0
$\begingroup$

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.

$\endgroup$

migrated from math.stackexchange.com Nov 17 '14 at 5:22

This question came from our site for people studying math at any level and professionals in related fields.

  • $\begingroup$ Maybe Engineering is a better home for this question actually. $\endgroup$ – ja72 Apr 8 '15 at 16:50
1
$\begingroup$

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:

$$-2\,\mathrm{Im}(\hat{\theta}_j)=-2\,\mathrm{Im}(\hat{\theta}_1)=-2\,\mathrm{Im}(\hat{\theta}_2)=A$$

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!

$\endgroup$
  • $\begingroup$ $Im(\theta_j) = -\theta_j\sin(\omega t) +\theta_j^*\sin(\omega t) = (\theta_j^* - \theta_j)\sin(\omega t)$ 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 '14 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 ..... $\endgroup$ – WetSavannaAnimal Nov 17 '14 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. $\endgroup$ – WetSavannaAnimal Nov 17 '14 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.