Question

A spring-mass system has a spring constant of \(3 \mathrm{N} / \mathrm{m}\). A mass of \(2 \mathrm{kg}\) is attached to the spring and the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude of the instantaneous velocity. If the system is driven by an external force of \(3 \cos 3 t-2 \sin 3 t \mathrm{N},\) determine the steady-state response. Express your answer in the form \(R \cos (\omega t-\delta)\)

Short answer

Question: Determine the steady-state response of the spring-mass system in the form \(R \cos (\omega t - \delta)\). Answer: The steady-state response of the system is \(x_p(t) = \frac{\sqrt{37}}{6} \cos(3t - \arctan(-\frac{1}{4}))\).

Step-by-step solution

Write down the equation of motion

The equation of motion for a spring-mass-damping system can be written using Newton's second law as: $$m\frac{d^2x(t)}{dt^2}+c\frac{dx(t)}{dt}+kx(t)=F(t)$$ Where, - \(m=2\,\text{kg}\) is the mass of the object - \(c\) is the viscous damping coefficient - \(k = 3\,\text{N/m}\) is the spring constant - \(x(t)\) is the displacement from the equilibrium position - \(F(t) = 3\cos 3t - 2\sin 3t\,\text{N}\) is the external force

Determine the viscous damping coefficient

The viscous fluid offers a resistance numerically equal to the magnitude of the instantaneous velocity. Thus, we can write: $$c\frac{dx(t)}{dt}=\frac{dx(t)}{dt}$$ Since resistance is not zero, \(c=1\,\text{kg/s}\).

Write down the equation of motion with given values

Now we can plug in the given values of \(m\), \(c\), \(k\) and \(F(t)\) into the equation of motion: $$2\frac{d^2x(t)}{dt^2}+\frac{dx(t)}{dt}+3x(t)=3\cos 3t - 2\sin 3t$$

Solve the equation for steady-state response

To find the steady-state response, we will look for a solution in the same form as the external force: $$x_p(t)=A\cos 3t + B\sin 3t$$ We will now differentiate \(x_p(t)\) twice with respect to \(t\) to plug it into the equation of motion: $$\frac{dx_p(t)}{dt}=-3A\sin 3t + 3B\cos 3t$$ $$\frac{d^2x_p(t)}{dt^2}=-9A\cos 3t - 9B\sin 3t$$ Now, substitute these expressions into the equation of motion: $$-18A\cos 3t - 18B\sin 3t + 3\cos 3t - 2\sin 3t = 3\cos 3t - 2\sin 3t$$ Comparing the coefficients of \(\cos 3t\) and \(\sin 3t\), we obtain: $$-18A+3=3$$ $$-18B+3=-2$$ Solving these equations, we get: $$A = -\frac{2}{3}, \quad B = \frac{1}{6}$$ Hence, the particular solution for the steady-state response is: $$x_p(t) = -\frac{2}{3}\cos 3t + \frac{1}{6}\sin 3t$$

Express the steady-state response in the desired form

Our goal is to express the steady-state response in the form \(R \cos(\omega t - \delta)\). We can rewrite the particular solution as: $$x_p(t) = R\cos(3t-\delta)$$ Where, \(R = \sqrt{A^2+B^2}\) and \(\delta = \arctan\frac{B}{A}\). Plugging the values of \(A\) and \(B\), we get: $$R = \sqrt{(-\frac{2}{3})^2+(\frac{1}{6})^2}=\frac{\sqrt{37}}{6}$$ $$\delta = \arctan\frac{1/6}{-2/3}=\arctan (-\frac{1}{4})$$ Hence, the steady-state response of the system is: $$x_p(t) = \frac{\sqrt{37}}{6} \cos(3t - \arctan(-\frac{1}{4}))$$

Key concepts

Spring-Mass-Damping System

Understanding a spring-mass-damping system is crucial for comprehending a range of physical phenomena, from vehicle suspension to the oscillations of a diving board. At its core, this model consists of a mass attached to a spring that can compress or stretch, with an additional damping component that represents resistance to motion, such as friction from air or, as in our example, resistance from a viscous fluid.

The system's motion is often described by how it responds to forces, which may include the spring's restoring force, which is directly proportional to the displacement (represented by Hooke's Law: \(F = -kx\)), the damping force, often proportional to the velocity of the mass, and any external forces applied to the system. In the given exercise, we analyze the response of such a system to an external oscillating force.

Differential Equations

Differential equations form the backbone of modeling physical systems in science and engineering. They relate functions to their derivatives, indicating how a particular quantity changes over time or space. In our case, the equation of motion for the spring-mass-damping system is a second-order linear differential equation where the highest derivative is squared.

The general process of solving such an equation involves identifying the type of equation it is, determining appropriate methods for solving it (such as characteristic equations or particular integrals), and applying initial or boundary conditions to find specific solutions. In the steady-state response scenario, we often look for solutions that match the form of the external force acting on the system.

Viscous Damping Coefficient

The viscous damping coefficient, denoted as \(c\), determines the magnitude of the damping force in a system. This coefficient is a measure of the resistance that a viscous fluid provides against the motion of an object. In essence, it dictates how much the motion of the mass is 'slowed down' or resisted due to this damping effect.

In engineering and physics, it's vital to know this value as it helps predict how a system will behave over time. For example, a higher viscous damping coefficient means that the system will return to its equilibrium position more slowly after being disturbed. In the context of the exercise, the coefficient is numerically equal to the resistance caused by the fluid, simplifying our calculations substantially.

Newton's Second Law

Newton's second law is fundamental to dynamics and states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (\( F=ma \)). This law forms the basis for deriving the equation of motion for our spring-mass-damping system exercise.

In practical terms, Newton's second law can be used to predict how an object will move under the influence of various forces. For our system in which a mass is attached to a spring and experiences damping, this law is what links the forces to the resulting motion through the differential equation. By manipulating this deep-rooted physical principle, we can glean insights into the complex movements of the system.