13.3: Free damped vibrations
In practice a vibration is never completely undamped, since there are always small practical imperfections that cause some kind of dissipation due to nonconservative forces. The most common type of damping, that is also most easily analysed, is by a velocity proportional force \(\overrightarrow{\boldsymbol{F}}_{c}=-c \overrightarrow{\boldsymbol{v}}\), that is generated by a piston or other linear damper element. The force points in the opposite direction from the velocity vector of the mass, and thus the power done by such a force on the mass is given by \(P=\overrightarrow{\boldsymbol{F}}_{c} \cdot \overrightarrow{\boldsymbol{v}}=-c|\overrightarrow{\boldsymbol{v}}|^{2}\), which is always negative. This causes the kinetic and potential energy of the mass-spring system to reduce to zero.
When connecting a linear damper to the mass, in parallel to the spring, as shown in Figure 13.5, we obtain a mass-spring-damper system, we have from Newton’s second law that \(\overrightarrow{\boldsymbol{F}}_{k}+\overrightarrow{\boldsymbol{F}}_{c}=m_{A} \overrightarrow{\boldsymbol{a}}_{A}\), which when projected along the \(y\)-axis, results in the following equation of motion:
\[\begin{align} \sum F_{y}=-k y_{A}-c \dot{y}_{A} & =m_{A} \ddot{y}_{A} \tag{13.29} \label{13.29}\\[4pt] m_{A} \ddot{y}_{A}+c \dot{y}_{A}+k y_{A} & =0 \tag{13.30} \label{13.30}\end{align}\]
Here we note that normally \(m_{A}, c\) and \(k\) are all positive real constants and in static equilibrium \(\dot{y}_{A}=0\), such that the damper does generate force and does not affect \(y_{A, \mathrm{st}}\).
To solve this equation of motion we propose the following complex trial function:
\[y_{A}(t)=\Re A_{c} e^{\lambda t} \tag{13.31} \label{13.31}\]
In this equation both \(A_{c}\) and \(\lambda\) are complex numbers with a nonzero real and imaginary part. Note that we only use a complex trial function and don’t use a trial function with trigonometric equations since working with complex numbers is a lot easier when working with damped vibrations. Substitution in Equation 13.30 results in:
\[\Re\left(\left[m_{A} \lambda^{2}+c \lambda+k\right] A_{c} e^{\lambda t}\right)=0 \tag{13.32} \label{13.32}\]
Since this equation has to hold for all times \(t\), the quadratic equation in the square brackets has to be zero, and can be solved, to determine \(\lambda\), using the quadratic formula or ’ \(A B C\) formula’, with \(A=m_{A}, B=c\) and \(C=k\) :
\[\begin{align} & m_{A} \lambda^{2}+c \lambda+k=0 \tag{13.33} \label{13.33}\\[4pt] \lambda= & \frac{1}{2 A}\left(-B \pm \sqrt{B^{2}-4 A C}\right) \tag{13.34} \label{13.34}\\[4pt] \lambda= & \frac{1}{2 m_{A}}\left(-c \pm \sqrt{c^{2}-4 m_{A} k}\right)=\frac{-c \pm \sqrt{\Delta}}{2 m_{A}} \tag{13.35} \label{13.35}\end{align}\]
The term \(\Delta=c^{2}-4 m_{A} k\) is called the discriminant and depending on the sign of \(\Delta\), we obtain three different values of \(\lambda\) that correspond to three quite different types of solutions:
- \(\Delta>0\) : Overdamped vibration, the values of \(\lambda\) are both real and negative numbers with two solutions \(\lambda_{o, \pm}=\frac{-c \pm \sqrt{\Delta}}{2 m_{A}}\).
- \(\Delta<0\) : Underdamped vibration, the values of \(\lambda\) are complex numbers with a negative real part \(-c /\left(2 m_{A}\right)\). When taking the square-root of a negative number we get \(\sqrt{-1}=i\), such that:
\[\frac{\sqrt{\Delta}}{2 m_{A}}=i \frac{\sqrt{4 m_{A} k-c^{2}}}{2 m_{A}}=i \sqrt{\omega_{n}^{2}-\frac{c^{2}}{4 m_{A}^{2}}} \equiv i \omega_{d} \tag{13.36} \label{13.36}\]
On the right side of this equation we have defined the damped angular resonance frequency \(\omega_{d} \equiv \sqrt{\omega_{n}^{2}-c^{2} /\left(4 m_{A}^{2}\right)}\), and it can be seen that for small damping \(c\), this damped resonance frequency is almost equal to the natural angular resonance frequency \(\left(\omega_{d} \approx \omega_{n}\right)\). We then obtain two values for \(\lambda\) that solve the EoM: \(\lambda_{u, \pm}=\frac{-c}{2 m_{A}} \pm i \omega_{d}\).
3. \(\Delta=0\) : Critically damped vibration. In this case there is only one value \(\lambda=-c /\left(2 m_{A}\right)\). The value of the damping coefficient that belongs to this situation \(\Delta=c_{c}^{2}-4 m_{A} k=0\) is called the critical damping coefficient \(c_{c}=2 \sqrt{m_{A} k}\). For \(c>c_{c}\) we have overdamped vibrations and for \(c<c_{c}\) we get underdamped vibrations.
Let us now determine the different solutions Equation 13.31 of the EoM with these values of \(\lambda\).
13.3.1 Overdamped motion
Since \(\lambda\) is a real number for \(\Delta>0\), we need to add the solutions of the EoM for both values \(\lambda_{o, \pm}=\frac{-c \pm \sqrt{\Delta}}{2 m_{A}}\) to get a function that satisfies the equation of motion, while introducing two real constants \(A_{+}\)and \(A_{-}\)that can be adjusted to satisfy the initial position and velocity conditions. We find:
\[\begin{gathered} y_{A, o}(t)=A_{+} e^{\lambda_{o,+} t}+A_{-} e^{\lambda_{o,-} t} \tag{13.37} \label{13.37}\\[4pt] y_{A, o}(t)=A_{+} e^{\frac{-c+\sqrt{\Delta}}{2 m_{A}} t}+A_{-} e^{\frac{-c-\sqrt{\Delta}}{2 m_{A}} t} \tag{13.38} \label{13.38}\end{gathered}\]
13.3.2 Underdamped motion
For the case \(\Delta<0\), the complex number \(A_{c}=A e^{i \varphi_{0}}\) can be substituted in Equation 13.31. Using the positive value \({ }^{2} \lambda_{u,+}=\frac{-c}{2 m_{A}}+i \omega_{d}\) we obtain the solution:
\[y_{A, u}(t)=A e^{-\frac{c}{2 m_{A}} t} \Re e^{i\left(\omega_{d} t+\varphi_{0}\right)}=A e^{-\frac{c}{2 m_{A}} t} \cos \left(\omega_{d} t+\varphi_{0}\right) \tag{13.39} \label{13.39}\]
The motion consists of a fast oscillating cosine function with a slowly decaying amplitude given by the exponential function. Just like in Sec. 13.2.4 one might also choose to replace \(\cos \left(\omega_{d} t+\varphi_{0}\right)\) by the sum of a cosine and sine function, in which case the solution is written as:
\[y_{A, u}(t)=e^{-\frac{c}{2 m_{A}} t}\left[A_{1} \cos \left(\omega_{d} t\right)+A_{2} \sin \left(\omega_{d} t\right)\right] \tag{13.40} \label{13.40}\]
13.3.3 Critically damped motion*
For the general solution of critically damped motion we have a problem, since we have found only one value of \(\lambda\) that corresponds to the real solution \(y_{A}=A_{0} e^{-c /\left(2 m_{A}\right) t}\). For this specific case it can by shown, as can be checked by substitution in Equation 13.30, that the function \(y_{A}=A_{t} t e^{-c /\left(2 m_{A}\right) t}\) also obeys the EoM, such that the general solution for critically damped vibrations becomes:
\[y_{A, c}(t)=A_{0} e^{-\frac{c}{2 m_{A}} t}+A_{t} t e^{-\frac{c}{2 m_{A}} t} \tag{13.41} \label{13.41}\]
As can be seen in Figure 13.5 the critically damped vibrations approach zero faster than the overdamped vibrations with the same damping \(c / m_{A}\).
13.3.4 More complex free vibrations
Up to now we only dealt with mass-spring and mass-spring-damper systems. However, more complex systems can also result in the same equation of motion. For instance, applying Euler’s second law to rotating rigid bodies connected to springs leads to an EoM that is similar to that of the free vibration of a mass-spring-damper system. Even though the constants and variables in that EoM are different from \(y_{A}, m_{A}, k\) and \(c\), the EoM can be written similarly by introducing effective functions \(y_{e}, m_{e}, k_{e}\) and \(c_{e}\), such that the EoM becomes:
\[m_{e} \ddot{y}_{e}+c_{e} \dot{y}_{e}+k_{e} y_{e}=0 \tag{13.42} \label{13.42}\]
Here, the function \(y_{e}(t)\) can represent a position coordinate but can also be an angle, or any generalised coordinate that uniquely describes the position or orientation of the system. Mathematically, the motion of these more complex systems can be derived in exactly the same way as a mass-spring-damper system, like discussed in previous sections. Only the constants \(m_{e}, c_{e}\) and \(k_{e}\) and function name \(y_{e}(t)\) are different. So, everything we have derived for the mass-spring-damper system remains valid for these more complex systems.
13.3.5 Free vibrations: solution procedure
Finally, we discuss step-by-step the procedure for analysing free vibrations discussed in this section and the previous one.
- Sketch the system and coordinate system.
- Draw the free-body diagram and project the force and/or moment vectors on the coordinate axes. Determine the relevant constraint equations.
- Use Newton’s second law and Euler’s first law for translations and Euler’s second law for rotations to determine the scalar equation of motion along the relevant translation or rotation coordinate.
- Determine the static equilibrium position \(y_{A, \mathrm{st}}\) by setting all time derivatives in the EoM to zero \(\ddot{y}_{A, \text { tot }}=\dot{y}_{A, \text { tot }}=0\) and solving for \(y_{A}\).
- Define the time dependent part of the motion as \(y_{A}(t)=y_{A, \mathrm{tot}}-y_{A, \mathrm{st}}\).
- The system will only show free vibration if at least one of the EoMs can be written in the form \(m_{e} \ddot{y}_{e}+c_{e} \dot{y}_{e}+k_{e} y_{e}=0\).
- Determine the equations for the variable \(y_{e}\) and constants \(m_{e}, c_{e}\) and \(k_{e}\). Now one can either solve the ODE, like in the next step 8 , or directly go to step 9 and use the known solutions.
- To solve the ODE, substitute the trial function \(y_{e}=\Re A_{c} e^{\lambda t}\) and its time-derivatives into the EoM to obtain \(-m_{e} \lambda^{2}+c_{e} \lambda+k_{e}=0\). Solve for \(\lambda\) and substitute it back into the trial function to find the motion \(y_{e}(t)\) like explained in the previous sections.
- If \(c_{e}=0\), then we are dealing with free undamped vibration. In that case the solution of the EoM is \(y_{e}(t)=\Re A e^{i \omega_{n} t+\varphi_{0}}=A \cos \left(\omega_{n} t+\varphi_{0}\right)\), with natural resonance frequency \(\omega_{n}=\sqrt{k_{e} / m_{e}}\).
- If \(c_{e} \neq 0\) we are dealing with free damped vibration.
Determine the discriminant \(\Delta=c_{e}^{2}-4 m_{e} k_{e}\).
- If \(\Delta>0\) we are dealing with overdamped free vibration with \(y_{e}=A_{+} e^{\lambda_{+} t}+A_{-} e^{\lambda-t}\) and \(\lambda_{ \pm}=\frac{-c_{e} \pm \sqrt{\Delta}}{2 m_{e}}\).
- If \(\Delta<0\) we are dealing with underdamped free vibration with \(y_{e}=A e^{-\frac{c_{e}}{2 m_{e}} t} \cos \left(\omega_{d} t+\varphi_{0}\right)\) and \(\omega_{d}=\sqrt{\omega_{n}^{2}-c_{e}^{2} /\left(4 m_{e}^{2}\right)}\)
- If \(\Delta=0\) we are dealing with critically damped free vibration with \(y_{e}=A_{0} e^{-\frac{c_{e}}{2 m_{e}} t}+A_{t} t e^{-\frac{c_{e}}{2 m_{e}} t}\).
- Two initial conditions of the form \(y_{e}\left(t_{1}\right)=y_{1}\) or \(\dot{y}_{e}\left(t_{2}\right)=v_{2}\) need to be solved to determine the two unknown constants in the solution \(\left(A \& \varphi_{0}\right.\), \(A_{+} \& A_{-}\)or \(A_{0} \& A_{t}\) ). See Sec. 13.2.5.
- In some cases it is convenient to rewrite the cosine with phase-shift \(\varphi_{0}\) into a sum of a cosine and sine using the mathematical equation:
\[A \cos \left(\omega t+\varphi_{0}\right)=A_{1} \cos (\omega t)+A_{2} \sin (\omega t) \tag{13.43} \label{13.43}\]
where \(A^{2}=A_{1}^{2}+A_{2}^{2}\) and \(\tan \left(\varphi_{0}\right)=-A_{2} / A_{1}\).
Consider the car \(A\), in Figure 13.6 with a mass of \(m=1000 \mathrm{~kg}\) that can be considered as a point mass. It has a spring stiffness of \(k=50000 \mathrm{~N} / \mathrm{m}\) and a damping constant of \(c=1000 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}\). The car drops from a small height, such that it reaches its equilibrium position \(y_{A}(0)=0\) with a speed of \(v_{0}=-1 \mathrm{~m} / \mathrm{s}\) at \(t=0\). Determine the motion \(y_{A}(t)\) of the car.
Solution
We follow the steps from this section. The sketch and FBD are given in Figure 13.6.
So, we project the forces and write down the equation of motion along the \(y\)-axis:
\[\begin{align} \sum F_{y}=F_{k, y}+F_{c, y}+F_{g, y} & =m_{A} \ddot{y}_{A, \mathrm{tot}} \tag{13.44} \label{13.44}\\[4pt] -k\left(y_{A, \mathrm{tot}}-L_{0}\right)-c \dot{y}_{A, \mathrm{tot}}-m_{A} g & =m_{A} \ddot{y}_{A, \mathrm{tot}} \tag{13.45} \label{13.45}\\[4pt] -k y_{A}-c \dot{y}_{A} & =m_{A} \ddot{y}_{A} \tag{13.46} \label{13.46}\\[4pt] m_{A} \ddot{y}_{A}+k y_{A}+c \dot{y}_{A} & =0 \tag{13.47} \label{13.47}\end{align}\]
Here we used that \(y_{A, \mathrm{st}}=-\frac{m_{A} g}{k}+L_{0}\) like in Equation 13.2, and \(y_{A, \text { tot }}=y_{A}+y_{A, \mathrm{st}}\) to eliminate the static particular solution. Although, since we are only asked for the motion with respect to the equilibrium position, we could also have skipped this step, and directly eliminate the static terms. Now we see that the equation of motion Equation 13.47 can be written in the form of Equation 13.42, with \(y_{e}=y_{A}, m_{e}=m_{A}\), \(c_{e}=c\) and \(k_{e}=k\) (we will encounter more complex cases later). Now we can solve this EoM, like in Sec. 13.3. We will not repeat the derivation and directly substitute the values of \(m_{e}, c_{e}\) and \(k_{e}\) in the solution. We first determine that the discriminant \(\Delta=c^{2}-4 m_{A} k=-199 \times 10^{6} \mathrm{~kg}^{2} / \mathrm{s}^{2}\) is negative, such that we deal with an underdamped vibration. Then we determine the damped resonance frequency as \(\omega_{d}=\sqrt{k_{e} / m_{e}-c_{e} /\left(4 m_{e}^{2}\right)}\). The solution can then be written as:
\[y_{A}(t)=A e^{-\left(c_{e} / 2 m_{e}\right) t} \cos \left(\omega_{d} t+\varphi_{0}\right) \tag{13.48} \label{13.48}\]
Finally we need to determine the constants \(A\) and \(\varphi_{0}\) from the initial conditions. Since \(y_{A}(t=0)=0\), we find that \(\varphi_{0}=\pi / 2\). By taking the time derivative of \(y_{A}\), with the product rule and using that \(\dot{y}_{A}(0)=v_{0}\) we find:
\[\begin{align} \dot{y}_{A}(0) & =\left[-\left(c_{e} / 2 m_{e}\right) \cos \left(\omega_{d} 0+\pi / 2\right)-\omega_{d} \sin \left(\omega_{d} 0+\pi / 2\right)\right] A e^{-\left(c_{e} / 2 m_{e}\right) 0}=v_{0} \\[4pt] & =-\omega_{d} A=v_{0}=-1 \mathrm{~m} / \mathrm{s} \tag{13.49} \label{13.49}\end{align}\]
from which we find \(A\). Thus we have fully determined all constants in Equation 13.48, and fully determined the motion \(y_{A}(t)\).
A Example 13.3 For the car in the previous example we now design the optimal damping, that reduces the time for the amplitude to go to zero. The initial speed \(v_{0}\) and kinetic energy of the car is kept constant and the damping coefficient \(c\) is varied from underdamped \(c=\frac{1}{2} c_{c}\), to critically damped \(c=c_{c}\) to overdamped \(c=2 c_{c}\).
Plot the motion of the car for these three situations, and determine the optimal value of \(c\).
First we determine the constants \(\omega_{d}=\sqrt{\frac{k}{m}-\frac{c^{2}}{4 m^{2}}}, c_{c}=2 \sqrt{k m}\) and \(\Delta=c^{2}-4 m k\), and define \(\omega_{d o}=\sqrt{\Delta} /(2 m)\). Then we use the initial conditions \(y_{A}(0)=0\) and \(\dot{y}_{A}(0)=v_{0}=-1 \mathrm{~m} / \mathrm{s}\), to determine the constants \(A_{*}\) and \(\varphi_{0}\) for the underdamped, critically damped and overdamped motion. Check with some math that \(\phi_{0}=\Pi / 2\) this results in:
\[\begin{align} & y_{A, u}(t)=v_{0} t e^{-\frac{c}{2 m} t \frac{\sin \left(\omega_{d} t\right)}{\omega_{d} t}} \tag{13.50} \label{13.50}\\[4pt] & y_{A, c}(t)=v_{0} t e^{-\frac{c}{2 m} t} \tag{13.51} \label{13.51}\\[4pt] & y_{A, o}(t)=v_{0} t e^{-\frac{c}{2 m} t} \frac{\sinh \left(\omega_{d o} t\right)}{\omega_{d o} t} \tag{13.52} \label{13.52}\end{align}\]
We see that for a damping coefficient \(c=c_{c}\) the frequencies \(\omega_{d}\) and \(\omega_{d o}\) approach zero and because for small \(x\) we have \(\sin x \approx x\) and \(\sinh x \approx x\), the three expressions become equal. In Figure 13.7 it can be seen the that the solution approaches zero fastest when the damper is chosen to be critically damped with \(c=c_{c}\). The underdamped motion damps slower because of the smaller value of \(c\) in the exponent, and the overdamped motion is slower due to the slow decay of the \(A_{+}\)term in Equation 13.38.