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

The concept of modes and their corresponding natural frequencies is useful in understanding the behavior of lightly damped systems. Before adding the complication of damping to our models, we will consider several applications to industrial situations.

Example

The model of the automobile shown is used to predict the low frequency modes of vibrations as these are responsible for the low frequency noise in the passenger compartment. Knowing these models can assist the designer in controlling them. In addition, it helps the designer in predicting the response of the frame to various inputs from road irregularities. As an initial estimate, the three modes of vibration of the vehicle may be considered as uncoupled. For the specific case below, the assumption of uncoupled modes is to be compared to the coupled model.

a) For the longitudinal model of the vehicle, specialize the equations of motion for the case of h = 0 (see previous section).
b) If any of the modes are uncoupled determine an expression of the natural frequency. For the coupled modes derive the equations of motion for free vibration starting from Newton’s laws.
c) Calculate the natural frequencies and mode shapes for the two coupled modes of vibration. The weight of the vehicle is 3550 \ \text{lbs}.\ k_1 = 2200 \ \text{lbs/ft}, k_2 = 2700\ \text{lbs/ft}, a = 4.5\ \text{ft}, b = 5.5\ \text{ft} and the radius of gyration on the vehicle about G is 4\ \text{ft}.

Solution

a)

    \[ \begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & J_G \end{bmatrix}\begin{Bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{\theta} \end{Bmatrix} + \begin{bmatrix} \ell_1 + \ell_2 & 0 & 0 \\ 0 & k_1 + k_2 & k_2b - k_1a \\ 0 & k_2b-k_1a & k_1a^2 + k_2b^2 \end{bmatrix}\begin{Bmatrix} x \\ y \\ \theta \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \\ 0 \end{Bmatrix} (+)\]

b)

For horizontal motion:

    \[ m\ddot{x} + (\ell_1 + \ell_2)x = 0 \qquad \mathrm{ (uncoupled) }\]

Therefore:

    \[ p_H = \sqrt{\frac{\ell_1 + \ell_2}{m}} \]

    \[ \begin{split} + \uparrow \sum F_y = & m\ddot{y} = -k_2(y+b\theta) - k_1(y-a\theta) \\ & m\ddot{y} + (k_1 + k_2)y + (k_2b-k_1a)\theta = 0 \qquad (I) \end{split}\]

    \[ \begin{split}+\circlearrowleft \sum M_G =& J_G\ddot{\theta} = k_1a(y - a\theta) - k_2b(y+b\theta) \\ & J_G\ddot{\theta} + (k_2b-k_1a)y + (k_1a^2+k_2b^2)\theta \qquad (II) \end{split}\]

(I) & (II) are the same as (+).

c)

    \[ \omega = 3550 \ \text{lbs} \]

    \[a = 4.5\ \text{ft}\]

    \[b = 5.5 \ \text{ft}\]

    \[\bar{r} = 4 \ \text{ft}\]

    \[k_1 = 2200 \ \text{lb/ft}\]

    \[k_2 = 2700 \ \text{lb/ft}\]

Therefore:

    \[\begin{bmatrix} m & 0 \\ 0 & J_G \end{bmatrix} \begin{Bmatrix} \ddot{x} \\ \ddot{\theta} \end{Bmatrix} + \begin{bmatrix} k_1 + k_2 & k_2b - k_1a \\ k_2b - k_1a & k_1a^2 +k_2b^2 \end{bmatrix} \begin{Bmatrix} x \\ \theta \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} \]

Assume \begin{Bmatrix} x \\ \theta \end{Bmatrix} = \begin{Bmatrix} X \\ \Theta \end{Bmatrix}\sin{(pt + \phi)}:

    \[\begin{bmatrix} (k_1 + k_2) - mp^2 & k_2b-k_1a \\ k_2b - k_1a & k_1a^2 + k_2b^2 - J_Gp^2 \end{bmatrix} \begin{Bmatrix} X \\ \Theta \end{Bmatrix} = 0 \]

    \[ [k_1 + k_2 -mp^2][k_1a^2+k_2b^2-J_Gp^2] - (k_2b - k_1a)^2 = 0 \]

    \[(k_1+k_2)(k_1a^2+k_2b^2)-p^2\big[J_G(k_1+k_2)+(k_1a^2+k_2b^2)m\big]+mJ_Gp^4-(k_2b-k_1a)^2 =0 \]

    \[mJ_Gp^4-mp^2\big[\vec{r}^2(k_1+k_2)+(k_1a^2+k_2b^2)\big]+[(k_1+k_2)(k_1a^2+k_2b^2)]-(k_2b-k_1a)^2 = 0 \]

    \[p^4-p^2\bigg[\frac{k_1+k_2}{m}+\frac{(k_1a^2+k_2b^2)}{m\vec{r}^2}\bigg]+\bigg(\frac{k_1+k_2}{m}\bigg)\bigg(\frac{k_1a^2+k_2b^2}{m\vec{r}^2}\bigg)-\frac{(k_2b-k_1a)^2}{mm\vec{r}^2} = 0 \]

    \[p^4-p^2[44.45+74.99]+(44.45)(74.99)-126.0 =0 \]

    \[p^4-119.44p^2+3207 = 0 \]

    \[p^2 =\frac{119.44}{2}\pm \sqrt{\left(\frac{119.44}{2}\right)^2-3207} = 59.7 \pm 18.96 \]

    \[\begin{split} p_1^2 &=40.74 \\ p_2^2 &=78.65 \\ p_{1,2} &= 6.38 \ \text{rads/s}, 8.87 \ \text{rads/s} \\ &= \underline{\underline{1.01 \ \text{Hz}}}, \underline{\underline{1.41 \ \text{Hz}}} \end{split}\]

    \[(k_1+k_2-mp^2)X + (k_2b-k_1a)\Theta = 0 \]

    \[\begin{split} \frac{X}{\Theta} &= \frac{(k_1a-k_2b)}{k_1+k_2-mp^2} \\ & \\ \Big(\frac{X}{\Theta}\Big)_1 &= \frac{(k_1a-k_2b)/m}{\frac{(k_1+k_2)}{m}-p^2} \\ &= \frac{[(2200)(45)-2700(5.5)]\frac{32.2}{3550}}{(44.45)-40.74} \\ &= \underline{\underline{-12.1 \ \text{ft/rad}}} \\ & \\ \Big(\frac{X}{\Theta}\Big)_2 &= \frac{-44.90}{44.45-78.65} \\ &= \underline{\underline{1.31}} \text{ft/rad}\end{split}\]

If we “uncoupled” the modes

Assume vertical node is

    \[\begin{split} p_1 &= \sqrt{\frac{k_1+k_2}{m}} \\ &= \sqrt{\frac{4900*32.3}{3550}} \\ &= 6.67 \ \text{rads/s} \\ &= \underline{\underline{1.06 \ \text{Hz}}} \end{split}\]

Rotational Mode

    \[\begin{split} p_2 &=\sqrt{\frac{k_1a^2+k_2b^2}{m\vec{r}^2}} \\ &= \sqrt{\frac{[(2200)(4.5)^2+(2700)(5.5)^2]32.2}{3550(4)^2}} \\ &= 8.46 \ \text{rads/s} \\ &= \underline{\underline{1.35 \ \text{Hz}}} \end{split}\]

If the horizontal stiffness, l_1, l_2, were the same as the vertical then the third frequency, p_H, would be the same as the vertical.

The assumption of uncoupling gives essentially the same answer.

Example

The illustration shows the detail of the proposed isolation system for the reciprocating compressor and motor considered previously and modelled as a single dof system. The actual system does not have the center of gravity symmetrically located on the supporting structure and isolation springs. In situations like these the industrial installation guides recommend that the isolation springs be selected so that the vertical static deflection of all springs is the same. The reason stated is that this will tend to reduce the frequency of the “rocking motions”. This exercise is to investigate this idea as well as the differences found for the isolation of the system compared with the single degree approximation used earlier. More details of the installation are given below. The stiffnesses shown are representative of both springs at the ends of the supports with the lateral stiffnesses assumed to be equal to the vertical values.

a) If the vertical static deflection of the spring is the same what is the ratio of k_2 and k_1 ?

    \[ S_1 = \frac{F_1}{k_1}, \ S_2=\frac{F_2}{k_2} \]

For static equilibrium

    \[\begin{split} F_1a &=F_2b \\ \frac{F_1}{k_1} &= \frac{F_1(\frac{a}{b})}{k_2} \end{split} \]

therefore:

    \[\begin{split} \frac{k_2}{k_1} &=\frac{a}{b} \\ k_2b &= k_1a \end{split} \]

(b) Under this assumption what happens to the equations of motion?

    \[\begin{bmatrix} m & 0 & 0 \\0 & m & 0 \\ 0 & 0 & m\vec{r}^2 \end{bmatrix}\begin{Bmatrix}\Ddot{x} \\ \Ddot{y} \\ \Ddot{\theta}\end{Bmatrix} + \begin{bmatrix} \ell_1 + \ell_2 & 0 & (\ell_1+\ell_2)h \\0 & k_1+k_2 & 0 \\(\ell_1+\ell_2)h & 0 & h^2(\ell_1+\ell_2)+k_1a^2+k_2b^2\end{bmatrix}\begin{Bmatrix}x \\ y \\ \theta\end{Bmatrix} = \begin{Bmatrix}0\end{Bmatrix}\]

as the vertical mode is uncoupled from the other two

    \[ p_v = \sqrt{\frac{k_1+k_2}{m}} \]


becomes

    \[\begin{bmatrix}m & 0 \\ 0 & mr^2\end{bmatrix}\begin{Bmatrix}\Ddot{x} \\ \Ddot{\theta}\end{Bmatrix} + \begin{bmatrix}\ell_1+\ell_2 & (\ell_1+\ell_2)h \\(\ell_1+\ell_2)h & h^2(\ell_1+\ell_2)+k_1a^2+k_2b^2\end{bmatrix}\begin{Bmatrix}x \\ \theta\end{Bmatrix}=\begin{Bmatrix}0\end{Bmatrix}\]

(c) Noting that the original had an operating speed of 1750 RPM and the original transmissibility was 1/15, calculate k_1+k_2 for this installation assuming k_1a=k_2b

    \[ \frac{1}{(\frac{\omega}{p_v})^2-1}=\frac{1}{15} \]


therefore:

    \[ \begin{split} \frac{\omega}{p_v} &=4 \\ p_v &=\frac{1750}{4} \\ &=437.5 \ \text{RPM} \\ &= \frac{437.5*2\pi}{60} \\ &=45.8 \ \text{rads/s} \\ &=7.3 \ \text{Hz} \end{split} \]

therefore:

    \[\begin{split} \sqrt{\frac{(k_1+k_2)32.2}{4600}} &=45.8 \\ k_1 + k_2 &= (45.8)^2\frac{4600}{32.2} \\ &=300,000 \text{ lb/ft} \end{split}\]

therefore:

    \[\begin{split}k_1+k_1\frac{a}{b} &=k_1\Big(1+\frac{3.5}{6.9}\Big) \\ &=300,000 \text{ lbs/ft} \end{split}\]

    \[k_1=\frac{300,000}{1.65217}=181,600 \ \text{lbs/ft}\]

    \[k_2 = 0.65217\cdot181,600 = \ \underline{\underline{\text{118,400 lbs/ft}}}\]

(d) Assuming SSHM

    \[\begin{Bmatrix}x \\ \theta\end{Bmatrix}= \begin{bmatrix}X \\ \Theta\end{bmatrix}\sin(pt+\phi)\]


    \[ \begin{bmatrix}\ell_1+\ell_2-mp^2 & (\ell_1+\ell_2)h \\(\ell_1+\ell_2)h & h^2(\ell_1+\ell_2)+k_1a^2+k_2b^2-mr^2p^2\end{bmatrix}\begin{Bmatrix}X \\ \Theta\end{Bmatrix}= \begin{Bmatrix}0\end{Bmatrix}\]

    \[(r = \text{radius of gyration})\]

therefore:

    \[ \Big[\ell_1+\ell_2-mp^2\Big]\Big[h^2(\ell_1+\ell_2)+k_1a^2+k_2b^2-mr^2p^2\Big]-(\ell_1+\ell_2)^2h^2 = 0\]


Divide by m^2r^2 to get

    \[ \biggr[\frac{\ell_1+\ell_2}{m}-p^2\biggr]\biggr[\frac{(\ell_1+\ell_2)}{m}\frac{h^2}{r^2}+\frac{k_1}{m}\frac{a^2}{r^2}+\frac{k_2}{m}\frac{b^2}{r^2}-p^2\biggr]-\frac{(\ell_1+\ell_2)^2}{m^2}\frac{h^2}{r^2} = 0\]


therefore:

    \[\begin{split}p^4 &-p^2\biggr[\frac{\ell_1+\ell_2}{m}+\frac{\ell_1+\ell_2}{m}\frac{h^2}{r^2}+\frac{k_1}{m}\frac{a^2}{r^2}+\frac{k_2}{m}\frac{b^2}{r^2}\biggr] \\ &+ \frac{(\ell_1+\ell_2)^2}{m^2}\frac{h^2}{r^2} + \frac{(\ell_1+\ell_2)}{m}(\frac{k_1}{m}\frac{a^2}{r^2}+\frac{k_2}{m}\frac{b^2}{r^2}) - \frac{(\ell_1+\ell_2)^2}{m^2}\frac{h^2}{r^2} = 0 \end{split}\]

    \[p^4-p^2\biggr[\Big(\frac{\ell_1+\ell_2}{m}\Big)\Big(1+\frac{h^2}{r^2}\Big)+\frac{(k_1a^2+k_2b^2)}{mr^2}\biggr] + \frac{(\ell_1+\ell_2)(k_1a^2+k_2b^2)}{m^2r^2} = 0\]

    \[ \begin{split}p^4 &-p^2\biggr[\frac{(300,000)}{4600}\Big(1+\Big(\frac{0.8}{2.1}\Big)\strut^2\Big)32.2+\frac{181,600}{4600}\Big(\frac{3.5}{2.1}\Big)\strut^2\Big(32.2\Big)+\frac{118,400}{4600}\Big(\frac{6.9}{2.1}\Big)\strut^2 32.2\biggr] \\ &+ \frac{300,000}{(4600)^2}\biggr[181,600\Big(\frac{3.5}{2.1}\Big)\strut^2+118,400\Big(\frac{6.9}{2.1}\Big)\strut^2\biggr](32.2)^2 = 0\end{split} \]

    \[ \begin{split} p^2 &= 7470 \pm 5429 \\ & \\ p_1 &= 45.2 \ \text{rads/s (7.2 Hz)} \\ p_2 &= 113.6 \ \text{rad/s (18.1 Hz)} \end{split}\]

(e)

    \[ \begin{split} p_v &= 45.8 \ \text{rads/s (7.3 Hz)} \\ p_H &= 45.8 \text{rads/s (7.3 Hz)} \end{split} \]

The rotational mode can be calculated starting from a FBD

    \[\begin{split} \Sigma M_G &= m\vec{r}^2\Ddot{\theta} \\ &= -ka^2\theta - ka^2\theta \end{split} \]


therefore:

    \[ \begin{split} P^2_{\theta} &= \frac{2ka^2}{mr^2} \\ &= \frac{300,000(5.2)^2(32.2)}{4600(2.1)^2} \end{split}\]


    \[ \begin{split} P_{\theta} &= 113.5 \ \text{rads/s} \\ &= (18.1 \ \text{Hz}) \end{split} \]

(f) Do the calculation using a program

(g) From d)

    \[ \begin{split} p_v &=7.3 \ \text{Hz} \\ (p_H)_1 &= 7.2 \ \text{Hz} \\ (p_{\theta})_2 &= 18.1 \ \text{Hz} \end{split} \]

from e)

    \[ \begin{split} p_v &= 7.3 Hz \\ p_H &= 7.3 Hz \\ p_{\theta} &= 18.1 Hz \end{split} \]


from (f)

    \[ \begin{split} (p)_{v_1} &= ? \\ (p_H)_2 &= ? \\ (p_{\theta})_3 &= ? \end{split} \]

The results show that because h is small compared to a and b that there is little difference. The “rotational” (p_{\theta}, p_3) is very much higher than the other two and closer to the excitation frequency of 29.2 Hz (1750 RPM).

To lower it, the spacing of the supports would have to be reduced. For example if it were brought to 1/2 of original (5.1′ instead of a total of 10.4′) then the rotational natural frequency would be

    \[ p_{\theta} = \sqrt{\frac{300,000(2.6)^2(32.2)}{4600(2.1)^2}} = 56.7 \ \text{rads/s} \]

It is obvious from this calculation that the rule of thumb should be that the spacing should be no more than twice the radius of gyration of the total mass!

Example

A commercial front loading washing machine is being used by a spa located adjacent to a medical clinic in a strip mall. During the spin cycle of the washing machine the fluctuating force due to the imbalance of the load causes the floor to vibrate and induces unacceptable levels of noise in the clinic. To reduce this disturbance, an engineering consultant recommends using spring isolation between the washing machine and the floor to reduce the dynamic force transmitted to 20% of the exciting force. The specifications of the washing machine are summarized below. Assume only vertical vibration and neglect damping.

Net weight1049 lbs
Maximum load255lbs
Maximum Dynamic Force212 lbs
Spin Speed (Frequency of Dynamic Force)950 RPM (15.8 Hz)
Cylinder Diameter27.6 inch


a) Calculate the appropriate spring constant for each of four springs used to mount the washing machine to meet the specification for the transmissibility and select them from the enclosed table.

b) This washing machine can also be run at a spin speed of 750 RPM. Calculate the Dynamic Force at this RPM and then calculate the force transmitted to the supporting structure compared with the force transmitted at 950 RPM.

c) Using the size of the rotating wash cylinder estimate the unbalanced weight in the maximum load.

Mount
No.		Mount
Constant
(lb/in.)		Load at
2-in. Deflection
(lb)
SLF-40150100
SLF-40270140
SLF-40390180
SLF-404115230
SLF-405150300
SLF-406200400
SLF-407305610
SLF-408365730
SLF-4095001,000
SLF-4106501,300
SLF-4117401,880
SLF-4121,1502,300
SLF-4131,5003,000
SLF-4142,0004,000
SLF-4152,6805,360
SLF-4163,5007,000
SLF-4174,7509,500
SLF-4186,25012,500
SLF-4198,20016,400
SLF-42011,25022,500
SLF-42115,00030,000
SLF-42220,00040,000

a)

    \[M = \frac{1049+255}{386}\]

    \[\begin{split} T &= 0.2 \\&= \frac{1}{(\frac{\omega}{p})^2-1} \end{split}\]

Therefore,

    \[\frac{\omega^{2}}{p^{2}} = 5\]

    \[\begin{split} \frac{\omega}{p} &= \sqrt{6} \\&= 2.44 \end{split}\]

    \[\begin{split} p &= \frac{\omega}{\sqrt{6}} \\&= \frac{950\cdot(2 \pi)}{60 \sqrt{6}} \\&= 40.61 \ \text{rad/s} \end{split}\]

    \[\begin{split} k &= p^{2}M = (40.61)^{2} \\&= (40.61)^{2}\Big(\frac{1304}{386}\Big) \\&= 5571.3 \ \text{lb/in} \end{split}\]

Therefore, each spring has

    \[\begin{split} k &= \frac{5571.3}{4} \\&= 1393 \ \text{lbs/in} \end{split}\]

Therefore, SLF 412 as 413 would not meet the transmissibility spec.

b)

    \[\omega_{950} = \sqrt{6}p\]

    \[\omega_{750} = \sqrt{6}p \Big(\frac{750}{950}\Big)\]

    \[\frac{\omega}{p}_{750} = \sqrt{6}\Big(\frac{750}{950}\Big)\]

    \[\begin{split} T_{750} &= \frac{1}{6\big(\frac{75}{95}\big)^{2}-1} \\&= 0.365 \end{split}\]

    \[T_{950} = 0.2\]

    \[F_{o-950} = 212\ \text{lbs}\]

    \[\begin{split} F_{o-750} &= 212\Big(\frac{75}{96}\Big)^{2} \\&= 132\ \text{lbs} \end{split}\]

    \[\begin{split} F_{T-950} &= 212\cdot0.2 \\&= 42 \ \text{lbs} \end{split}\]

    \[\begin{split} F_{T-750} &= 132\cdot0.365 \\&= 48 \ \text{lbs} \end{split}\]

Therefore slightly higher than at 950 RPM

c)

    \[\begin{split} \text{Dynamic Force} &= \tilde{m}e\omega^{2} \\&= \frac{W_{im}}{g}e\omega^{2} \\&= \frac{W_{im}}{386}(13.8)(15.2\cdot2\pi)^{2} \\&= 212 \ \text{lbs} \end{split}\]

    \[\begin{split} W_{im} &= \frac{(212)(386)}{(13.8)(15.2\cdot2\pi)^{2}} \\&= 0.60 \ \text{lbs} \end{split}\]

Perhaps a wet hand towel out of balance

The commercial washer above was considered under the assumption that the vibration was only a vertical mode. However the excitation due to rotating imbalance would cause excitation both horizontally and vertically that would excite coupled modes as well. As a result it is necessary to consider all the planar modes of vibration to investigate if they are near the excitation frequency. The diagram of the dryer is shown below. The lateral stiffness of the supporting springs is estimated to be 0.8 of the vertical stiffness. The excitation frequency is 950 rpm.

    \[W = 1049+255 \ \text{lbs}\]

    \[\bar{r} = 15 \text{ in (radius of gyration)}\]

    \[k_1 = k_2 = k = 1150 \ \text{lbs/in}\]

    \[\ell_1 = \ell_2 = 0.8k \text{ (4 springs)}\]

    \[h = 28 \ \text{in}\]

    \[a = 19 \ \text{in}\]

a) Specialize the equations of motion for the case shown in the above diagram

b) Determine if any of the modes are uncoupled. For the coupled modes derive the equations of motion for free vibration starting from Newton’s laws.

c) Calculate the natural frequencies and mode shapes for the dryer.

d) If the supports for the dryer were redesigned to make loading easier with h = 0 (as illustrated) what are the natural frequencies compared to the original design. How do they compare with the excitation frequency? 

a)

      \begin{equation*} \begin{bmatrix} m & 0 & 0\\ 0 & m & 0\\ 0 & 0 & J_G\\ \end{bmatrix} \begin{Bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{\theta} \\ \end{Bmatrix} + \begin{bmatrix} 2\ell & 0 & 2\ell h\\ 0 & 2k & 0\\ 2\ell h & 0 & 2\ell h^{2}+2ka^{2}\\ \end{bmatrix} \begin{Bmatrix} x\\ y \\ \theta \\ \end{Bmatrix} = \begin{Bmatrix} 0\\ 0 \\ 0 \\ \end{Bmatrix} \end{equation*}

b) Therefore m\ddot{y} + 2ky = 0,p_v= \sqrt{\frac{2k}{m}}

The vertical mode is uncoupled

    \[\begin{split} \underrightarrow{+} \sum F_x &= m\ddot{x} \\&= -4\ell(x+h\theta) \end{split}\]

    \[\begin{split} + \circlearrowleft \sum M_G &= J_G \ddot{\theta} \\&= -4\ell(x+h\theta)h-2k(a\theta)a - 2k(a\theta)a \end{split}\]

Therefore

    \[m\ddot{x} + 4kx + 4\ell h\theta = 0\]

    \[J\ddot{\theta} + 4\ell hx+ (4\ell h^{2}+4ka^{2})\theta = 0\]

Which matches the reduction of the general case where k has been replaced by 2k and \ell by 2\ell

c) Calculate the natural frequencies and mode shapes for the dryer

The vertical mode is as previously

    \[\begin{split} p_v &= \sqrt{\frac{2k}{m}}_\text{eff} \\&= \sqrt{\frac{4\cdot1150\cdot386}{(1049+255)}} \\&= 36.90 \ \text{rad/s} \\&= 5.98 \ \text{Hz} \end{split}\]

which is well below the excitation frequency of 950RPM (15.8Hz)

      \begin{equation*} \begin{bmatrix} m & 0 \\ 0 & J_G \\ \end{bmatrix} \begin{Bmatrix} \ddot{x}\\ \ddot{\theta}\\ \end{Bmatrix} + \begin{bmatrix} 4\ell & 4\ell h\\ 4\ell h & 4\ell h^{2}+4ka^{2}\\ \end{bmatrix} \begin{Bmatrix} x\\ \theta\\ \end{Bmatrix} = \begin{Bmatrix} 0\\ 0\\ \end{Bmatrix} \end{equation*}

Assume that

      \begin{equation*} \begin{Bmatrix} x\\ \theta\\ \end{Bmatrix} = \begin{Bmatrix} X\\ \Theta\\ \end{Bmatrix} \sin(pt+\phi) \end{equation*}
      \begin{equation*} \begin{bmatrix} 4\ell-mp^{2} & 4\ell h \\ 4\ell h & 4\ell h^{2} + 4ka^{2} - m\bar{r}^{2}p^{2}\\ \end{bmatrix} =0 \end{equation*}

    \[(4\ell-mp^{2})(4\ell h^{2} + 4ka^{2} - m\bar{r}^{2}p^{2}) - 16\ell^{2}h^{2} = 0\]

    \[16\ell^{2}h^{2} + 16k\ell a^{2} - 4\ell \bar{r}^{2}p^{2}m- 4\ell h^{2}mp^{2} - 4ka^{2}mp^{2} + m\bar{r}^{2}p^{4} -16\ell^{2}h^{2}= 0\]

    \[m^{2}\bar{r}^{2}p^{4}-mp^{4} -mp^{2}(4\ell \bar{r}^{2}+4\ell h^{2}+4ka^{2})+16k\ell a^2=0\]

Therefore

    \[p^{4}-p^{2}\biggr[\frac{4\ell}{m}+ \frac{4\ell h^{2}}{m\bar{r}^{2}+\frac{4ka^{2}}{m\bar{r}^{2}}}\biggr]+ \frac{16k\ell}{m^{2}} \frac{a^{2}}{\bar{r}^{2}}\]

    \[p^{4}- p^{2}\big[1089+3796+2185\big]+ 2379851\]

    \[p^{4}-p^{2}[7070]+2379851=0\]

    \[\begin{split} p^{2}_{1,2} &= -\frac{7070}{2} \pm \sqrt{\Big(\frac{7070}{2}\Big)^{2}-2379851} \\&= 3535 \pm 3180 \\&= 354,6715 \end{split}\]

    \[p_{1,2} = 18.8, 81.9 \ \text{rad/s} \quad (2.99\text{ Hz}) (13\text{ Hz})\]

Mode Shapes

    \[(4\ell-mp^{2})X+4\ellh\Theta = 0\]

Therefore

    \[\begin{split} \frac{X}{\Theta} &= \frac{-4\ellh}{(4\ell-mp^{2})} \\&= \frac{-h}{1-\frac{m}{4\ell}p^{2}} \end{split}\]

    \[\begin{split} \frac{X}{\Theta}\Big|_{1} &= \frac{-28}{1-\frac{(1049+255)(354)}{386(4)(0.8)(1150)}} \\&= -28.3 \end{split}\]

    \[\begin{split} \frac{X}{\Theta}\Big|_{2} &= \frac{-28}{1-\frac{(1304)(6715)}{386(4)(0.8)(1150)}} \\&= 5.42 \end{split}\]

d)

    \[p_x = \sqrt{\frac{4\ell}{m}}\]

    \[p_y = \sqrt{\frac{4k}{m}}\]

    \[p_\theta = \sqrt{\frac{4ka^{2}}{m\bar{r}^{2}}}\]

    \[p_x = \sqrt{\frac{4(0.8)(1150)}{1304}}\cdot386 = 33.0 \ \text{rad/s } (5.25 \text{ Hz})\]

    \[p_y = \sqrt{\frac{4(1150)(386)}{1304}} = 36.9 \ \text{rad/s } (5.87 \text{ Hz})\]

    \[p_x = \sqrt{\frac{4(1150)(19)^{2}(386)}{1304(15)^{2}}}= 46.7 \ \text{rad/s } (7.44 \text{ Hz})\]

now all natural frequencies are all below the excitation frequency of 158Hz (950RPM)

Semi-Definite Systems – Rigid Body Modes

So far we have dealt with systems that have positive definite stiffness and mass matrices. This means the eigenvalues are positive & real. We can then define an orthonormal basic system. However consider the example

      \begin{equation*}[m]=\begin{bmatrix}m_1 & 0 & 0  \\0 & m_2 & 0 \\0 & 0 & m_3 \\\end{bmatrix}\end{equation*}
      \begin{equation*}[k]=\begin{bmatrix}k_1 & -k_1 & 0  \\-k_1 & k_1+ k_2 & -k_2 \\0 & -k_2 & k_2 \\\end{bmatrix}\end{equation*}

but the stiffness matrix is not positive definitive. It is singular. This is because one “natural frequency” is zero.

    \[p_o = 0\]

      \begin{equation*}\{u_o\}=\begin{pmatrix}1  \\1   \\1  \\\end{pmatrix}\end{equation*}

u_o is called a zeroth mode

      \begin{equation*} [k] \{u\}= \begin{bmatrix} k_1 & -k_1 & 0 \\ -k_1 & k_1+ k_2 & -k_2 \\ 0 & -k_2 & k_2 \\ \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix} =\{u\}^{T}[k] \end{equation*}

Therefore

      \begin{equation*}\{u\}^{T}[m]\begin{Bmatrix}\ddot{x}_1\\\ddot{x}_2\\\ddot{x}_3\\\end{Bmatrix}=0\end{equation*}

but since [m] is positive definite \{u\}^{T}[m] \neq 0, therefore \{\ddot{x}\} = 0, \{x\} = \alpha + \beta t

and the system is simply moving off in one direction. To suppress this mode we set:

    \[\{u_o\}^{T}[m]\{x\}= 0\]

This is the constraint equation:

      \begin{equation*}\begin{bmatrix}1 & 1 & 1 \\\end{bmatrix}\begin{bmatrix}m_1 & 0 & 0 \\0 & m_2 & 0 \\0 & 0 & m_3 \\\end{bmatrix}\begin{Bmatrix}x_1\\x_2\\x_3\\\end{Bmatrix}= 0\end{equation*}

    \[m_1 x_1 + m_2 x_2 + m_3 x_3 = 0\]

or

    \[m_1 \dot{x}_1 + m_2 \dot{x}_2 + m_3 \dot{x}_3 = 0\]

Where this expression says the momentum of the 3 masses must be zero. This essentially says this is a 2 DOF system.

Note: If we tried to find a flexibility matrix it is not defined.

We can write the constraint in a compound form as:

      \begin{equation*}\begin{Bmatrix}x_1\\x_2\\x_3\\\end{Bmatrix}=\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\\frac{-m_1}{m_3} & \frac{-m_2}{m_3} & 0 \\\end{bmatrix}\begin{Bmatrix}x_1\\x_2\\x_3\\\end{Bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\frac{-m_1}{m_3} & \frac{-m_2}{m_3} \\\end{bmatrix}\begin{Bmatrix}x_1\\x_2\\\end{Bmatrix}=[C]\begin{Bmatrix}x_1\\x_2\\\end{Bmatrix}\end{equation*}

We can now solve the eigenvalue problem using

    \[ [k^{'}] = [C]^{T}[k][C] \]

    \[ [m^{'}] = [C]^{T}[m][C] \]

so that

    \[ [k^{'}]\{u\} = p^{2}[m^{'}]\{u\} \]

For the example above set k_1 = k_2 = k, m_1 = m_2 = m_3 = m

      \begin{equation*}[k']=\begin{bmatrix}1 & 0 & -1 \\0 & 1 & -1 \\\end{bmatrix}\begin{bmatrix}k & -k & 0 \\-k & 2k & -k \\0 & -k & k \\\end{bmatrix}\begin{bmatrix}1 & 0 \\0 & 1 \\-1 & -1 \\\end{bmatrix}=\begin{bmatrix}1 & 0 & -1 \\0 & 1 & -1 \\\end{bmatrix}\begin{bmatrix}k & -k \\0 & 3k \\-k & -2k \\\end{bmatrix}=\begin{bmatrix}2k & k \\k & 5k \\\end{bmatrix}\end{equation*}
      \begin{equation*}[m'] = m\begin{bmatrix}2 & 1 \\1 & 2 \\\end{bmatrix}\end{equation*}
      \begin{equation*}\begin{bmatrix}2 & 1 \\1 & 5 \\\end{bmatrix}\begin{Bmatrix}u_1 \\u_2 \\\end{Bmatrix}-\frac{p^{2}m}{k}\begin{bmatrix}2 & 1 \\1 & 2 \\\end{bmatrix}\begin{Bmatrix}u_1 \\u_2 \\\end{Bmatrix}= 0\end{equation*}
      \begin{equation*}\begin{bmatrix}2-2\frac{p^{2}m}{k} & 1-\frac{p^{2}m}{k} \\1-\frac{p^{2}m}{k} & 5-2\frac{p^{2}m}{k} \\\end{bmatrix}\begin{Bmatrix}u_1 \\u_2 \\\end{Bmatrix}={0}\end{equation*}

    \[\left(2-2\frac{p^{2}m}{k}\right)\left(5-2\frac{p^{2}m}{k}\right) - \left(1 - \frac{p^{2}m}{k}\right)^{2} = 0\]

    \[10-14\frac{p^{2}m}{k} + 4\left(\frac{p^{2}m}{k}\right)^{2} - [1 - 2\frac{p^{2}m}{k} + \frac{p^{4}m^{2}}{k^{2}}] = 0 \]

    \[9 - 12 \frac{p^{2}m}{k} + 3 \frac{p^{4}m^{2}}{k^{2}} = 0\]

    \[\frac{p^{4}m^{2}}{k^{2}} - \frac{4p^{2}m}{k} + 3 = 0\]

    \[p^{2} = \{2 \pm \sqrt{1}\} \frac{k}{m}\]

    \[p^{2}_{1,2} = \frac{k}{m}, \frac{3k}{m}\]

      \begin{equation*}\{u^{'}_1\}=\begin{Bmatrix}1\\0\\\end{Bmatrix}\end{equation*}
      \begin{equation*} \{u^{'}_2\}= \begin{Bmatrix} \frac{1}{2}\\ -1\\ \end{Bmatrix} \end{equation*}

    \[x_3 = -x_1 - x_2\]

      \begin{equation*}\{u_1\}=\begin{Bmatrix}1\\0\\-1\\\end{Bmatrix}\end{equation*}
      \begin{equation*} \{u_1\}= \begin{Bmatrix} \frac{1}{2}\\ -1\\ \frac{1}{2}\\ \end{Bmatrix} \end{equation*}

A second example of a 2DOF system is the vibration absorber as it can provide quite spectacular results when applied correctly. We will consider it first as a 2DOF undamped system, then include viscous damping. The closed form solution for the case of damping is much more involved.

Undamped Vibration Absorber

    \[\begin{split} + \uparrow \sum F_{x_1} &= m_1 \ddot{x_1} \\&= -k_1x_1 + k_2(x_2 - x_1) + F_o\sin\omega t \quad (1) \end{split} \]

    \[ \begin{split} + \uparrow \sum F_{x_2} &= m_2\ddot{x_2} \\&= -k_2(x_2 - x_1) \quad (2) \end{split} \]

    \[ m_1\ddot{x_1} + (k_1+k_2)x_1 - k_2x_2 = F_o \sin\omega t \]

    \[ m_2\ddot{x_2} - k_2x_1 + k_2x_2 = 0 \]

    \[ \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \begin{Bmatrix} \ddot{x_1} \\ \ddot{x_2} \end{Bmatrix} + \begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} F_o\sin\omega t \\ 0 \end{Bmatrix} \]

For S.S. solution, assume:

    \[\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} \sin \omega t \]

Under the steady state assumption, the equations of motion become:

    \[ [-m_1\omega^2 + k_1 + k_2]X_1 - k_2X_2 = F_o \quad (*) \]

    \[-k_2X_1 + (k_2 - m_2\omega^2)X_2 = 0 \quad (**) \]

Therefore, from (**):

    \[ X_2 = \frac{k_2X_1}{(k_2-m_2\omega^2)}\]

And then from (*):

    \[(k_1 +k_2-m_1\omega^2)X_1 - \frac{k_2^2X_1}{k_2-m_2\omega^2} = F_o\]

    \[\frac{\Big\{[(k_1+k_2)-m_1\omega^2](k_2-m_2\omega^2)-k_2^2\Big\}X_1}{k_2-m_2\omega^2} = F_o \]

Therefore:

    \[X_1 = \frac{F_o(k_2-m_2\omega^2)}{[(k_2-m_2\omega^2)(k_1+k_2-m_1\omega^2) - k_2^2]} \]

    \[X_2 = \frac{k_2F_o}{[(k_2 - m_2\omega^2)(k_1+k_2-m_1\omega^2)-k_2^2]} \]

For use in vibration absorber applications, we define:

    \[p_{11} := \sqrt{\frac{k_1}{m_1}},\ p_{22} := \sqrt{\frac{k_2}{m_2}},\ \mu := \frac{m_2}{m_1} \]

Then divide numerator and denominator by k_1k_2.

For the 2DOF system shown with the forcing appiled to the mass m_1, the response of the masses m_1 and m_2 is:

    \[X_1 = \frac{F_o(k_2-m_2\omega^2)}{(k_2-m_2\omega^2)(k_1+k_2-m_1\omega^2)-k_2^2} \quad (*) \]

    \[X_2 = \frac{F_ok_2}{(k_2-m_2\omega^2)(k_1+k_2-m_1\omega^2)-k_2^2} \quad (**) \]

It is seen that from (*) that if k_2 -m_2\omega^2 = 0 or \frac{k_2}{m_2} = \omega^2, then the mass m_1 is stationary. This leads to the idea that the motion of m_1 can theoretically be reduced to zero if we employ k_2, m_2 at the operating frequency \omega. For this application. the subsystem k_2, m_2 is called a vibration absorber and at the operating frequency:

    \[X_1 = 0 \text{ and } X_2 = \frac{-F_o}{k_2} \]

For this application it is useful to modify (*) and (**) to reflect what we can call – the original system k_1, m_1, the absorber system k_2, m_2 using the definitions:

    \[p_{11} := \sqrt{\frac{k_1}{m_1}},\ p_{22} := \sqrt{\frac{k_2}{m_2}},\ \mu := \frac{m_2}{m_1} \]

Using those definitions show that (*) and (**) become:

    \[X_1 = \frac{\frac{F_o}{k_1}\left[1-\left(\frac{\omega}{p_{22}}\right)^2\right]}{\left{\left[1-\left(\frac{\omega}{p_{22}}\right)^2\right]\left[1+\mu\left(\frac{p_{22}}{p_{11}}\right)^2-\left(\frac{\omega^2}{p_{11}}\right)^2\right] - \mu\left(\frac{p_{22}}{p_{11}}\right)^2\right}} \]

    \[X_2 = \frac{\frac{F_o}{k_1}}{\left{\left[1-\left(\frac{\omega}{p_{22}}\right)^2\right]\left[1+\mu\left(\frac{p_{22}}{p_{11}}\right)^2-\left(\frac{\omega^2}{p_{11}}\right)^2\right] - \mu\left(\frac{p_{22}}{p_{11}}\right)^2\right}} \]

And when \omega^2 = \frac{k_2}{m_2}:

    \[X_1 = 0,\ X_2 = -\frac{F_o}{k_2}\]

This shows that when the subsystem k_2, m_2 is timed so \frac{k_2}{m_2} = p_{22}^2 that m_1 is motionless as the force in the spring k_2 exactly opposes the excitation force F_o.

The natural frequencies of the combined system are found from setting the denominator of the responses X_1, X_2 to zero. Therefore, the natural frequencies are found when \omega is replaced by p and the expression becomes:

    \[ \Big[1 - \Big(\frac{p}{p_{22}}\Big)^2\Big]\Big[1 + \mu\Big(\frac{p_{22}}{p_{11}}\Big)^2-\Big(\frac{p}{p_{11}}\Big)^2\Big] - \mu\Big(\frac{p_{22}}{p_{11}}\Big)^2 = 0\]

    \[1 + \mu\Big(\frac{p_{22}}{p_{11}}\Big)^2-\Big(\frac{p}{p_{11}}\Big)^2-\Big(\frac{p}{p_{22}}\Big)^2-\mu\Big(\frac{p_{22}}{p_{11}}\Big)^2\Big(\frac{p}{p_{22}}\Big)^2+\Big(\frac{p}{p_{11}}\Big)^2\Big(\frac{p}{p_{22}}\Big)^2 -\mu\Big(\frac{p_{22}}{p_{11}}\Big)^2 = 0\]

Therefore:

    \[\Big(\frac{p}{p_{22}}\Big)^4\Big(\frac{p_{22}}{p_{11}}\Big)^2-\Big(\frac{p}{p_{22}}\Big)^2\Big[1+\mu\Big(\frac{p_{22}}{p_{11}}\Big)^2 + \Big(\frac{p_{22}}{p_{11}}\Big)^2\Big] + 1= 0 \]

    \[\Big(\frac{p}{p_{22}}\Big)^4-\Big(\frac{p}{p_{22}}\Big)^2\Big[1+\mu+\Big(\frac{p_{11}}{p_{22}}\Big)^2\Big] + \Big(\frac{p_{11}}{p_{22}}\Big)^2 = 0 \quad (*) \]

If p_{11} = p_{22}, this becomes:

    \[\Big(\frac{p}{p_{22}}\Big)^4 - \Big(\frac{p}{p_{22}}\Big)^2[2+\mu] + 1 = 0\]

If instead of the mass m_1 being excited by the force F_o\sin\omega t, the exication is through the base as x_o = X_o\sin\omega t, what is the difference in the response of X_1 and X_2 of masses m_1 and m_2 ?

The mass ratio, \mu, essentially determines the spread of the two resonances:

    \[\mu = 0.1, \quad \frac{\omega}{p_{22}} = 0.85,\ 1.17\]

    \[\mu = 0.2, \quad \frac{\omega}{p_{22}} = 0.79,\ 1.25\]

    \[\mu = 0.05, \quad \frac{\omega}{p_{22}} = 0.89,\ 1.122\]

(\text{TAIPEI TOWER:} \quad m_2 = 700\ \text{TONS},\ \mu \approx 0.05)

    \[\begin{split} \sum F_{x_1} &= m_1\ddot{x_1} \\&= -k_1(x_1-x_o) + k_2(x_2-x_1) \quad (1) \end{split} \]

    \[\begin{split} \sum F_{x_2} &= m_2\ddot{x_2} \\&=-k_2(x_2-x_1) \quad (2) \end{split}\]

Therefore, the equations are identical except that F_o is replaced by k_1X_o. The response of x_1 and x_2 for the vibration absorber is:

    \[X_1= \frac{X_0[1-(\frac{\omega}{p_{22}})^2]}{\{[1-(\frac{\omega}{p_{22}})^2][1+\mu(\frac{p_{22}}{p_{11}})^2-(\frac{\omega}{p_{11}})^2]-\mu(\frac{p_{22}}{p_{11}})^2\}}\]

    \[X_2= \frac{X_0}{\{[1-(\frac{\omega}{p_{22}})^2][1+\mu(\frac{p_{22}}{p_{11}})^2-(\frac{\omega}{p_{11}})^2]-\mu(\frac{p_{22}}{p_{11}})^2\}}\]

Again when \frac{k_2}{m_2} := p_{22}^2 = \omega^2:

    \[X_1 = 0\]

    \[\begin{split} X_2 &= -X_o\frac{1}{\mu}\Big(\frac{p_{11}}{p_{22}}\Big)^2 \\&= -X_o\frac{m_1}{m_2}\frac{k_1}{m_1}\frac{m_2}{k_2} \\&= -X_o\frac{k_1}{k_2} \end{split} \]

Therefore, k_2X_2 = -k_1X_1 so that at any time:

And the net dynamic force on m_1 is zero!

To solve the system with damping it is more advantageous to use complex numbers:

    \[\{x\} =\{X\}e^{i\omega t}\]

Or:

    \[\{x\} =\{\zeta\}e^{st}\]

There are a number of special cases of the general forced damped analysis that are useful.

Damped Forced 2DOF System

Consider a general damped system that can be specialized for special applications.

    \[f_1(t) = f_1e^{st}\]

    \[f_2(t) = f_2e^{st}\]

    \[m\ddot{x_1} = k_2(x_2 - x_1) + c_2(\dot{x_2}-\dot{x_1}) - k_1x_1 - c_1\dot{x_1} + f_1(t) \]

    \[m\ddot{x_1} + (k_1+k_2)x_1 + (c_1+c_2)\dot{x_1} - k_2x_2 - c_2\dot{x_2} = f_1(t) \]

    \[m_2\ddot{x_2} = f_2(t) - k_2(x_2-x_1)-c_2(\dot{x_2}-\dot{x_1})\]

Assume a solution

    \[\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} \zeta_1 \\ \zeta_2 \end{Bmatrix} e^{st}\]

Thus:

    \[[m_1s^2 + (c_1+c_2)s + (k_1+k_2)]\zeta_1 - [c_2s+k_2]\zeta_2 = f_1\]

    \[-(c_2s + k_2)\zeta_1 + [m_2s^2+c_2s+k_2]\zeta_2 = f_2\]

Therefore:

    \[V_{11} = m_1s^2 + (c_1 + c_2)s + k_1 + k_2 \]

    \[V_{12} = -(c_2s + k_2) = V_{21} \]

    \[V_{22} = m_2s^2 + c_2s + k_2 \]

    \[V_{11}\zeta_1 + V_{12}\zeta_2 = f_1\]

    \[V_{21}\zeta_1 + V_{22}\zeta_2 = f_2\]

And we wish to find \zeta_1 and \zeta_2:

    \[ \begin{bmatrix} V_11 & V_12 \\ V_21 & V_22 \end{bmatrix} \begin{bmatrix} \zeta_1 \\ \zeta_2 \end{bmatrix} = \begin{bmatrix} f_1 \\ f_2 \end{bmatrix} \]

    \[ [V]^{-1} \equiv [Y] \]

    \[Y_{11}(s) = \frac{V_{22}(s)}{\Delta},\ Y_{12} = -\frac{V_{21}(s)}{\Delta},\ Y_{22} = \frac{V_{11}(s)}{\Delta}\]

Where \Delta = |V|. Thus:

    \[\zeta_1 = Y_{11}(s)f_1 + Y_{12}(s)f_2\]

    \[\zeta_2 = Y_{21}(s)f_1 + Y_{22}(s)f_2\]

There is a lot hidden in these symbols so we can consider some special cases.

1. Untuned Viscous Vibration Absorber
(Houdaille damper or viscous Lanchester damper)

If we look at our general solution then:

    \[c_2 = c,\ c_1 = 0,\ k_2 = 0,\ f_2 = 0 \]

    \[V_{11} = m_1s^2 + cs + k_1 \]

    \[V_{12} = -cs = V_{21} \]

    \[V_{22} = m_2s^2 + cs \]

    \[\zeta_1 = Y_{11}(s) f_1,\quad Y_{11}(s) = \frac{V_{22}(s)}{\Delta} \]

    \[\zeta_2 = Y_{21}(s) f_1,\quad Y_{21}(s) = -\frac{V_{21}(s)}{\Delta} \]

Therefore:

    \[ \Delta = [m_1s^2 + cs + k][m_2s^2 + cs] - (cs)^2 \]

Put s =i\omega:

    \[\begin{split} \Delta &= [k-m_1\omega^2 + ic\omega][-m_2\omega^2 +ic\omega]+c^2\omega^2 \\&= -m_2\omega^2(k-m_1\omega^2) + ic\omega[-m_2\omega^2 + (k-m\omega^2) \end{split} \]

Therefore:

    \[ \zeta_1 = \frac{(-m_2\omega^2 + ic\omega)F_o}{-m_2\omega^2(k-m_1\omega^2) +ic\omega[-m_2\omega^2 + (k-m\omega^2)]}\]

    \[ \zeta_2 = \frac{ic\omega}{-m_2\omega^2(k-m_1\omega^2) + ic\omega[-m_2\omega^2 + (k-m\omega^2)]} \]

We are really interested in the amplitudes. Use the result that if:

    \[\begin{split} \zeta &= \frac{B+iC}{D+iE} \cdot \frac{D-iE}{D-iE} \\&=\frac{BD+CE+i(CD-BE)}{D^2+E^2} \end{split} \]

We wish to write \zeta = Ae^{i\alpha}.

    \[ \begin{split} A &= \frac{\sqrt{(BD+CE)^2 + (CD-BE)^2}}{D^2 + E^2} \\&= \frac{\sqrt{(BD)^2 + (CE)^2 + 2BDCE + (CD)^2 + (BE)^2 - 2BCDE}}{D^2+E^2} \\&= \frac{\sqrt{B^2(D^2+E^2) + C^2(D^2+E^2)}}{D^2+E^2} \\&= \sqrt{\frac{B^2+C^2}{D^2+E^2}} \end{split} \]

    \[ \alpha = \tan^{-1} \frac{CD-BE}{BD+CE} \]

    \[X_1 = \frac{F_o\sqrt{(m_2\omega^2)^2+(c\omega)^2}}{\sqrt{[m_2\omega^2(k-m_1\omega^2)]^2+(c\omega)^2[m_2\omega^2 - (k-m_1\omega^2)]^2}} \]

    \[X_o = \frac{F_o}{k},\ p_1^2 = \frac{k_1}{m_1},\ \nu = \frac{c}{2m_1p_1},\ \mu = \frac{m_2}{m_1},\ r = \frac{\omega}{p_1} \]

Thus:

    \[ \frac{X_1}{X_o} = \frac{\sqrt{(\frac{m_2}{k}\omega^2)^2 + (\frac{c\omega}{k})^2}}{\sqrt{[\frac{m_2\omega^2}{k}(1-\frac{m_1\omega^2}{k})]^2 + (\frac{c\omega}{k})^2[\frac{m_2\omega^2}{k} - (1- \frac{m_1\omega^2}{k})]^2}} \]

    \[\frac{m_2}{m_1} \frac{m_1}{k} \omega^2 = \mu r^2 = \frac{m_2}{k}\omega^2\]

    \[\frac{c\omega}{k} = \frac{cm_1}{m_1k}\omega = \frac{c}{m_1p_1^2} \]

    \[\omega = 2\nu r\]

Thus:

    \[\begin{split} \frac{X_1}{X_o} &= \frac{\sqrt{\mu^2r^4 + (2\nu r)^2}}{\sqrt{[\mu r^2(1-r^2)]^2 + (2\nu r)^2[\mu r^2 - (1-r^2)]^2}} \\& = \frac{\sqrt{\mu^2r^2 + 4\nu ^2}}{\sqrt{[\mu r(1-r^2)]^2 + 4\nu ^2[\mu r^2 - (1-r^2)]^2}} \\&= \sqrt{\frac{N}{D}} \end{split} \]

We wish to find the optimum ratio \nu which minimizes this amplitude;

    \[\begin{split} \frac{\text{d}\frac{X_1}{X_o}}{\text{d}\nu} &= \frac{D^{\frac{1}{2}}\frac{N^{-\frac{1}{2}}}{2}(8\nu) - N^{\frac{1}{2}}\frac{D^{\frac{-1}{2}}}{2}(8\nu)[\mu r^2 - (1-r^2)]^2}{D} \\&= 0 \end{split} \]

Multiply by D^{\frac{1}{2}}N^{\frac{1}{2}}. Therefore:

    \[ D-N[\mu r^2 - (1-r^2)]^2 = 0 \]

    \[[\mu r(1-r^2)]^2 + 4\nu^2[\mu r^2 - (1-r^2)]^2 = (\mu^2 r^2 + 4\nu^2)[\mu r^2 - (1-r^2)]^2 \]

    \[[\mu r(1-r^2)]^2 = (\mu r)^2[\mu r^2 - (1-r^2)]^2 \]

    \[(1-r^2) = \mu r^2 - (1-r^2) \]

    \[ 2(1-r^2) = \mu r^2 \]

    \[ 2 = r^2(2+\mu) \]

Therefore:

    \[r = \sqrt{\frac{2}{2+\mu}},\quad \mu = \frac{2(1-r^2)}{r^2} \]

Therefore, the minimum amplitude for optimum \nu occurs when r = \sqrt{\frac{2}{2+\mu}} = 0.894\ (\mu = \frac{1}{2}). To find the value of \nu, use this value of r.

    \[ \begin{split} \frac{X_1}{X_o} &= \frac{\sqrt{\frac{\mu^2 2}{2+\mu} +4\nu^2}}{\sqrt{\frac{\mu^2 2}{2+\mu} (1- \frac{2}{2+\mu})^2 + 4\nu^2[\mu\frac{2}{2+\mu} - (1-\frac{2}{2+\mu})]^2}} \\&= \frac{\sqrt{\frac{2\mu^2}{2+\mu} + 4\nu^2}}{\sqrt{\frac{2\mu^2}{2+\mu}\frac{\mu^2}{(2+\mu)^2} + 4\nu^2[\frac{2\mu}{2+\mu} - (\frac{\mu}{2+\mu}) ]^2}} \\&= \frac{\sqrt{\frac{2\mu^2 + 4\nu(2+\mu)}{2+\mu}}}{\sqrt{\frac{2\mu^4}{(2+\mu)^3} + \frac{4\nu^2\mu^2}{(2+\mu)^2}}} \\&= \frac{\sqrt{2\mu^2 +4\nu^2(2+\mu)}}{\mu\sqrt{\frac{2\mu^2 + 4\nu^2(2+\mu)}{(2+\mu)^2}}} \\&= \frac{2+\mu}{\mu} = \frac{X_1}{X_o}\Big|_{\text{MAX}} \\&= 5\quad (\mu = \frac{1}{2}) \end{split} \]

The amplitude at this value of r is independent of \nu!

    \[ \begin{split} \frac{X_1}{X_o} &= \frac{\sqrt{\mu^2r^2 + 4\nu^2}}{\sqrt{\mu^2r^2(1-2r^2+r^4)+4\nu^2[r^2(1+\mu)-1]^2}} \\&= \frac{\sqrt{\mu^2r^2 + 4\nu^2}}{\sqrt{\mu^2(r^2-2r^4+r^6) + 4\nu^2[r^4(1+\mu)^2-2r^2(1+\mu)+1]}} \\&= \sqrt{\frac{N}{D}} \end{split} \]

    \[ \begin{split} \frac{\text{d}(\frac{X_1}{X_o})}{\text{d}r} &= \frac{D^{\frac{1}{2}}\frac{N^{-\frac{1}{2}}}{2}2\mu^2r-N^{\frac{1}{2}}\frac{D^{-\frac{1}{2}}}{2}[\mu^2(2r+8r^3+6r^5) + 4\nu^2(4r^3(1+\mu)^2 - 4r(1+\mu))]}{D} \\& = 0 \end{split} \]

Multiply by D^{\frac{1}{2}}N^{-\frac{1}{2}}. Then:

    \[ 0 = D(\mu^2r)-N[\mu^2(r-4r^3+3r^5) + 2\nu^2(4r^3(1+\mu)^2-4r(1+\mu))] \]

Now evaluate this for r = \sqrt{\frac{2}{2+\mu}},\ r^2 = \frac{2}{2+\mu}.

    \[D\mu^2 = N[\mu^2(1-4r^2+3r^4)+2\nu^2(4r^2(1+\mu)^2-4(1+\mu))] \]

Note that:

    \[ \begin{split} \frac{X_1}{X_0} &= \sqrt{\frac{N}{D}} \\&= \frac{2+\mu}{\mu} \end{split}\]

    \[ \frac{N}{D} = \frac{(2+\mu)^2}{\mu^2}\]

    \[D = \frac{\mu^2}{(2+\mu)^2}N\]

    \[ \frac{\mu^4}{(2+\mu)^2} = \mu^2\bigg(1 - \frac{4(2)}{2+\mu} + \frac{3(4)}{(2+\mu)^2}\bigg) + 2\nu^2\bigg[\frac{8}{(2+\mu)}(1+\mu)^2 - 4(1+\mu)\bigg] \]

    \[\mu^4 = \mu^2\bigg[ (2+\mu)^2 - 8 ( 2+ \mu) + 12 \bigg] + 2\nu^2\bigg[8(1+\mu)^2(2+\mu) - 4 ( 1 + \mu)(2+\mu)^2\bigg] \]

    \[\mu^4 = \mu^2 \bigg[ 4 + 4 \mu + \mu^2 - 16 - 8 \mu + 12 \bigg ] + 2\nu^2(1+\mu)(2+\mu)\bigg[8(1 + \mu ) - 4(2 + \mu) \bigg] \]

    \[ \mu^4 = \mu^2\big[\mu^2 - 4\mu\big] + 2\nu^2(1+\mu)(2+\mu)4\mu\]

    \[ \mu^4 = \mu^4 - 4\mu^3 + 2\nu^2(1 + \mu)(2+ \mu) 4 \mu \]

    \[ 4\mu^3 = 2\nu^2(1+\mu)(2+\mu)4\mu \]

    \[\nu^2 = \frac{\mu^2}{2(1+\mu)(2+\mu)} \]

    \[ \nu_\text{OPT} = \frac{\mu}{\sqrt{2(1+\mu)(2+\mu)}} \]

    \[ \mu = \frac{1}{2} ,\ \nu_\text{OPT} = 0.183\]

When r = 1:

    \[\frac{X_1}{X_0} = \frac{\sqrt{\mu^2 + 4\nu^2}}{\sqrt{4\nu^2\mu^2}} \]

When \mu = \frac{1}{2},\ \nu = 0.365:

    \[ \frac{X_1}{X_0} = \frac{\sqrt{ 0.25 + 4(0.365)^2}}{\sqrt{4(0.365)^2(\frac{1}{4})}} = 2.42\]

When \nu = 1.0,\ \mu = \frac{1}{2}:

    \[\frac{X_1}{X_0} = \frac{\sqrt{0.25 + 4}}{\sqrt{4(1)(\frac{1}{4})}} = 2.06\]

When \nu = 0.1,\ \mu = \frac{1}{2}:

    \[\frac{X_1}{X_0} = \frac{\sqrt{0.25 + 0.04}}{\sqrt{4(0.01)(\frac{1}{4})}} = 5.39\]

When \nu = 0.1825,\ \mu = \frac{1}{2}:

    \[\frac{X_1}{X_0} = \sqrt{\frac{0.25 + 4(0.1825)^2}{0.1825^2}} = 3.39 \]

Example

A second very practical application is the viciously damped vibration absorber.

Solution

From the general solution:

    \[ c_1 = 0,\ c_2 = C \]

Therefore:

    \[V_{11} = m_1s^2 + cs + k_1 + k_2 \]

    \[V_{12} = - (cs + k_2) = V_{21}\]

    \[V_{22} = m_2s^2 + cs + k_2 \]

    \[ \zeta_1 = Y_{11}f_1 ,\ \zeta_2 = Y_{21}f_1 \]

    \[Y_{11} = \frac{V_{22}}{\Delta},\ Y_{21} = -\frac{V_{12}}{\Delta} \]

    \[\Delta = \big[ - m_1\omega^2 + ic\omega +k_1 + k_2\big] \big[-m_2\omega^2 + ic\omega + k_2 \big] - \big[k_2 + ic\omega\big]^2\]

    \[\implies \Delta = \big[(-m_1\omega^2 + k_1)(-m_2\omega^2 + k_2) - m_2\omega^2 k_2\big] + ic\omega[-m_1\omega^2 + k_1 - m_2\omega^2]\]

    \[ \zeta_1 = Y_{11}F_0e^{i\omega t} = \frac{(-m_2\omega^2 + ic\omega + k_2)}{\Delta}F_0e^{i\omega t}\]

If we set \zeta_1 = X_1e^{i\theta_1},\ \zeta_2 = X_2e^{i\theta_2} :

    \[ (\frac{X_1}{F_0})^2 = \frac{(k_2 - m_2\omega^2)^2 +\omega^2c^2}{\Delta \bar{\Delta}}\]

    \[\begin{split}\zeta_2 &= Y_{21}f_1 \\&= \frac{k_2 + ic\omega}{\Delta \bar{\Delta}} \\&= X_2e^{i\theta_2} \end{split}\]

    \[(\frac{X_2}{F_0})^2 = \frac{k_2^2+c^2\omega^2}{\Delta \bar{\Delta}}\]

In order to apply this to various situations, it is useful to non-dimensionalize these relationships similar to that of the undamped vibration absorber.

Set:

    \[X_0 = \frac{F_0}{k_1},\ \nu = \frac{c}{2m_2p_{11}} \]

    \[p_{11} = \sqrt{\frac{k_1}{m_1}},\ p_{22} = \sqrt{\frac{k_2}{m_2}} \]

NOTE:

    \[(\omega c)^2 = \bigg( \frac{\omega}{p_{11}^2}c\frac{k_1}{m_1}\bigg)^2\]

    \[ r = \frac{\omega}{p_{11}},\ \mu = \frac{m_2}{m_1} \]

Therefore:

    \[ (\omega c)^2 = \bigg[ \frac{\omega}{p_{11}}2\frac{c}{2m_2p_{11}}\mu^2k_1\frac{m_1}{m_2}\bigg]^2 = k_1^2\mu^2[2\nu r]^2 \]

    \[ \begin{split}(k_2 - m_2\omega^2)^2 &= k_1^2\bigg[\frac{k_2}{k_1} - \frac{m_2\omega^2}{k_1} \bigg] ^2 \\&= \bigg[\frac{m_1}{m_2}\frac{k_2}{k_1}\frac{m_2}{m_1} - \frac{m_2}{m_1}\omega^2 \frac{m_1}{k_1}\bigg]^2k_1^2 \\&= \mu^2k_1^2[r^2 - g^2]^2 \end{split}\]

    \[ g = \frac{p_{22}}{p_{11}} \]

    \[ \begin{split}\Delta \bar{\Delta} &= \big[(-m_1\omega^2 + k_1)(-m_2\omega^2 + k_2) - m_2\omega^2 k_2 \big] ^2 + c^2\omega^2[-m_1\omega^2 + k_1 - m_2 \omega^2 \big] ^2 \\&= (2\nu r)^2[r^2 - 1 + \mu r^2]^2 + \big[\mu g^2r^2 - (r^2-1)(r^2-g^2)\big]^2\end{split}\]

    \[ \frac{X_1}{X_0} = \sqrt{\frac{(2\nu r)^2 + ( r^2 - g^2)^2}{(2\nu r)^2(r^2 - 1 + \mu r^2)^2 + \big[\mu g^2r^2 - (r^2-1)(r^2-g^2)\big]^2}} \]

    \[ \mu = \frac{m_2}{m_1} \quad \text{(mass ratio)}\]

    \[ \nu = \frac{c}{2m_2p_{11}} \quad \text{(damping ratio)} \]

    \[g = \frac{p_{22}}{p_{11}} \quad \text{(frequency ratio of subsystems)}\]

    \[r = \frac{\omega}{p_{11}} \]

    \[ \frac{X_2}{X_0} = \frac{\mu r\sqrt{1+4\nu^2}}{\sqrt{(2\nu r)^2(r^2 - 1 +\mu r^2)^2 + \big[\mu g^2r^2 - ( r^2 - 1)(r^2 - g^2)\big]^2}}\]

CHECK
  1. If \nu = 0:

        \[\frac{X_1}{X_0} = \frac{r^2 - g^2}{\mu g^2r^2 - (r^2 - 1)(r^2 - g^2)} \]

  2. If r = g, then X_1 = 0
    Note that X_1 \rightarrow 0 only when c \rightarrow 0
    If \nu \rightarrow \infty:

        \[ \begin{split}\frac{X_1}{X_0} &= \frac{1}{r^2 - 1 + \mu r^2} \\&= \frac{1}{(1+\mu)r^2-1} \\&= \frac{1}{\frac{m_1 + m_2}{k_1}\omega^2 -1} \end{split}\]

As a result somewhere between 0 and \nu, c will give an optimal result.

Note that all the curves intersect at P and Q. That is these points are independent of the damping. If we calculate their location, our problem is essentially solved as we must find the curve that passes through them with a horizontal tangent. Also by changing the “tuning” g = \frac{p_{11}}{p_{22}} the two points can be shifted up and down the c = 0 curve. We can do this until the two points have the same height and the horizontal tangent through one of them.

To find the two points where the amplitude is independent of the damping \frac{X_1}{X_0} = \frac{\sqrt{A\omega^2 + B}}{C\omega^2 + D} and this is independent when \frac{A}{C} = \frac{B}{D}:

    \[\bigg(\frac{1}{r^2 - 1 + \mu^2r^2}\bigg)^2 = \bigg[ \frac{(r^2-g^2)^2}{\mu g^2r^2 - (r^2-1)(r^2 - g^2)}\bigg]^2 \]

This leads to:

    \[ \mu g^2r^2 - (r^2-1)(r^2-g^2) = \pm(r^2-g^2)(r^2 - 1 + \mu r^2) \]

The negative gives r^2 = 0 while the positive gives r^4 - 2r^2 \frac{(1+g^2 + \mu g^2)}{2+\mu} + \frac{2g^2}{2+\mu} = 0.

The two answers for the forcing frequency ratio are functions of g and \mu but not of the damping \nu.

We now wish to adjust g so that P & Q are equal. Since their magnitude is independent of the damping, choose c \rightarrow \infty then the magnitude:

    \[\frac{X_1}{X_0} = \frac{1}{\mu r^2 + r^2 - 1} \]

For r_1^2 and r_2^2 (the two roots), we want the same magnitude:

    \[ \frac{1}{\mu r_1^2 + r_1^2 - 1} = -\frac{1}{\mu r_2^2 + r_2^2 - 1}\]

Therefore:

    \[1 - (\mu + 1) r_2^2 = -1 + (\mu + 1) r_1^2\]

    \[ \frac{2}{1+\mu} = r_1^2 + r_2^2 \]

Now it is not necessary to solve for r_1^2 and r_2^2 if we remember the sum of the roots in a quadratic in the negative coefficient of the middle term.

Therefore:

    \[ \frac{2}{1+\mu} = \frac{2(1+g^2+\mu g^2)}{2+\mu} \]

    \[ \frac{2+\mu}{1+\mu} = 1+ g^2(1+\mu) \]

    \[\begin{split} \implies g^2(1+\mu) &= \frac{2+\mu - 1 - \omega}{1+ \mu} \\&= \frac{1}{1+\mu} \end{split}\]

Therefore:

    \[g = \frac{1}{1+\mu} \]

Amplitudes of the main mass for various values of absorber damping. The absorber is twenty times as small as the main machine and is tuned to the same frequency. All curves pass through the fixed points P and Q.

Resonance curves for the motion of the main mass fitted with the most favorably tuned vibration-absorber system of one-forth of the size of the main machine.

(a)

(b)

(c)

(a) Peak amplitudes of the main mass as a function of the ratio m/M for various absorbers attached to the main mass. (b) Peak relative amplitudes between the masses M and m for various absorbers. (c) Damping constants required for the most favorable operation of the absorber, i.e., for obtaining the results of (a) and (b).

Curve 1 for the most favorably tuned and damped absorber; curve 2 for the most favorably damped absorber tuned to the frequency of the main system; curve 3 of the most favorably damped viscous Lanchester damper; curve 4 for the most favorably damped Coulomb Lanchester damper.

    \[ \bigg(\frac{X_1}{X_0}\bigg)_\text{MAX} = \frac{1}{1-r^2(1+\mu)}\]

and using one of the roots:

    \[ \bigg(\frac{X_1}{X_0}\bigg)_\text{MAX} = \sqrt{1+ \frac{2}{\mu}} \]

while the optimum damping is given by:

    \[ \nu = \sqrt{\frac{3\mu}{8(1+\mu)^3}} \]

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