13.2: Free undamped vibrations

A mass-spring system is one of the simplest systems in which vibrations can occur. It consists of a point mass \(A\), at position coordinate \(y_{A, \text { tot }}\) that is hanging from the ceiling by a spring with spring constant \(k\). Such a system is a type of mechanical resonator. The motion of such a system is called free undamped vibration. It is free or undriven because there is no external periodic or time-dependent driving force, and it is undamped because there is no damping force acting on the mass.

Figure 13.1: A mass \(A\) is suspended by a spring. When the spring is relaxed it has a length \(L_{0}\). When it is in static equilibrium with the gravitational force, the spring has a length \(L_{0}+y_{A, \mathrm{st}}\). The displacement of the mass with respect to this static equilibrium position is measured by position coordinate \(y_{A}(t)\).

Figure 13.1 shows a sketch of what happens if a mass is suspended by a spring. First of all the spring has a relaxed length \(L_{0}\) even if there are no forces acting on it. Then if the mass \(A\) is released it moves down by a distance \(y_{A, \mathrm{st}}\) to a static equilibrium position for which the sum of forces is zero. Finally, if the mass is displaced from that position by an amount \(y_{A}\) the sum of forces is not zero such that the mass accelerates.

We determine the free vibrating motion of a mechanical resonator similarly to other kinetic problems, using the methods of Ch. 6. First we determine the EoMs and then we solve them. The sketch and coordinate system are in Figure 13.1 and we draw the FBD in Figure 13.2 to obtain the EoM:

\[\sum F_{y}=m_{A} g-k\left(y_{A, \mathrm{tot}}-L_{0}\right)=m_{A} \ddot{y}_{A, \mathrm{tot}} \tag{13.1} \label{13.1}\]

This EoM is a so-called second-order ordinary differential equation because it contains the second time-derivative \(\ddot{y}_{A, \text { tot }}\). In contrast to ODEs that we encountered in Ch. 5 it cannot easily be solved by integration. Instead, we solve the ODE by first finding its particular solution and then its homogeneous solutions.

Figure 13.2: Free body diagram of the mass in a mass-spring system from Figure 13.1.

Conveniently, an undriven mechanical resonator always has a static solution with \(\ddot{y}_{A, \text { tot }}=0\) and \(y_{A, \text { tot }}=\) constant as particular solution. Therefore, to solve free vibration problems, we always first find this static particular solution and then determine the dynamic part of the motion by solving the homogeneous ODE. This proceeds along the following steps:

1. Set the acceleration, velocity and other terms to zero and solve the EoM for the static equilibrium position \(y_{A, \text { tot }}=y_{A, \mathrm{st}}\). This particular solution is identical to the statics solution. If there are time-dependent functions set them to their time-averaged value (so a term \(\cos \omega t\) becomes zero).
2. Introduce a new coordinate \(y_{A}(t)\) which has its datum (zero point) at the static equilibrium point.
3. Substitute \(y_{A, \text { tot }}=y_{A, \text { st }}+y_{A}(t)\) into the EoM. Then all static terms will cancel out, and you obtain a ’dynamic’ homogeneous ODE which only contains \(y_{A}(t)\) and can be solved.
4. If you already know the equilibrium position, you can skip steps 1-3 above and define the coordinate \(y_{A}\) with respect to that position, and directly write down the dynamic EoM without the constant terms, since you know that they will cancel out.

Let us illustrate this procedure.

13.2.1 Determining the static equilibrium position

We first solve for the particular static solution \(y_{A, \text { tot }}(t)=y_{A, \text { st }}\) for which \(\ddot{y}_{A, \text { tot }}=0\) and \(y_{A, \mathrm{st}}\) is a constant. Substitution in Equation 13.1 results in:

\[\begin{align} m_{A} g-k\left(y_{A, \mathrm{st}}-L_{0}\right) & =0 \tag{13.2} \label{13.2}\\[4pt] y_{A, \mathrm{st}} & =\frac{m_{A} g}{k}+L_{0} \tag{13.3} \label{13.3}\end{align}\]

Thus we have found the static, time-independent solution \(y_{A, \text { tot }}(t)=y_{A, \mathrm{st}}\) of the ODE for which the \(y\)-coordinate of the mass is \(y_{A, \mathrm{st}}\) at all times. This position is called the static equilibrium, equilibrium or rest position of the mass. In this case, the static \(y\)-coordinate is the sum of the distance \(\frac{m_{A} g}{k}\) the spring is elongated by the gravitational force and the relaxed length \(L_{0}\) of the spring.

13.2.2 Equation of motion with respect to equilibrium

The static solution with \(\ddot{y}_{A, \text { tot }}=0\), is a particular solution of the ODE. However, it is not the only solution of the EoM, so let us now look for time-dependent solutions of the ODE. We define a time-dependent function \(y_{A}(t)\) which represents the displacement of the mass with respect to the static equilibrium position. The function \(y_{A}(t)\) is added to the found static solution such that the total displacement can be written as:

\[y_{A, \mathrm{tot}}=y_{A}(t)+y_{A, \mathrm{st}} \tag{13.4} \label{13.4}\]

If we substitute this function \(y_{A, \text { tot }}\) into equation Equation 13.1, we find:

\[\begin{align} m_{A} g-k\left(y_{A}+y_{A, \mathrm{st}}-L_{0}\right) & =m_{A} \ddot{y}_{A} \tag{13.5} \label{13.5}\\[4pt] -k y_{A} & =m_{A} \ddot{y}_{A} \tag{13.6} \label{13.6}\end{align}\]

Here we used Equation 13.2 to eliminate \(y_{A, \mathrm{st}}\). All constant, time-independent, terms in the equation always sum up to zero, like they do in static equilibrium. By rewriting Equation 13.6 we find the equation of motion for free undamped vibrations in the form of the following ordinary differential equation (ODE):

Concept: Equation of motion for free undamped vibrations

The equation of motion is a homogeneous second order linear ODE, that holds at all times.

\[m_{A} \ddot{y}_{A}+k y_{A}=0 \tag{13.7} \label{13.7}\]

13.2.3 Solving the equation of motion

In Ch. 5 we discussed how first-order ordinary differential equations (ODE) can be solved by integration. The ODE in Equation 13.7 can however cannot be easily solved by integration. Instead the strategy \({ }^{1}\) to solve this ODE is to find a trial solution \(x_{A}(t)\) of the right form with unknown parameters, substitute it into the ODE and use the ODE and initial conditions to determine the parameters. Since the ODE in Equation 13.7 contains \(y_{A}\) and its second derivative \(\ddot{y}_{A}\) it is logical to look for a trial solution that is a constant times its own second derivative, like \(\cos \left(\omega_{n} t\right)\) or \(e^{\lambda t}\).

Before starting with the solution, it is important to note that Equation 13.7 is a homogeneous linear differential equation, because it only sums linear functions of \(y_{A}\) and its time-derivatives, which are equal to zero. These homogeneous linear ODE have the following useful property: if we find two different solutions of this ODE, \(y_{A, 1}(t)\) and \(y_{A, 2}(t)\), then substituting each of them individually in Equation 13.7 gives zero, so their sum \(y_{A}=y_{A, 1}+y_{A, 2}\) will also be a solution of the ODE, and summing many times is the same as multiplying, so \(y_{A}=c_{1} \times y_{A, 1}\) is also a solution. These summative and multiplicative properties of the solutions of the ODE simplify the analysis of the motion significantly. Moreover, they allow solving the EoM with complex numbers, since if a complex function \(y_{A}=y_{A, r}+i y_{A, i}\) is a solution of the ODE, the functions \(y_{A, r}\) and \(y_{A, i}\), which are the real and imaginary part of \(y_{A}\), are also solutions of the ODE individually.

These ODEs can either be solved by trigonometric functions, or by complex functions. If you are familiar with working with complex numbers, they often make it simpler to solve the ODE. However, to make the reader familiar with both methodologies, we provide both the trigonometric and complex solutions in the derivation.

Derivation: Free undamped vibration of a mass spring system

Since we look for a function that is proportional to its own second derivative, we use a trial solution of the form \(y_{A}(t)=A \cos \left(\omega_{n} t+\varphi_{0}\right)=\Re A_{c} e^{i \omega_{n} t}\) where \(\Re\) stands for the real part of the function and \(A_{c}=A e^{i \varphi_{0}}\). The value of the parameters \(A, \varphi_{0}\) and \(\omega_{n}\) are unknown at this stage. Substituting this trial solution into the ODE Equation 13.1 we obtain:

\[\begin{align} & m_{A} \ddot{y}_{A}+k y_{A}=0 \tag{13.8} \label{13.8}\\[4pt] &\left(-m_{A} \omega_{n}^{2}+k\right) A \cos \left(\omega_{n} t+\varphi_{0}\right)=0 \tag{13.9} \label{13.9}\\[4pt] & \text { Trigonometric method } \\[4pt] &\left(-m_{A} \omega_{n}^{2}+k\right) \Re A_{c} e^{i \omega_{n} t}=0 \tag{13.10} \label{13.10}\\[4pt] & \text { Complex method } \tag{13.11} \label{13.11}\\[4pt] &\left(k-\omega_{n}^{2} m_{A}\right) y_{A}=0\end{align}\]

Here we used that \(\dot{y}_{A}=-A \omega_{n} \sin \left(\omega_{n} t+\varphi_{0}\right)\) and \(\ddot{y}_{A}=-A \omega_{n}^{2} \cos \left(\omega_{n} t+\varphi_{0}\right)\), and similarly \(\dot{y}_{A}=\Re\left[i A_{c} \omega_{n} e^{i \omega_{n} t}\right]\) and \(\ddot{y}_{A}=\Re\left[-A_{c} \omega_{n}^{2} e^{i \omega_{n} t}\right]\). From Equation 13.11 we see that the proposed trial function is a solution if \(y_{A}=0\), and if \((k-\) \(\left.m_{A} \omega_{n}^{2}\right)=0\). The solution \(y_{A}=0\) is the static particular equilibrium solution that we derived before, so we now find the solution of the EoM.

Concept: General Solution

General solution \(y_{A}(t)\) of the equation of motion for free undamped vibration:

\[\begin{align} y_{A}(t) & =A \cos \left(\omega_{n} t+\varphi_{0}\right)=\Re A_{c} e^{i \omega_{n} t} \tag{13.12} \label{13.12}\\[4pt] k-\omega_{n}^{2} m_{A} & =0 \tag{13.13} \label{13.13}\\[4pt] \omega_{n} & =\sqrt{\frac{k}{m_{A}}} \quad \text { Natural angular resonance freq. } \tag{13.14} \label{13.14}\end{align}\]

The motion for free undamped vibration \(y_{A}(t)\) is plotted in Figure 13.3.

13.2.4 Resonance frequency, period and phase

So we have found that the trial solution only satisfies the ODE for a specific value of \(\omega_{n}=\sqrt{k / m_{A}}\), which is called the natural angular resonance frequency of the system and tells us something about the period of the vibration. The period \(T\) after which the vibration repeats itself is found from \(\cos \left(\omega_{n} t+2 \pi\right)=\) \(\cos \left(\omega_{n}(t+T)\right)\) from we find

\[\begin{align} T & =\frac{2 \pi}{\omega_{n}} \quad \text { Period } \tag{13.15} \label{13.15}\\[4pt] f_{n} & =\frac{1}{T}=\frac{\omega_{n}}{2 \pi} \quad \text { Natural resonance frequency } \tag{13.16} \label{13.16}\end{align}\]

Note that the found expression \(y_{A}\) is a solution of the ODE for any value of the complex amplitude \(A_{c}\), which can be written as \(A_{c}=A e^{i \varphi_{0}}\), where the real number \(A=\left|A_{c}\right|\) is the amplitude or magnitude of the vibration and the real number \(\varphi_{0}\) is its phase at \(t=0\). In some cases it is more convenient to write the solution as a sum of a cosine and sine function with prefactors \(A_{1}\) and \(A_{2}\) as follows:

\[y_{A}(t)=A_{1} \cos \left(\omega_{n} t\right)+A_{2} \sin \left(\omega_{n} t\right) \tag{13.17} \label{13.17}\]

Figure 13.3: The free undamped vibration \(y_{A}(t)\) of a mass-spring system and its period \(T\), amplitude \(A\) and the constants \(A_{1}\) and \(A_{2}\) indicated.

It can be shown that this equation is identical to that found in Equation 13.12 by substituting \(A_{c}=e^{i \varphi_{0}}\) and using Euler’s equation \(e^{i x}=\cos x+i \sin x\) and taking the real part of the complex function:

\[\begin{align} y_{A}(t) & =\Re A\left[\cos \left(\varphi_{0}\right)+i \sin \left(\varphi_{0}\right)\right]\left[\cos \left(\omega_{n} t\right)+i \sin \left(\omega_{n} t\right)\right] \tag{13.18} \label{13.18}\\[4pt] A_{1} & =A \cos \left(\varphi_{0}\right) \tag{13.19} \label{13.19}\\[4pt] A_{2} & =-A \sin \left(\varphi_{0}\right) \tag{13.20} \label{13.20}\\[4pt] A & =\sqrt{A_{1}^{2}+A_{2}^{2}} \quad \text { Amplitude } \tag{13.21} \label{13.21}\\[4pt] \varphi_{0} & =-\arctan \left(A_{2} / A_{1}\right) \quad \text { Phase at } t=0 \tag{13.22} \label{13.22}\end{align}\]

The last four equations, were obtained by comparing Equation 13.17 and Equation 13.18. They can be used to relate the amplitude \(A\) and phase \(\varphi_{0}\) to \(A_{1}\) and \(A_{2}\) and vice versa. In Figure 13.4 the relations between the motion \(y_{A}(t)\) and these four constants are visualised by a point moving along a circle with radius \(A\).

Figure 13.4: The free undamped vibration \(y_{A}(t)\) of a mass-spring system visualised by a point moving at angular velocity \(\omega_{n}\) along a circle with radius \(A\) (green arrow). The initial angle \(\phi_{0}\) at \(t=0\) and its relation to \(A_{1}\) and \(A_{2}\) is indicated.

13.2.5 Initial conditions

In the previous section we have found the solution of the ODE, however there are still two unknown constants \(A\) and \(\varphi_{0}\) (or \(A_{1}\) and \(A_{2}\) ) that need to be found to know \(y_{A}(t)\) at all times. Similar to the time integration in Equation 5.34, we now also need initial conditions \(y_{A}\left(t_{1}\right)=y_{1}\) and \(\dot{y}_{A}\left(t_{1}\right)=v_{1}\) to determine the unknown constants as follows:

1. Determine the function \(y_{A}(t)\) and its time derivative \(\dot{y}_{A}\).
2. Substitute \(t_{1}\) and write down the two equations \(y_{A}\left(t_{1}\right)=y_{1}\) and \(\dot{y}_{A}\left(t_{1}\right)=\) \(v_{1}\).
3. Solve the two equations for the two unknowns, \(A_{1}\) and \(A_{2}\) or \(A\) and \(\varphi_{0}\).

S Example 13.1 As an example let us consider the case where the initial conditions, position \(y_{A}(0)=y_{0}\) and velocity \(\dot{y}_{A}(0)=v_{0}\) at \(t_{1}=0\) are known. We choose to use Equation 13.17, \(y_{A}(t)=A_{1} \cos \omega_{n} t+A_{2} \sin \omega_{n} t\) and obtain the constants \(A_{1}\) and \(A_{2}\) as
follows:

\[\begin{align} y_{A}(0) & =A_{1}=y_{0} \tag{13.23} \label{13.23}\\[4pt] \dot{y}_{A}(t) & =-\omega_{n} A_{1} \sin \omega_{n} t+\omega_{n} A_{2} \cos \omega_{n} t \tag{13.24} \label{13.24}\\[4pt] \dot{y}_{A}(0) & =\omega_{n} A_{2}=v_{0} \tag{13.25} \label{13.25}\\[4pt] A_{2} & =v_{0} / \omega_{n} \tag{13.26} \label{13.26}\end{align}\]

By combining this with Equation 13.4 and Equation 13.17 we find for the case that the initial position \(y_{0}\) and velocity \(v_{0}\) are known that:

\[\begin{align} & y_{A, \text { tot }}(t)=A_{1} \cos \left(\omega_{n} t\right)+A_{2} \sin \left(\omega_{n} t\right)+y_{A, \mathrm{st}} \tag{13.27} \label{13.27}\\[4pt] & y_{A, \text { tot }}(t)=y_{0} \cos \left(\omega_{n} t\right)+v_{0} / \omega_{n} \sin \left(\omega_{n} t\right)+y_{A, \mathrm{st}} \tag{13.28} \label{13.28}\end{align}\]

If required \(A\) and \(\phi_{0}\) can be determined using Equation 13.21 and Equation 13.22