2 Single Degree of Freedom Systems
Learning Outcomes
By the end of this chapter, students will be able to:
- Describe the free vibration motion of a single degree of freedom system
- Describe the natural frequency and damping ratio of a single degree of freedom system
- Describe the harmonically forced vibration motion of a single degree of freedom system
- Describe the homogeneous and particular solutions of a single degree of freedom system
- Describe the response amplitude operator of a single degree of freedom system
2.1 Motivation
A ship or an offshore platform operating in the open ocean is free to more in all six degrees of freedom and experiences forces due to the random waves, wind and currents in the ocean. While we embark on this journey to be able to analyze this problem by the end of the course, let’s start small by understanding the dynamics of a simple spring-mass-damper system. We will later see that some of the concepts we learn with respect to this simplified system transfer over to our ship or floating platform dyanmics too.
2.2 Spring mass system - Free vibration
Consider a mass on a flat surface connected to a massless spring as shown in Figure 2.1. The mass is free to move in the horizontal direction. One end of the spring is attached to the mass and the other end of the spring is attached to a fixed point. We will assume that the surface is frictionless. We denote the displacement of the mass from its equilibrium position as \(x(t)\).
The free body diagram of the spring mass system is shown in Figure 2.2. The equilibrium of forces along horizontal and vertical axes yields:
\[\begin{align} m \frac{d^2x}{dt^2} + kx = 0 \\ mg = N \end{align}\]
The first equation can be recast as:
\[\begin{align} \frac{d^2x}{dt^2} + \omega_n^2 x = 0 \label{eq-sdof-spring-mass-eom} \end{align}\]
where \(\omega_n\) is the natural frequency of the system and is denoted by:
\[\begin{align} \omega_n = \sqrt{\frac{k}{m}} \end{align}\]
It is easy to see that the solution to \(\eqref{eq-sdof-spring-mass-eom}\) is:
\[\begin{align} x(t) = A \cos(\omega_n t) + B \sin(\omega_n t) \label{eq-sdof-undamped-gen-sol} \end{align}\]
The constants \(A\) and \(B\) can be determined from the specified initial conditions. If the initial displacement is \(x(0) = x_0\) and initial velocity is \(\dot{x}(0) = v_0\), then the solution in terms of the initial conditions is given by:
\[\begin{align} x(t) = x_0 \cos(\omega_n t) + \frac{v_0}{\omega_n} \sin(\omega_n t) \end{align}\]
We can see that the solution persists indefinitely for a non-zero initial condition.
Common Pitfall
Note that the initial condition has no influence on the natural frequency of the system. The natural frequency is only dependent on the mass \(m\) and stiffness \(k\).
Interactive Simulation
Use the tool below to explore how the mass \(m\), stiffness \(k\), initial displacement \(x_0\) and initial velocity \(v_0\) affect the free vibration response of the spring mass system. The 3D animation shows the physical motion of the mass, while the plots track the displacement time history, velocity time history, and the phase space trajectory (\(x\) vs. \(\dot{x}\)).
Displacement vs Time
Velocity vs Time
Phase Space
2.3 Complex response
We saw that in the previous section, both \(x_1(t) = \cos(\omega_n t)\) and \(x_2(t) = \sin(\omega_n t)\) satisfy the differential equation \(\eqref{eq-sdof-spring-mass-eom}\). Both these solutions can be combined together using the Euler’s identity as:
\[\begin{align} x_c(t) = x_1(t) + i x_2(t) = \cos(\omega_n t) + i \sin(\omega_n t) = e^{i \omega_n t} \end{align}\]
We can see that the complex response \(x_c(t)\) also satisfies the differential equation \(\eqref{eq-sdof-spring-mass-eom}\). The notion of a complex response for a spring-mass system may seem odd at first. However, we will see its usefulness in a bit. Let’s assume a general solution of the form:
\[\begin{align} x_c(t) = A_c e^{i \omega_n t} \end{align}\]
where \(A_c\) is complex valued amplitude. The complex amplitude \(A_c\) can be related back to the real coefficients \(A\) and \(B\) as:
\[\begin{align} A_c &= \sqrt{A^2 + B^2} e^{i\phi} \quad \text{where} \quad \phi = \tan^{-1}{\left(\frac{-B}{A}\right)} \\ &= A - iB \end{align}\]
\(\phi\) is called the phase angle and its importance will be evident when we discuss the forced vibration of a spring mass system. It can be seen that the real part of this complex response is:
\[\begin{align} x_r(t) &= \operatorname{Re}(x_c(t)) = \operatorname{Re}\left(\sqrt{A^2 + B^2} e^{i(\omega_n t + \phi)}\right) = \sqrt{A^2 + B^2} \cos{(\omega_n t + \phi)} \nonumber \\ &= \sqrt{A^2 + B^2} \left(\cos(\omega_n t) \cos(\phi) - \sin(\omega_n t) \sin(\phi) \right) \nonumber \\ &= A \cos(\omega_n t) + B \sin(\omega_n t) \end{align}\]
the same as we saw in \(\eqref{eq-sdof-undamped-gen-sol}\). The complex response in terms of the initial conditions can be expressed as:
\[\begin{align} x_c(t) = \left(x_0 - \frac{iv_0}{\omega_n} \right) e^{i \omega_n t} \end{align}\]
2.4 Spring mass damper system - Free vibration
Now let’s add a dashpot to our system that provides damping to the system as seen in Figure 2.3. The corresponding free body diagram is now shown in Figure 2.4.
The equilibrium of forces along the horizontal direction gives us the equation of motion as:
\[\begin{align} m \frac{d^2x}{dt^2} + b \dot{x} + kx = 0 \\ \end{align}\]
This equation can be recast as:
\[\begin{align} \frac{d^2x}{dt^2} + 2\zeta\omega_n\frac{dx}{dt} + \omega_n^2 x = 0 \label{eq-sdof-smd-eom} \end{align}\]
where \(\omega_n\) is the natural frequency of the system and \(\zeta\) is the damping ratio denoted by:
\[\begin{align} \zeta = \frac{b}{2\sqrt{mk}} \end{align}\]
It is easy to see now that \(e^{i\omega_n t}\) no longer satisfies the differential equation \(\eqref{eq-sdof-smd-eom}\). Thus, we assume a general solution of the form:
\[\begin{align} x_c(t) = A_c e^{i \omega t} \end{align}\]
where we do not yet know the \(\omega\). Note that we choose the exponential function as the differential equation \(\eqref{eq-sdof-smd-eom}\) is still linear. Substituting into the differential equation yields:
\[\begin{align} \left(-\omega^2 + i2\zeta\omega_n\omega + \omega_n^2\right) A_c e^{i \omega t} = 0 \end{align}\]
We are interested in the non-trivial solution (\(A_c \neq 0\)) and that leads to the quadratic equation:
\[\begin{align} -\omega^2 + i2\zeta\omega_n\omega + \omega_n^2 = 0 \end{align}\]
with the roots:
\[\begin{align} \omega_{1,2} = i\zeta \omega_n \pm \omega_n \sqrt{1 - \zeta^2} \end{align}\]
Thus the general solution for the spring mass damper system is given by:
\[\begin{align} x_c(t) &= A_c e^{i\omega_1 t} + B_c e^{i\omega_2 t} \\ &= e^{-\zeta \omega_n t} \left(A_c e^{i\omega_nt\sqrt{1 - \zeta^2}} + B_c e^{-i\omega_nt\sqrt{1 - \zeta^2}} \right) \end{align}\]
Now there are three possible scenarios:
- Overdamped scenario \(\zeta > 1\)
- Critically damped scenario \(\zeta = 1\)
- Underdamped scenario \(\zeta < 1\)
Overdamped case
In this case, we see that the \(\sqrt{1 - \zeta^2} = i\sqrt{\zeta^2 - 1}\) and the general solution takes the form:
\[\begin{align} x_c(t) &= A_c e^{\omega_nt(-\zeta - \sqrt{\zeta^2 - 1})} + B_c e^{\omega_nt(-\zeta + \sqrt{\zeta^2 -1})} \end{align}\]
The coefficients of time for both exponents are negative and real. This leads to the solution \(x_c(t) \rightarrow 0\) as \(t \rightarrow 0\).
Critically damped case
In this case we have repeated roots \(\omega_1 = \omega_2 = i\zeta \omega_n\). The solution \(x_c(t)\) in this case takes the form:
\[\begin{align} x_c(t) = A_c e^{-\zeta\omega_nt} + B_c te^{-\zeta\omega_nt} \end{align}\]
Here too both terms have negative time coefficients in the exponential terms and the solution \(x_c(t) \rightarrow 0\) as \(t \rightarrow 0\).
Underdamped case
In this case the system exhibits an oscillatory response. The response reduces to the form:
\[\begin{align} x_c(t) &= e^{-\zeta \omega_n t} \left(A_c e^{i\omega_dt} + B_c e^{-i\omega_dt}\right) \end{align}\]
where \(\omega_d = \omega_n \sqrt{1-\zeta^2}\) is the damped natural frequency. The real part of the solution can be expressed as:
\[\begin{align} x_r(t) &= e^{-\zeta \omega_n t} \left(A \cos(\omega_dt) + B \sin(\omega_dt) \right) \end{align}\]
The coefficients \(A\) and \(B\) are determined as before from the initial conditions. If the initial conditions are \(x_r(0) = x_0\) and \(\dot{x}_r = v_0\), then the solution in terms of the initial conditions can be expressed as:
\[\begin{align} x_r(t) = e^{-\zeta \omega_n t} \left(x_0 \cos(\omega_dt) + \frac{v_0 + \zeta\omega_nx_0}{\omega_d} \sin(\omega_dt) \right) \end{align}\]
Interactive Simulation
Use the tool below to explore how the natural frequency \(\omega_n\) and damping ratio \(\zeta\) affect the free vibration response of the spring mass damper system. Drag the damping ratio slider across \(\zeta = 1\) to move between the underdamped, critically damped and overdamped regimes. The 3D animation shows the physical motion of the mass, while the plots track the displacement time history (with its decay envelope shown as a dashed line when underdamped), velocity time history, and the phase space trajectory (\(x\) vs. \(\dot{x}\)).
Displacement vs Time
Velocity vs Time
Phase Space
2.5 Spring mass damper system - Forced vibration
Now let’s consider that an external force \(F(t) = F_0 \cos(\omega t) = \operatorname{Re}(F_0 e^{i\omega t})\) acting on the mass as seen in Figure 2.5. The corresponding free body diagram is now shown in Figure 2.6.
The equilibrium of forces along the horizontal direction gives us the equation of motion as:
\[\begin{align} m \frac{d^2x}{dt^2} + b \dot{x} + kx = F_0 \cos(\omega t) \\ \end{align}\]
This equation can be recast as:
\[\begin{align} \frac{d^2x}{dt^2} + 2\zeta\omega_n\frac{dx}{dt} + \omega_n^2 x = f_0 \cos(\omega t) \label{eq-sdof-smd-forced-eom} \end{align}\]
where \(f_0 = F_0/m\). Assuming a particular solution of the form \(x_c(t) = A_c e^{i\omega t}\) and substituting it into \(\eqref{eq-sdof-smd-forced-eom}\) yields:
\[\begin{align} \left(-\omega^2 + i2\zeta\omega_n\omega + \omega_n^2\right)A_c e^{i\omega t} = f_0 e^{i\omega t} \end{align}\]
For a non-trivial solution (\(A_c \neq 0\)):
\[\begin{align} A_c = \frac{F_0/k}{\left(1 - \frac{\omega^2}{\omega_n^2}\right) + i 2 \zeta \frac{\omega}{\omega_n}} \end{align}\]
The ratio of the amplitude of the response to the amplitude of the static displacement \(F_0/k\) is defined as the response amplitude operator (RAO) or the dynamic amplification factor (DAF) and is given by:
\[\begin{align} \frac{A_c}{F_0/k} = \frac{1}{\left(1 - \frac{\omega^2}{\omega_n^2}\right) + i 2 \zeta \frac{\omega}{\omega_n}} \end{align}\]
The plot of the RAO for various values of damping ratios is shown in Figure 2.7. We can see that for a low damping ratio, the amplitude of the response amplifies significantly when \(\omega/\omega_n \approx 1\). When \(\omega/\omega_n \rightarrow 0\), the RAO tends to unity and when \(\omega/\omega_n \rightarrow \infty\), the RAO tends to zero.
For \(\zeta \ll 1\) and \(\omega/\omega_n < 1\), the response tends to be in phase with the excitation. However, for \(\omega/\omega_n > 1\) the response tends to be out of phase (\(-180^\circ\)) with the excitation. For larger damping ratios, this progression is more gradual. For all damping cases, the response phase is \(-90^\circ\) with respect to the excitation at \(\omega/\omega_n = 1\).
The total solution is the combination of the homogeneous solution (free vibration solution) and the particular solution (forced vibration solution) and is given by:
\[\begin{align} x_c(t) = e^{-\zeta \omega_n t} \left(A_h e^{i\omega_dt} + B_h e^{-i\omega_dt}\right) + \frac{f_0/\omega_n^2}{\left(1 - \frac{\omega^2}{\omega_n^2}\right) + i 2 \zeta \frac{\omega}{\omega_n}} e^{i\omega t} \end{align}\]
where \(A_h\) and \(B_h\) are determined from the initial conditions.
Interactive Simulation
Use the tool below to explore how the natural frequency \(\omega_n\), damping ratio \(\zeta\), frequency ratio \(\omega/\omega_n\) and excitation amplitude \(f_0\) affect the forced vibration response of the spring mass damper system. The first plot overlays the excitation \(f(t)/f_0 = \cos(\omega t)\) (right axis) and the non-dimensional response \(x(t)/(f_0/\omega_n^2)\) (left axis) on a common time axis so that the phase difference between them is visible directly. The second plot shows the full RAO curve for the current damping ratio, with a dot marking the current operating point \(\omega/\omega_n\). The third plot shows the phase space trajectory (\(x\) vs. \(\dot{x}\)) of the total response, including the initial transient.
Excitation & Response vs Time
Response Amplitude Operator
Phase Space
2.6 Heave motion of a spar platform
One of the popular designs of an offshore platform is the classic spar platform. The spar is primarily a large cylinder with a very deep draft.