13.4: Forced vibrations
After having considered free vibrations we now turn to forced vibrations. In a forced vibration there is an additional time-dependent external force \(\overrightarrow{\boldsymbol{F}}(t)\) acting on the mass. These forced vibrations are important in examples like clocks, vibrations in cars driving over a bumpy road, the beating of your heart, audio speakers, and electromechanical frequency generators for data transmission in your mobile phone.
Figure 13.8 shows a sketch and FBD of a mass-spring-damper system driven by a time-dependent force \(F(t)\) in the \(y\)-direction:
\[\overrightarrow{\boldsymbol{F}}(t)=F_{0} \Re e^{i \omega t} \hat{\boldsymbol{\jmath}}=F_{0} \cos (\omega t) \hat{\boldsymbol{\jmath}} \tag{13.53} \label{13.53}\]
Similar to Equation 13.1 we obtain the equation of motion by projection of the forces on the \(y\)-axis and using Newton’s second law:
\[\sum F_{y}=m_{A} g-k\left(y_{A, \text { tot }}-L_{0}\right)-c \dot{y}_{A, \text { tot }}+F_{0} \Re e^{i \omega t}=m_{A} \ddot{y}_{A, \text { tot }} \tag{13.54} \label{13.54}\]
13.4.1 Determining the equilibrium position
Like for the free vibration we first determine the static equilibrium position by setting all time derivatives to zero: \(\ddot{y}_{A, \text { tot }}=\dot{y}_{A, \text { tot }}=0\). Note, that if the average value of the external force \(F_{\text {avg }}\) is not zero, it needs to be included in the determination of the static equilibrium position. However, for a cosine function \(F_{\text {avg }}=0\). We find for the static equilibrium position in the presence of vibrations:
\[\begin{align} m_{A} g-k\left(y_{A, \mathrm{st}}-L_{0}\right)+F_{\text {avg }} & =0 \tag{13.55} \label{13.55}\\[4pt] y_{A, \mathrm{st}} & =\frac{m_{A} g}{k}+L_{0}+\frac{F_{\text {avg }}}{k} \tag{13.56} \label{13.56}\end{align}\]
So for \(F(t)=F_{0} \cos \omega t\), the static equilibrium position \(y_{A, \mathrm{st}}\) is identical to that found for free vibrations.
13.4.2 Equation of motion with respect to equilibrium
Similar to the case of free vibrations, we substitute \(y_{A, \text { tot }}(t)=y_{A}(t)+y_{A, \mathrm{st}}\) into Equation 13.54, such that all static terms cancel and we obtain the EoM for the displacement \(y_{A}(t)\) of the mass with respect to the equilibrium position:
\[\Re\left[m_{A} \ddot{y}_{A}+c \dot{y}_{A}+k y_{A}=F_{0} e^{i \omega t}\right] \tag{13.57} \label{13.57}\]
We see that the only difference between the EoM for forced and free vibrations is the forcing term \(F_{0} \Re e^{i \omega t}\).
13.4.3 Particular and homogeneous solutions of the EoM
Let us assume we find a particular solution \(y_{A, f}(t)\) of the differential equation (13.57). Then we can add a solution \(y_{A, h}(t)\) of the corresponding homogeneous \(\mathrm{EoM}, m_{A} \ddot{y}_{A}+c \dot{y}_{A}+k y_{A}=0\). And then the resulting function \(y_{A}(t)=\) \(y_{A, f}(t)+y_{A, h}(t)\) is still a solution of the EoM Equation 13.57 as can be seen here:
\[\Re\left[m_{A}\left(\ddot{y}_{A, f}+\ddot{y}_{A, h}\right)+c\left(\dot{y}_{A, f}+\dot{y}_{A, h}\right)+k\left(y_{A, f}+y_{A, h}\right)=F_{0} e^{i \omega t}\right] \tag{13.58} \label{13.58}\]
The terms with \(y_{A, f}(t)\) add up to \(F_{0} e^{i \omega t}\) and the terms with \(y_{A, h}\) add up to zero. The function \(y_{A, f}(t)\) is called the particular solution of the ODE and \(y_{A, h}(t)\) is the homogeneous (or complementary) solution of the EoM for forced vibrations. We see that the differential equation for \(y_{A, h}(t)\) is identical to that of a free damped vibration. So, we can just use the methods from the previous section to analyse these solutions. Moreover, due to the damping, these vibrations damp out after waiting for a sufficiently long time after which we are left with \(y_{A, f}(t)\), which is therefore also called the steady-state solution of the forced damped vibration EoM, while \(y_{A, h}(t)\) is called the transient (part of the) solution. We note that \(y_{A, h}(t)\) has two free constants \(A_{h}\) and \(\varphi_{0, h}\) that still need to be determined using the initial conditions. The procedure to determine these constants using the initial conditions is identical to that explained in section Sec. 13.2.5, although the equations for position and velocity are longer, since they contain both the steady-state and transient part of the solution.
13.4.4 Solving the EoM for forced vibrations
Since the transient solutions \(y_{A, h}(t)\) were already found from the free vibration analysis, we only need to solve the particular, steady-state solution \(y_{A, f}(t)\).
Derivation. Forced damped vibration
To solve the EoM we are looking for an equation that is proportional to its own time-derivatives and also proportional to the forcing function. Therefore we attempt using a trial solution of the following form:
\[y_{A}(t)=\Re A_{c} e^{i \omega t} \tag{13.59} \label{13.59}\]
Note that in contrast to the case of free vibrations, we do not solve for an exponent \(\lambda\) that gives us a (damped) natural frequency, but instead we take the forcing frequency \(\omega\) directly into the trial function \(y_{A}\). Let us now show that this trial function is indeed a solution of the differential equation and determine the complex amplitude \(A_{c}\) by substituting it into the EoM Equation 13.57:
\[\Re\left[A_{c} e^{i \omega t}\left(-\omega^{2} m_{A}+i \omega c+k\right)=F_{0} e^{i \omega t}\right] \tag{13.60} \label{13.60}\]
We see from this equation that our choice to choose \(e^{i \omega t}\) as exponential function was correct, since we can now divide both sides of this equation by \(e^{i \omega t}\) to eliminate it. Then the complex amplitude \(A_{c}\) needs to obey this equation to ensure \(y_{A}(t)\) is a solution at all times:
\[A_{c}\left(k-\omega^{2} m_{A}+i \omega c\right)=F_{0} \tag{13.61} \label{13.61}\]
By dividing both sides by the term between brackets we find the complex amplitude \(A_{c}\) of the vibration:
\[A_{c}=\frac{F_{0}}{k-\omega^{2} m_{A}+i \omega c} \tag{13.62} \label{13.62}\]
Now we would like to express the complex amplitude as a product of a real amplitude \(A\) and phase \(\varphi_{0}\), such that \(A_{c}=A e^{i \varphi_{0}}\). We find for the amplitude:
\[\begin{align} A & =\left|A_{c}\right|=\sqrt{A_{c} A_{c}^{*}} \tag{13.63} \label{13.63}\\[4pt] & =\left[\frac{F_{0}}{\left(k-\omega^{2} m_{A}\right)+i \omega c} \cdot \frac{F_{0}}{\left(k-\omega^{2} m_{A}\right)-i \omega c}\right]^{1 / 2} \tag{13.64} \label{13.64}\\[4pt] & =\frac{F_{0}}{\sqrt{\left(k-\omega^{2} m_{A}\right)^{2}+(\omega c)^{2}}}=\frac{F_{0} / m_{A}}{\sqrt{\left(\omega_{n}^{2}-\omega^{2}\right)^{2}+\frac{\omega^{2} \omega_{n}^{2}}{Q^{2}}}} \tag{13.65} \label{13.65}\end{align}\]
where we used Equation 2.34 and \(Q=k /\left(\omega_{n} c\right)\). From this equation, that is plotted in Figure 13.10 we see that for small values of damping \(c\), the amplitude approximately has a maximum if the driving frequency \(\omega\) is equal to the natural frequency \(\omega_{n}=\sqrt{k / m_{A}}\), in this situation the system is driven at resonance.
The amplitude when driving at its natural frequency \(\omega=\omega_{n}\) is \(A=F_{0} Q / k\), while it is \(A=F_{0} / k\) at low frequencies. The ratio between these two numbers is called the quality factor \(Q=k /\left(\omega_{n} c\right)\) and is a measure of how many periods it takes for the resonant vibrations to dampen out if the driving force is removed. At critical damping \(Q_{c}=\sqrt{k m_{A}} / c_{c}=\frac{1}{2}\) and for the undamped system with \(c=0\) the amplitude and \(Q\)-factor becomes infinite when it is driven at resonance. At very high driving frequencies \(\omega\) the amplitude tends to zero because the inertia of the mass prevents its motion to follow the driving force.
Near and above the resonance frequency \(\omega_{n}\), a phase difference \(\varphi_{0}\) develops, such that the motion is delayed with respect to the driving force, as indicated in Figure 13.8 and Figure 13.11. The value of this phase shift can be found by taking the ratio of the imaginary and real parts of \(A_{c}=A e^{i \varphi_{0}}\) and Equation 13.62:
\[\tan \left(\varphi_{0}\right)=\frac{\Im A_{c}}{\Re A_{c}}=\frac{-\omega c}{k-\omega^{2} m_{A}} \tag{13.66} \label{13.66}\]
We see that at low driving frequencies \(\omega\) the motion is in phase with the force because the phase difference \(\varphi_{0}=0\). At larger driving frequencies the phase difference \(\varphi_{0}\) becomes negative, indicating that the motion lags behind the force. This continues up to the natural resonance frequency, where \(\omega^{2}=\omega_{n}^{2}=k / m_{A}\) and \(\varphi_{0}=-\pi / 2\), for which the motion is -90 degrees out of phase, and the velocity is in phase with the force. At even higher frequencies, the phase lag between motion and force increases more, until it reaches a phase difference close to -180 degrees with \(\varphi_{0}=-\pi\) at very high driving frequencies \(\omega \gg \omega_{n}\).
Steady-state solution for damped forced vibrations
We have now fully determined the complex amplitude \(A_{c}=A e^{i \varphi_{0}}\), with \(A\) being given by Equation 13.65 and \(\varphi_{0}\) by Equation 13.66, and thus we have found the function \(y_{A, f}(t)\) in Equation 13.59 and proven that it is a solution of the EoM in Equation 13.57.
\[\begin{align} y_{A, f}(t) & =\Re A e^{i\left(\omega t+\varphi_{0}\right)}=A \cos \left(\omega t+\varphi_{0}\right) \tag{13.67} \label{13.67}\\[4pt] A & =\frac{F_{0} / m_{A}}{\sqrt{\left(\omega_{n}^{2}-\omega^{2}\right)^{2}+\frac{\omega^{2} c^{2}}{m_{A}^{2}}}} \tag{13.68} \label{13.68}\\[4pt] \varphi_{0} & =\arctan \frac{-\omega c}{k-\omega^{2} m_{A}} \tag{13.69} \label{13.69}\end{align}\]
In contrast to free damped vibrations, the steady-state solution Equation 13.67 is the same for underdamped, overdamped and critically damped systems.
Steady state solution for forced undamped vibrations
An important special case is the situation in which there is no damper connected to the mass, such that \(c=0\). In this situation, the system will exhibit undamped forced vibrations. The solution of this EoM can be easily obtained from our previous analysis by just setting \(c=0\). It is seen from Equation 13.65 and Equation 13.66, that then \(A=\frac{F_{0}}{m_{A}}\left|\omega_{n}^{2}-\omega^{2}\right|^{-1}\) and \(\varphi_{0}=0\) for \(\omega<\omega_{n}\) and \(\varphi_{0}=-\pi\) for \(\omega>\omega_{n}\). We therefore have for the solution of the EoM for forced undamped vibrations:
\[y_{A, f u}(t)=\frac{F_{0}}{m_{A}\left(\omega_{n}^{2}-\omega^{2}\right)} \cos (\omega t) \tag{13.70} \label{13.70}\]
A special feature of undamped vibration is that its amplitude goes to infinity if the driving frequency equals the natural resonance frequency \(\left(\omega=\omega_{n}\right)\), see Figure 13.10 for \(Q=\infty\). In practice such infinite amplitudes never occur and a small damping constant \(c\) is always present.
13.4.5 More complex systems involving forced vibrations
The same methodology we have just derived for mass-spring-damper systems, can be extended to rigid bodies and other more complex systems for which the kinetic analysis yields an EoM that has the form of Equation 13.57:
\[\Re\left[m_{e} \ddot{y}_{e}+c_{e} \dot{y}_{e}+k_{e} y_{e}=F_{0, e} e^{i \omega t}\right] \tag{13.71} \label{13.71}\]
For analyzing such systems one first determines the equations for \(y_{e}, F_{0, e}\), \(m_{e}, c_{e}\) and \(k_{e}\) by comparing Equation 13.71 to the EoM and then analyses the vibrations in an exactly identical way as for the mass-spring-damper system (see also Sec. 13.3.4).
13.4.6 Forced vibration: solution procedure
Based on the previous analysis, let us summarise the step-by-step procedure for analysing forced vibrations and obtaining their motion.
- 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 and simplify the scalar equations of motion, reducing it to a single scalar EoM.
- Determine the static equilibrium position \(y_{A, \mathrm{st}}\) by setting all time derivatives in the EoM to zero \(\ddot{y}_{A, \text { tot }}=\dot{y}_{A, \text { tot }}=0\), setting the forcing term \(F(t)\) to its time-averaged value \(F_{\text {avg }}\) and solving for \(y_{A}\).
- Define the time dependent part of the motion as \(y_{A}(t)=y_{A, \text { tot }}-y_{A, \mathrm{st}}\).
- The system will show forced vibration if the EoM can be written in the form \(m_{e} \ddot{y}_{e}+c_{e} \dot{y}_{e}+k_{e} y_{e}=F_{0, e} \cos (\omega t)\).
- Determine the equations for \(y_{e}, F_{0, e}, m_{e}, c_{e}\) and \(k_{e}\) in terms of the given variables, constants or numerical values.
- Substitute a trial function of the form \(y_{e}=\Re A_{c} e^{i \omega t}\) into the EoM by taking the first and second derivative.
- Obtain the equation \(A_{c}\left(-m_{e} \omega^{2}+i c_{e} \omega+k_{e}\right)=F_{0, e}\) and determine the complex steady-state amplitude \(A_{c}\).
- Use \(A_{c}=A e^{i \varphi_{0}}\) to convert the complex prefactor to a real amplitude \(A\) and phase \(\varphi_{0}\).
- If \(c_{e}=0\), then we are dealing with forced undamped vibration. In that case \(\varphi_{0}=0\) such that the steady-state solution of the EoM is \(y_{e}(t)=A \cos (\omega t)\), with \(A=F_{0, e} /\left(k_{e}-m_{e} \omega^{2}\right)\).
- If \(c_{e} \neq 0\) we are dealing with forced damped vibration. The complex amplitude can be converted into a real amplitude \(A\) and phase \(\varphi_{0}\) using the relation \(A_{c}=A e^{i \varphi_{0}}\).
The steady-state solution of the EoM is: \(y_{e, f}(t)=A \cos \left(\omega t+\varphi_{0}\right)\).
The amplitude of the steady-state solution is:
\(A=F_{0, e} / \sqrt{m_{e}^{2}\left(\omega_{n}^{2}-\omega^{2}\right)^{2}+\omega^{2} c_{e}^{2}}\).
The phase difference between force and motion is:
\(\varphi_{0}=\arctan \left[-\omega c_{e} /\left(k_{e}-\omega^{2} m_{e}\right)\right]\).
- Solutions \(y_{e, h}(t)\) of the homogeneous EoM
\(m_{e} \ddot{y}_{e, h}+c_{e} \dot{y}_{A, h}+k_{e} y_{A, h}=0\) for free vibrations can be summed to the steady state solution \(y_{A, f}(t)\) to obtain new solutions of the EoM. For damped vibrations \((c>0)\), the function \(y_{A, h}(t)\) reduces to zero after sufficient time, it is therefore called the transient part of the solution. In general, any solution of the forced EoM can always be expressed as \(y_{A}(t)=y_{A, f}(t)+y_{A, h}(t)\).
- Two initial conditions of the form \(y_{A}\left(t_{1}\right)=y_{1}\) or \(\dot{y}_{A}\left(t_{2}\right)=v_{2}\) need to be solved to determine the two unknown constants in the transient part of the solution \(y_{A, h}(t)\) (see Sec. 13.2.5). Note that for forced vibrations the full motion \(y_{A}(t)\) including both the steady-state and transient parts needs to be used to determine these constants.
13.4.7 More complex driving forces*
Up to now we have only dealt with a single cosine shaped forcing function. However, in practice more complex periodic forcing functions can act, which are a sum of multiple cosine functions, each with a different driving frequency \(\omega_{j}\) and a different complex amplitude \(F_{j}\), and with the total forcing function \(F(t)=\Re \sum_{j=1}^{j=N} F_{j} e^{i \omega_{j} t}\), where there are \(N\) terms such that \(j\) runs from \(j=1 \ldots N\). The methodology derived in this section can be used to find a solution \(y_{A, j}\) for each of these driving frequencies and then add all solutions to find a solution for the full EoM. To satisfy the EoM each of these functions \(y_{A, j}\) needs to satisfy an ODE of the form:
\[m_{A} \ddot{y}_{A, j}+c \dot{y}_{A, j}+k y_{A, j}=F_{j} e^{i \omega_{j} t} \tag{13.72} \label{13.72}\]
From each of these \(N\) ODEs, with \(y_{A, j}=A_{c, j} e^{i \omega_{j} t}\) we determine the complex motion amplitude \(A_{c, j}\) using the corresponding complex force amplitude \(F_{j}\) and frequency \(\omega_{j}\) with Equation 13.62, \(A_{c, j}=F_{j} /\left(k-\omega_{j}^{2} m_{A}+i \omega_{j} c\right)\). By summing we find a steady-state solution of the form:
\[y_{A, f}(t)=\Re\left[\sum_{j} A_{c, j} e^{i \omega_{j} t}\right] \tag{13.73} \label{13.73}\]
Since, according to Fourier analysis, any periodic time-dependent function can be written as \(F(t)=\Re \sum_{j} F_{j} e^{i \omega_{j} t}\), this procedure allows to analyse the vibrations and motion of a system in response to any periodic force. Note that driving with two or more frequencies can result in beating, like shown in Figure 2.1.
13.4.8 Vibrations of rotating rigid bodies
As mentioned in Sec. 13.4.5, everything discussed in this chapter can also be used to analyse vibrations of rotating rigid bodies. We illustrate this by an example which includes a massive pulley as rigid body and also shows how to reduce the equation of motion of multiple objects to a single one by constraint equations.
The sketch of the problem is given in Figure 13.12. A point mass \(B\) with mass \(m_{B}\) is suspended by a rope via a pulley \(A\) that has moment of inertia \(I_{G}\), radius \(R\) and mass \(m_{A}\) (so it is different from the massless pulleys we have considered earlier). The other side of the rope is connected to a spring with stiffness \(k\). The rope does not slip and is always tight under tension. When the system is in static equilibrium, the mass \(B\) is at height \(y_{B}=h_{e}\). We define \(y_{e}=y_{B}-h_{e}\) as the displacement of \(B\) with respect to this equilibrium position. \(\theta_{e}\) is the angular displacement of the pulley with respect to the equilibrium position. \(y_{C}\) is the displacement of point \(C\) with respect to the equilibrium position.
Question: Determine the rotational motion \(\theta_{e}(t)\) of the pulley \(A\) with respect to the equilibrium position when the system is released from rest at \(\theta_{e}=\theta_{e 0}\).
Solution
- Draw the FBD of \(A\) and \(B\) shown in Figure 13.13. Don’t forget the normal force \(\overrightarrow{\boldsymbol{F}}_{N}\) and gravitational force \(\overrightarrow{\boldsymbol{F}}_{g}\) on the pulley. The rope generates forces \(\overrightarrow{\boldsymbol{F}}_{r B}\) and \(\overrightarrow{\boldsymbol{F}}_{r C}\) on points \(B\) and \(C\) respectively and by Newton’s third law the opposite forces act on the pulley. Ask yourself: why are the forces \(\overrightarrow{\boldsymbol{F}}_{r B}\) and \(\overrightarrow{\boldsymbol{F}}_{r C}\) different? The answer is: they are different because a moment is needed to provide the angular acceleration of the disk of the pulley according to Euler’s second law, since the pulley is not massless. If the forces would be equal this moment would be zero. This angular acceleration is provided via the static friction force between the disk and the rope, which does not slip. In the FBD we directly provide the equations to expand the vectors into their scalar components along the \(y\)-axis.
- Determine the constraint equations from the rope. Since the motion with respect to the equilibrium position is asked, we do not have to determine the static equilibrium position anymore. Since the rope has constant length and does not slip, and because in equilibrium all coordinates are zero \(y_{e}=y_{C}=\) \(\theta_{e}=0\), we have 2 constraint equations. We also write down the equation for the spring force \(F_{r C}\) of the rope on point \(C\) :
\[\begin{align} y_{e} & =-\theta_{e} R \tag{13.74} \label{13.74}\\[4pt] y_{C} & =+\theta_{e} R \tag{13.75} \label{13.75}\\[4pt] F_{r C} & =k y_{C} \tag{13.76} \label{13.76}\end{align}\]
- Now we write down the equation of motion for mass \(B\) along the \(y\)-axis, with respect to equilibrium:
\[\sum F_{B y}=F_{r B}=m_{B} \ddot{y}_{e} \tag{13.77} \label{13.77}\]
Note that the gravitational force is excluded, since it does not contribute to the time-dependent part of the force.
-
It is seen that the motion of the pulley is a pure rotation around its centre. Due to the pin around which it rotates, there is no translation of its CoM, such that we only need to consider the rotational EoM of the pulley, i.e. Euler’s
second law. As reference point we choose the centre of the pulley \(G\).
\[\begin{align} \sum M_{A z} & =I_{G} \ddot{\theta}_{e} \hat{\boldsymbol{k}} \tag{13.78} \label{13.78}\\[4pt] \overrightarrow{\boldsymbol{r}}_{B / G} \times\left(-\overrightarrow{\boldsymbol{F}}_{r B}\right)+\overrightarrow{\boldsymbol{r}}_{C / G} \times\left(-\overrightarrow{\boldsymbol{F}}_{r C}\right) & =I_{G} \ddot{\theta}_{e} \hat{\boldsymbol{k}} \tag{13.79} \label{13.79}\\[4pt] (-R \hat{\boldsymbol{\imath}}) \times\left(-F_{r B} \hat{\boldsymbol{\jmath}}\right)+(R \hat{\boldsymbol{\imath}}) \times\left(-F_{r C} \hat{\boldsymbol{\jmath}}\right) & =I_{G} \ddot{\theta}_{e} \hat{\boldsymbol{k}} \tag{13.80} \label{13.80}\end{align}\]
Now we use Eqs. (13.74)-(13.77) to simplify the EoM to a single scalar equation in \(\theta_{e}\) :
\[\begin{align} \left(R m_{B} \ddot{y}_{e}-R k y_{C}\right) \hat{\boldsymbol{k}} & =I_{G} \ddot{\theta}_{e} \hat{\boldsymbol{k}} \tag{13.81} \label{13.81}\\[4pt] -R^{2} m_{B} \ddot{\theta}_{e}-R^{2} k \theta_{e} & =I_{G} \ddot{\theta}_{e} \tag{13.82} \label{13.82}\\[4pt] \left(\frac{1}{2} m_{A}+m_{B}\right) R^{2} \ddot{\theta}_{e}+R^{2} k \theta_{e} & =0 \tag{13.83} \label{13.83}\end{align}\]
In the last step we used that the moment of inertia of a disk around its centre is \(I_{G}=\frac{1}{2} m R^{2}\).
- We see that the EoM resembles a free undamped vibration \(m_{e} \ddot{y}_{e}+k_{e} y_{e}=0\), with \(y_{e}=\theta_{e}, m_{e}=\left(\frac{1}{2} m_{A}+m_{B}\right) R^{2}\) and \(k_{e}=R^{2} k\). So, as derived in Equation 13.12 we have as the solution:
\[\begin{align} \theta_{e}(t) & =A \cos \left(\omega_{n} t+\varphi_{0}\right) \tag{13.84} \label{13.84}\\[4pt] \omega_{n} & =\sqrt{\frac{k_{e}}{m_{e}}}=\sqrt{\frac{k}{\frac{1}{2} m_{A}+m_{B}}} \tag{13.85} \label{13.85}\end{align}\]
- Since the vibration starts from rest, we need \(\dot{\theta}_{e}(0)=0\), from which we find \(\varphi_{0}=0\). Then the initial angle is found to be \(A=\theta_{e 0}\) because \(\cos (0)=1\), obtaining the final solution:
\[\theta_{e}(t)=\theta_{e 0} \cos \left(\sqrt{\frac{k}{\frac{1}{2} m_{A}+m_{B}}} t\right) \tag{13.86} \label{13.86}\]