Open Educational Resources

Review of single and multi-degree of freedom (mdof) systems: Introduction and Review

We are interested in small motions about some equilibrium positions. For SDOF systems, x is usually measured from this static equilibrium position.

This image has an empty alt attribute; its file name is part1fig1-3-scaled.jpg

    \[\begin{split}+\downarrow \sum{F_x} &= m \ddot{x} \\&= -k(x+\delta_\text{static})-c\dot{x} + F(t)+mg\end{split}\]

Let \dot{x}, \ddot{x} = 0 for static equilibrium. Therefore:

    \[-k\delta_{ST.}+mg = 0\]

Let \delta_\text{static} = \frac{mg}{k}. Therefore:

    \[m\ddot{x}+c\dot{x} +kx = F(t)\]

All SDOF systems can be reduced to this form (assuming viscous damping):

    \[m_e\ddot{x}+c_e\dot{x} +k_ex = F(t) \quad\]

Where e means equivalent. To determine the nature of the motion, first consider this linearized equation of free vibration (Note: May have to linearize).

Assume x_H = Ae^{st} (homogeneous), where A is constant and s is to be determined.

    \[m_es^2 + c_es+k_e = 0\]

    \[\begin{split}s_{1,2} &= -\frac{c_e}{2m_e} \pm \sqrt{\Big(\frac{c_e}{2m_e}\Big)^2-\frac{k_e}{m_e}}\\&= -\frac{\tilde{c}}{2} \pm \sqrt{\Big(\frac{\tilde{c}}{2}\Big)^2-\tilde{k}}\end{split}\]

If s_{1,2} are real and negative, then x\rightarrow 0 as t \rightarrow\infty. If either of s_1 or s_2 are real and positive, then the solution increases with time.

If s_{1,2} are complex, then they are complex conjugates and the nature of the motion depends on the real part. If the real part is negative, x\rightarrow 0 as t\rightarrow \infty. If positive, x\rightarrow \infty as t \rightarrow \infty. If the real part is zero, the motion is oscillatory and this is the borderline case between stability and instability.

Consider all cases:

  1. \tilde{c} > 0 \quad \tilde{k} > 0. The roots have a negative real part and is stable regardless of \tilde{k}. If |\tilde{k}| is larger than |\frac{\tilde{c}}{2}|^2 , there will be an oscillatory part, but x\rightarrow0 as t \rightarrow \infty. This is called asymptotically stable.
  2. \tilde{c} = 0 \quad \tilde{k}>0. The roots are imaginary, and the system is borderline stable.
  3. \tilde{k}<0 \quad\tilde{c}\le 0. The roots have a positive real part, or are complex conjugates. Therefore, the system is unstable.
  4. \tilde{k}<0 \quad\tilde{c}>0 One root has a positive real part. Therefore the system is unstable. The damping cannot overcome the negative spring.
  5. \tilde{k}>0 \quad\tilde{c}<0. The roots have a positive real part, and therefore the system is unstable. Negative damping feeds energy into the system.

Consider the solution for free damped vibration:

    \[m_e \ddot{x}+c_e\dot{x} +k_e x=0\]

Drop e, but remember that m,c,k are equivalent. Thus:

    \[\ddot{x}+\frac{c}{m}\dot{x}+\frac{k}{m}x = 0\]

Again, assume that x=Ae^{st}. Thus:

    \[s_{1,2}= \frac{-c}{2m}\pm\sqrt{\Big(\frac{c}{2m}\Big)^2-\frac{k}{m}}\]

The critical value is then when:

    \[\Big(\frac{c_c}{2m}\Big)^2=\frac{k}{m} := p^2\]


    \[c_c = 2mp\]

Define \zeta := \frac{c}{c_c}. Then:

    \[\frac{c}{2m} = \frac{c}{c_c}\frac{2mp}{2m} = \zeta p\]


    \[s_{1,2} = \Big(-\zeta \pm \sqrt{\zeta ^2-1}\Big)p\]

Where \zeta is the damping ratio (system parameter).

Assuming 0<\zeta<1:

    \[x(t) = e^{-\zeta pt}\Big[C_1e^{i\sqrt{1-\zeta^2}}pt+C_2e^{-i\sqrt{1-\zeta^2}}pt\Big]\]

And using e^{\pm i\theta}=\cos{\theta}\pm i\sin{\theta}:

    \[x(t) = e^{-\zeta pt}\Big[A\cos{\sqrt{1-\zeta^2}pt+B\sin{\sqrt{1-\zeta^2}pt}}\Big]\]

Using initial conditions x(0) = x_0, \dot{x}(0) = v_0, we get the following:

    \[x(t)=e^{-\zeta pt}\Big[x_0 \cos\sqrt{1-\zeta^2} pt + \frac{\nu_0 + \zeta p x_0}{\sqrt{1-\zeta^2}}\sin{pt\sqrt{1-\zeta^2}pt}\Big]\]

We can write this as:

    \[x(t) = e^{-\zeta pt}\Big[\sqrt{A^2+B^2}\sin{(\sqrt{1-\zeta^2}pt + \phi)}\Big]\]

Where \phi = \tan^{-1}{\frac{A}{B}}, A = x_0, and B = \frac{v_0 + \zeta p x_0}{\sqrt{1-\zeta ^2}}.

In some problems, we have to find the equilibrium (static) configurations and then determine the motion about these positions.


Find the motion about the equilibrium configuration.

m slides on a wire with viscous friction c.

This image has an empty alt attribute; its file name is part1fig2-4.jpg
This image has an empty alt attribute; its file name is part1fig2_2.jpg

    \[\begin{split}+\circlearrowright \sum{M_0}& = mL^2\ddot{\theta} \\& = mgL\sin{\theta}-kL\sin{\theta}(L\cos{\theta}) - cL^2\ddot{\theta}\end{split}\]

    \[mL^2 \ddot{\theta}+kL^2 \sin{\theta}\cos{\theta}- mgL\sin{\theta}+cL^2 \dot{\theta}=0\]

    \[\ddot{\theta}+\frac{k}{m}\sin{\theta}\cos{\theta}-\frac{g}{L}\sin{\theta}+\frac{c}{m}\dot{\theta}=0 \quad (1)\]

We must find static equilibrium configurations and determine equation of motion for small perturbations about these static configurations.

Set \ddot{\theta} = \dot{\theta} = 0, and let \theta = \theta_0. Therefore:

    \[\frac{k}{m} \sin{\theta_0} \cos{\theta_0} = \frac{g}{L} \sin{\theta_0}\]

    \[\sin{\theta_0}\Big[\frac{k}{m}\cos{\theta_0}-\frac{g}{L}\Big] = 0\]

Therefore. the static configurations are given by:

    \[\theta_0 = 0, \pi \quad \sin{\theta_0}=0 \quad \cos{\theta_0} = \frac{mg}{kL}\]

Assuming \frac{mg}{kL}<1, there are 3 positions.

To find the motion, linearize (1) about \theta_0. Set \theta = \theta_0 + \hat{\theta}, where \hat{\theta} is a small angle.

    \[\sin \theta = \sin(\theta_o + \hat{\theta}) \approx \sin \theta_o +\hat{\theta} \cos \theta_o\]

    \[\cos \theta \approx \cos \theta_o - \hat{\theta}\sin\theta_o\]

Put the above into the original equation of motion to get:

    \[\ddot{\hat{\theta}} + \frac {k}{m}(\sin \theta_o + \hat{\theta} \cos \theta_o)(\cos \theta_o - \hat{\theta}\sin \theta_o) - \frac{g}{L}(\sin \theta_o + \hat{\theta}\cos \theta_o) = 0\]

Note that \dot{\theta_o} = \ddot{\theta_o} = 0.

    \[\ddot{\hat{\theta}} + \frac {k}{m}(\sin \theta_o \cos \theta_o + \hat{\theta} \cos^2\theta_o - \hat{\theta}\sin^2\theta_o-\hat{\theta}^2\sin\theta_o \cos\theta_o) - \frac{g}{L}(\sin \theta_o + \hat{\theta}\cos \theta_o) + \frac{c}{m}\dot{\hat{\theta}} = 0\]

Using the fact that the \hat{\theta}^2 term is small compared to 1, it becomes:

    \[\ddot{\hat{\theta}} + \Big(\frac{k}{m}\cos\theta_o - \frac{g}{L}\Big)\sin\theta_o + \hat{\theta}(\cos^2\theta_o-\sin^2\theta_o)\frac{k}{m} - \frac{g}{L}\hat{\theta}\cos\theta + \frac{c}{m}\dot{\hat{\theta}} = 0\]

    \[\ddot{\hat{\theta}} + \hat{\theta}\Big(\frac{k}{m}\cos2\theta_o - \frac{g}{L}\cos\theta_o\Big) + \frac{c}{m}\dot{\hat{\theta}} + \Big(\frac{k}{m}\cos\theta_o - \frac{g}{L}\Big)\sin\theta_o = 0\]

The above equation looks like the standard form except for a constant term that is zero for all conditions.

Now consider all three positions:

1. \theta_o = 0.

    \[\ddot{\hat{\theta}} + \Big(\frac{k}{m} - \frac{g}{L}\Big)\hat{\theta} + \frac{c}{m}\hat{\theta} = 0\]

Therefore, this is stable if \frac{k}{m} > \frac{g}{L} as \frac{c}{m} > 0, but unstable if \frac{k}{m} < \frac{g}{L}.

2.\theta_o = \pi.

    \[\ddot{\hat{\theta}} + \Big(\frac{k}{m} + \frac{g}{L}\Big)\hat{\theta} + \frac{c}{m} = 0\]

Therefore, this is always stable.

3.\cos\theta_o = \frac{mg}{kL}.

    \[\ddot{\hat{\theta}} + \frac{k}{m}\Big(2\cos^2\theta_o - 1 - \frac{gm}{kL}\cos\theta_o\Big)\hat{\theta} + \frac{c}{m}\dot{\hat{\theta}} = 0\]

    \[\ddot{\hat{\theta}} + \frac{k}{m}(\cos^2\theta_o - 1)\hat{\theta} + \frac{c}{m}\dot{\hat{\theta}} = 0\]

Therefore, this is always unstable, regardless of damping.


1.\frac{mg}{kL} = \frac{1}{2}, \theta_o = \pm 60^{\circ}, \frac{k}{m} = \frac{2g}{L}. Therefore:

    \[\frac{k}{m} > \frac{g}{L}\]

This image has an empty alt attribute; its file name is part1fig3-2-scaled.jpg

2.\frac{mg}{kL} = 2, \frac{k}{m} = \frac{1}{2}\frac{g}{L}. Therefore:

    \[\frac{k}{m} < \frac{g}{L}\]

This image has an empty alt attribute; its file name is part1fig3_2-2.jpg

Leave a Reply

Your email address will not be published. Required fields are marked *