Find the angular displacement $\theta_1(t)$ and $\theta_2(t)$ of the system shown in figure below for the initial conditions $\theta_1(0)$, $\theta_2(0)$, and $\dot{\theta}_1(0) = \dot{\theta}_2(0) = 0$. enter image description here

The equation of motions are (I am highly suspect of my equations of motion) \begin{align*} J_1\ddot{\theta}_1 &= k_t(\theta_2 - \theta_1)\\ J_2\ddot{\theta}_2 &= -k_t(\theta_2 - \theta_1) \end{align*} Therefore, the matrix equation is $$ \begin{bmatrix} k_t - \omega^2J_1 & -k_t\\ -k_t & k_t - \omega^2J_2 \end{bmatrix} \begin{bmatrix} \theta_1\\ \theta_2 \end{bmatrix} = \mathbf{0} $$ when we let $\theta_i(t) = \theta_i\cos(\omega t + \phi)$. The determinant of the matrix $$ \det\Biggl( \begin{bmatrix} k_t - \omega^2J_1 & -k_t\\ -k_t & k_t - \omega^2J_2 \end{bmatrix} \Biggr) = J_1J_2\omega^4 - k_t(J_1 + J_2)\omega^2 = 0 $$ Then the natural frequencies are $$ \omega_{1,2} = \pm\frac{\sqrt{k_t(J_2+J_1)}}{\sqrt{J_1J_2}} $$ but the natural frequency isn't negative so that only leads to $\omega_1 = \frac{\sqrt{k_t(J_2+J_1)}}{\sqrt{J_1J_2}}$ so this doesn't seem correct that I have only one natural frequency for two equations.

  • $\begingroup$ Maybe Engineering is a better home for this question actually. $\endgroup$ Apr 8, 2015 at 16:50

1 Answer 1


Your time equations of motion look good. Most of what you have done is right, there are some subtleties to the meaning of the characteristic equation $ J_1J_2\omega^4 - k_t(J_1 + J_2)\omega^2 = 0$ (which you've gotten right) and you need to look at the physics and its relation to the initial conditions.

Assuming the real, trigonometric form is a headache, because you need to assume two different phase angles for the two different $\theta$s, i.e. you need to assume $\theta_i(t) = \hat{\theta}_i\cos(\omega\,t+\phi_i)$. So you don't get the cosine functions dropping out of your equations quite like you imagine. What you need is to assume functions of the form $\theta_i(t) = \hat{\theta}_i\exp(-i\,\omega\,t)$, so that the phase angle can be encoded in the phase of the complex scaling constants $\hat{\theta}_i$. When you use this form, you then find two of your natural frequencies, exactly as you have derived. There is indeed both a positive and negative natural frequency; both must be present so that we can build real valued trigonometric functions out of the complex exponentials: any sum of solutions is also a solution for these linear equations. But there are also two solutions $\omega^2 = 0$. Thus there must be a constant term in the solution as well (corresponding to the zero frequency solution). Actually, the repeated root means that a linear function of time will fulfill the equation (try solutions of the form $\theta_j(t)=A_j\,t+B_j$; these are solutions as long as $A_1=A_2$ and $B_1 = B_2$). Because we know the solution must be real for all $t$, the solutions must be of the form:

$$\theta_1(t) = \hat{\theta}_1\,\exp(-i\,\omega_0\,t) + \hat{\theta}_1^*\,\exp(+i\,\omega_0\,t) + A\,t+B$$ $$\theta_2(t) = \hat{\theta}_2\,\exp(-i\,\omega_0\,t) + \hat{\theta}_2^*\,\exp(+i\,\omega_0\,t) + A\,t+B$$

where $\omega_0 = \sqrt{\frac{k_t\,(J_1+J_2)}{J_1\,J_2}}$ as you found and $A\,B$ are real. Notice that $B$ corresponds to a constant angular offset of the system, and $A$ a constant angular speed of the system about its axis: you can take any general solution you get and set the system in uniform rotational motion on top of that solution and the total motion will still be a general solution.

Since $\dot{\theta}_j(0) = 0$ we get

$$i\,\omega_0(\hat{\theta}_j^*-\hat{\theta}_j) + A=0$$

so that:


To find the value of $A$, we take heed that the system begins with an angular momentum of nought ($\dot{\theta}_1(0)=\dot{\theta}_2(0)=0$). Angular momentum must be conserved, therefore at all times, we must have:

$$\begin{array}{lcl}J_1\,\dot{\theta}_1(t) + J_2\,\dot{\theta}_2(t) &=& J_1\,\left(i\,\omega_0(\hat{\theta}_1^*\,\exp(i\,\omega_0\,t)-\hat{\theta}_1\,\exp(-i\,\omega_0\,t) + A\right)+J_2\,\left(i\,\omega_0(\hat{\theta}_2^*\,\exp(i\,\omega_0\,t)-\hat{\theta}_2\,\exp(-i\,\omega_0\,t) + A\right)\\ &=&J_1\,\left(A-2\,\omega_0\,|\hat{\theta}_1|\,\sin(\omega_0\,t+\arg\hat{\theta}_1)\right)+J_2\,\left(A-2\,\omega_0\,|\hat{\theta}_2|\,\sin(\omega_0\,t+\arg\hat{\theta}_2)\right)\\ &=&0\;\forall\,t>0\end{array}$$

and so we see that

$$A=0$$ $$|\hat{\theta}_1|\,J_1+|\hat{\theta}_2|\,J_2=0$$ $$\arg\hat{\theta}_2 = \pi + \arg\hat{\theta}_1$$

and since we have already found that $-2\,\mathrm{Im}(\hat{\theta}_j)=A$ we now know $\arg\hat{\theta}_1 = 0;\,\arg\hat{\theta}_1 = \pi$ and so

$$\theta_1(t) = \alpha\,J_2\,\cos(\omega_0\,t) + B$$ $$\theta_2(t) = -\alpha\,J_1\,\cos(\omega_0\,t) + B$$

where it remains to find the common real scaling constant $\alpha$ and the offset $B$. From our initial conditions, we get from the above equations:

$$B=\frac{J_1\,\theta_1(0)+J_2\,\theta_2(0)}{J_1+J_2}$$ $$\alpha = \frac{\theta_1(0)-\theta_2(0)}{J_1+J_2}$$

Phew! We're here at last!

  • $\begingroup$ $Im(\theta_j) = -\theta_j\sin(\omega t) +\theta_j^*\sin(\omega t) = (\theta_j^* - \theta_j)\sin(\omega t)$ but you have $2Im(\theta_j) = (\theta_j^* - \theta_j)\omega + A$. $A$ would have to be purely imaginary to stay in current form but the most we know is $A\in\mathbb{C}$ but why is $B\in\mathbb{R}$ only since it didn't survive taking the imaginary part? Also, I noticed you stated $A\in\mathbb{R}$ so there is no way it is picked up by taking the imaginary part. $\endgroup$
    – dustin
    Nov 17, 2014 at 15:09
  • $\begingroup$ @dustin Be careful: I think you may be mixing up $\hat{theta}_j$ (amplitude) and $\theta_j$, the latter a function of time. Also, you've left an $i$ off the second equation. $A$ and $B$ are real because $\theta_j(t)$ are real: there is no way for them to be otherwise and have $\theta_j$ real, since $A\,t+B$ is linearly independent of $\exp(\pm i\,\omega_0\,t)$. $\hat{\theta}_j$ can be complex and as long as the overall expression pairs conjugates, it stays real. Also, I didn't take an imaginary part of the whole equation: I used $\mathrm{Im}$ to simplify the expression ..... $\endgroup$ Nov 17, 2014 at 21:57
  • $\begingroup$ @dustin ... $-i\,\omega_0\,(\hat{\theta}_j-\hat{\theta}_j^*)$. Try putting the solutions back into your equations and hopefully see that they fulfill all the initial conditions and then work backwards to understand all the detailed steps. $\endgroup$ Nov 17, 2014 at 21:57

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