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The mass of the cylindrical polley is $M$ with radius $R$ and is hung from the ceiling with two springs, one of them in one side (left and right, as you can see in the picture below). What are the normal frequencies of oscillations for this mechanical system in the small oscillations approximation?

Consider a general physical motion with $3$ degrees of freedom, so the generalized coordinates are $\textbf{q}\equiv (x,y,\theta)$. Note that if you want to calculate the normal frequencies of oscillation the pendulum motion is irrelevant because it doesn't involve the springs.

enter image description here

The solutions have this form:

$$ \omega_i^2=a_i \dfrac{k}{M}, \qquad a_i \in \{0,1,2,3,4\} \quad i=1,2 $$

where $k=k_1=k_2$. What are the values of $a_i$? Hint: Use the secular equation $$\text{det } (\hat{K}-\omega^2_i \hat{M})=0.$$

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  • $\begingroup$ What have you tried yourself, if you include a summary of your calculations on your post, someone might be able to give you a hint. $\endgroup$ – user108787 Aug 14 '16 at 13:04
  • $\begingroup$ What count_t0_10 said and a diagram makes a question more attractive. $\endgroup$ – Gert Aug 14 '16 at 13:25
  • $\begingroup$ You also wrote: "The pulley can move both of axes: it exists 3 degrees of freedom (x,y,θ)(x,y,θ) but the pendulum motion is irrelevant, so the generalized coordinates are (y,θ)" You should make that more clear. $\endgroup$ – Gert Aug 14 '16 at 16:45
  • $\begingroup$ "with two springs at both sides." From your drawing I see one spring on the left hand side, one on the right hand side. Not two springs at both sides. As a student of physics you'll appreciate the importance of clear problem definition. :-) $\endgroup$ – Gert Aug 14 '16 at 16:55
  • $\begingroup$ @Gert : I spy a pendant... $\endgroup$ – sammy gerbil Aug 14 '16 at 19:47
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You might start by looking at what the physical modes of oscillation look like. I can see two. In one mode, the whole pulley moves up and down. For this to occur, there must be a fulcrum point in the middle of the pulley that distributes its weight in such a way that the two spring=mass systems (with different spring constants) have the same frequency of oscillation. That will involve one of the springs having a greater excursion than the other.

The other mode will have the pulley rotating back and forth with the springs moving opposite to each other. It probably doesn't rotate about its geometrical center, because again the two springs will have to have different excursions. You will need the small-angle approximation to keep the motion of the springs strictly vertical.

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  • $\begingroup$ Actually this is not a useful explanation, sorry. See my (possible) answer at the end. $\endgroup$ – Clare Francis Aug 14 '16 at 21:43
  • $\begingroup$ The difference between you and me is that you first write the equations and solve them, and then at the end consider what they actually mean. I do it the other way around. But I give you credit for at least taking the time to note the physical meaning at the very end. Although you are not quite correct when you say the first solution is all up-and-down: I'm sure there is a little bit of rotation superimposed on the up-and down because, as I explained, the two springs have different excursions. $\endgroup$ – Marty Green Aug 14 '16 at 21:51
  • $\begingroup$ Thanks @Marty Green, but I need formulas and mathematical expressions and mainly, the precise values. All corrections of my answer are welcomed. :) $\endgroup$ – Clare Francis Aug 14 '16 at 22:05
  • $\begingroup$ @Jonathan, the work you have done is excellent and much better than what I could have done. Actually, I didn't read the problem and I thought that k1 and k2 were in general different. Not correct...the problem specifies they are the same. So the first mode is the simple mass on the spring with w^2 = k/m, and that's what you've got. I'd be surprised if you were not also right on the rotational mode. And yes they are pure modes...in the general case with different spring constants, both modes are mixed rotational and translational, but of course I didn't read the problem. $\endgroup$ – Marty Green Aug 15 '16 at 2:34
  • $\begingroup$ @JonathanEstévez-Fernández : Marty is right, your question says $k=k_1=k_2$ but your solution says $2k=k_1+k_2$. Which one is correct? $\endgroup$ – sammy gerbil Aug 15 '16 at 20:55
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This is my finally answer: all the corrections are welcomed.

Let the lagrangian of the system with generalized coordinates $\textbf{q}\equiv (y,\theta)$ (consider, without loss of generality for our problem, retricted motion in $x-$axis)

$$ L(y,\theta,\dot{y},\dot{\theta}) \equiv T-V = \underbrace{\dfrac{1}{2} M \dot{y}^2}_{\text{Kinetic energy}}+\underbrace{\dfrac{1}{2}I\dot{\theta}^2}_{\text{Kinetic rot. energy}} -\underbrace{\dfrac{1}{2} k_T (y^2+R^2\theta^2)}_{\text{Harmonic potential}} \tag{1}$$

and $k_T$ is the total Hooke constant (cfr. parallel springs), so $k_T=k_1+k_2=2k$. Note that

$$I=\dfrac{1}{2}MR^2 $$

and obviously, there is not a kinetic traslational energy in $\theta$. So, the lagrangian $(1)$ takes the form

$$ L=\dfrac{1}{2} M(\dot{y}^2+\dfrac{1}{2}R^2 \dot{\theta}^2)-\dfrac{1}{2}k_T (y^2 +R^2 \theta^2) . \tag{2} $$

Hence, the Euler-Lagrange equations for $(2)$ are

$$ \dfrac{\text{d}}{\text{d}t} \left(\dfrac{\partial L}{\partial \dot{y}}\right) -\dfrac{\partial L}{\partial y} = \dfrac{\text{d}}{\text{d}t} \left(\dfrac{\partial L}{\partial \dot{\theta}}\right) -\dfrac{\partial L}{\partial \theta }=0 $$

that it is

$$ M\ddot{y}+k_Ty =\dfrac{1}{2}MR^2 \ddot{\theta}+k_TR^2\theta=0 .\tag{3}$$

Equations $(3)$ can be written as

$$ \begin{pmatrix} M & 0 \\ 0 & \dfrac{1}{2} MR^2 \end{pmatrix} \begin{pmatrix} \ddot{y}\\ \ddot{\theta}\end{pmatrix} +\begin{pmatrix} k_T & 0 \\ 0 & k_T R^2 \end{pmatrix} \begin{pmatrix} y \\ \theta \end{pmatrix} =0 \Longleftrightarrow \hat{M} \ddot{\textbf{q}} +\hat{K}\textbf{q} =0 \tag{4.1}$$

or, equivalently,

$$ \ddot{\textbf{q}} + \hat{\Lambda}^2 \textbf{q}=0, \qquad \hat{\Lambda}^2 \equiv \hat{M}^{-1} \hat{K} \tag{4.2}$$ where $\hat{\Lambda}$ is the frequency matrix of the generalized coordinates $(q_1,q_2)$.

As it is known, the general solution of Eq. $(4)$ is

$$ \textbf{q} = \sum_{i,j=1}^2 \delta_{ij}\textbf{a} \cos (\omega_i t + \varphi)\textbf{e}_i \tag{5}$$

where $\textbf{a}\equiv (a_1, a_2)$ is the amplitude vector of the oscillation and $\delta_{ij}$ is the Kronecker delta.

Substituting $(5)$ in $(4)$

$$ -\hat{M}\omega_i^2 \textbf{a} \cos (\omega_i t+\varphi)+\hat{K}\textbf{a}\cos(\omega_i t+\varphi)=0$$

then

$$ (\hat{K}-\omega_i^2 \hat{M})\textbf{a}=0 \tag{6}.$$

The Eq. $(6)$ has a solution for $\textbf{a}=0$ when the system is balanced up (equilibrium). The other solutions satisfy

$$ \text{det }(\hat{K}-\omega_i^2 \hat{M}) =0 \Longrightarrow \left| \begin{matrix} k_T -\omega_i^2 M & 0 \\ 0 & k_T R^2 -I\omega_i^2 \end{matrix}\right| =0 $$

so, the two normal frequencies are

$$ \boxed{\omega_1^2 = \dfrac{2k}{M}}, \qquad \boxed{\omega_2^2 = \dfrac{4k}{M}}$$

and the solution is $a_1 =2, a_2 =4.$

Now, take a look at the normal equations of motion:

Eigenvalues and eigenvectors:

  • For $\omega_1$ (vertical motion, $y$):

$$ \textbf{b}_1 =\text{ker }(\hat{K} -\omega_1^2 \hat{M}) = \text{ker }\begin{pmatrix} 0 & 0 \\ 0 & kR^2 \end{pmatrix} = \begin{pmatrix} \alpha \\ 0 \end{pmatrix}$$

with $\alpha \in \mathbb{R}.$ But $\alpha$ satisfy the quadratic form

$$ \textbf{b}_1^\top \hat{M}\textbf{ b}_1 =1 \Longrightarrow \alpha = \pm \dfrac{1}{\sqrt{M}}.$$

  • For $\omega_2$ (rotational motion, $\theta$):

$$ \textbf{b}_2 =\text{ker }(\hat{K} -\omega_2^2 \hat{M}) = \text{ker }\begin{pmatrix} -2k & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 \\ \beta\end{pmatrix}$$

with $\beta \in \mathbb{R}.$ But $\beta$ satisfy the quadratic form

$$ \textbf{b}_2^\top \hat{M}\textbf{ b}_2 =1 \Longrightarrow \beta = \dfrac{\sqrt{2}}{R} \alpha = \pm \dfrac{1}{\sqrt{I}}.$$

The Eigenvalues matrix $\hat{B}\equiv (\textbf{b}_1 \textbf{ b}_2 )$ is

$$ \hat{B} = \pm \dfrac{1}{\sqrt{M}} \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{\sqrt{2}}{R}\end{pmatrix}$$

and the normal solutions are

$$ \boxed{q_i = \textbf{b}_i \cos (\omega_i t +\varphi), \qquad i=1,2}. \tag{7} $$

Note that Eq. $(7)$ is equivalent to Eq. $(5)$ with $\textbf{a}\equiv \textbf{b}_i$.

Definitely, the normal equations of motions are

$$ y(t) = \pm \dfrac{1}{\sqrt{M}} \cos \left(\sqrt{\dfrac{2k}{M}}t+\varphi \right), \qquad \theta(t)= \pm \dfrac{1}{\sqrt{I}} \cos \left( \sqrt{\dfrac{4k}{M}}t+\varphi\right)$$

with the small oscillations condition

$$ y_\text{max}^2 \le \dfrac{1}{M}, \qquad \theta_\text{max}^2 \le \dfrac{1}{I}.\tag{8}$$

Note that $(8)$ is satisfied if $M,R \gg 1.$ All posible motions with the pendulum restriction (i.e. it cannot move in $x-$axis) is a linear combination of $q_i(t)$, $i=1,2$. The expression $q_1 (t)= y(t)$ describe the simultaneous up-down motion of the two springs, and $q_2 (t)=\theta(t)$ occur when one spring is up and the other is down and the pulley spins around.

[1] H. Goldstein (2001), Classical Mechanics, Pearson Education, 3rd edition, pp. 238-275 (chapter 6).

[2] C. Lanczos (1974), The Variational Principles of Mechanics, University of Toronto Press, 4th edn. Dover Paperback.

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