overshooting.

A higher damping constant usually means greater energy dissipation per cycle, reducing vibration more quickly. However, in a critically damped system, the damping is tuned perfectly to avoid both over-damping (slow return to equilibrium) and under-damping (continuous oscillations).

The relationship for a critically damped system is typically given by the equation:
\[\gamma^2 = 4mk\]
where \(k\) is the spring constant and \(m\) is the mass.

Spring Constant

The spring constant (\(k\)) is a parameter that determines the stiffness of the spring in a mass-spring-damper system.

A high spring constant indicates a spring that is more difficult to compress or extend, leading to a stronger force that opposes motion, while a low spring constant means the spring is easier to compress or extend, leading to a weaker force.

In the context of a critically damped system, the spring constant is related to the damping constant by the equation:
\[k = \frac{\gamma^2}{4m}\]
This relationship ensures the system returns to its equilibrium position without excessive oscillation, effectively leveraging both spring stiffness and damping to achieve stability.

Equation of Motion

In analyzing a mass-spring-damper system, the equation of motion is fundamental. It describes how the mass moves based on the forces applied by the spring and damper.

The general equation of motion for such a system is given by:
\[m\frac{d^2x}{dt^2} + \gamma\frac{dx}{dt} + kx = 0\]
In this equation:

• \(m\frac{d^2x}{dt^2}\) represents the inertial force from the mass.
• \(\gamma\frac{dx}{dt}\) accounts for the damping force.
• \(kx\) corresponds to the spring force, proportional to displacement.

In the case of critical damping, the equation transitions the system smoothly back to equilibrium. The differential equation helps determine how various parameters like mass, damping constant, and spring constant affect motion, allowing engineers and scientists to predict and optimize system behavior effectively.