© D.J.DUNN freestudy.co.uk 1

ENGINEERING COUNCIL

DYNAMICS OF MECHANICAL SYSTEMS D225

TUTORIAL 14 –NATURAL VIBRATIONS – TWO DEGREES OF FREEDOM

On completion of this tutorial you should be able to do the following.

• Find the natural frequencies for 2 DOF systems. • Define and determine the shape modes for 2 DOF systems. • Explain how to use Eigenvalues in the solution. • Describe higher DOF systems • Solve problems involving forced vibrations. • Explain the principles of vibration absorbers.

This subject required advanced mathematical techniques in particular the use of matrix Algebra. You are advised to study the tutorials on Matrix Algebra in the Maths tutorials before you start. You will find other tutorials on this subject with animations at the following links. http://www.mech.uwa.edu.au/bjs/Vibration/TwoDOF/Transient/default.html http://www.kettering.edu/~drussell/Demos/absorber/DynamicAbsorber.html http://www.efunda.com/formulae/vibrations/mdof_eom.cfm http://www.fsid.cvut.cz/en/U2052/node123.html

© D.J.DUNN freestudy.co.uk 2

1. MASS – SPRING SYSTEM k is the stiffness of the spring (N/m) m is the mass (kg) x is the displacement of a mass from its rest position When this system vibrates freely, there are two natural frequencies, one for each mode of vibration. Balancing forces on m1 we have

0)x(xkxkxm 1221111 =−−+&& Balancing forces on m2 we have

0)x(xkxm 12222 =−+&& Knowing the motion of each mass is sinusoidal we can substitute x = X ejωt = sin(ωt) and = ωx&& 2 X ejωt x1 = X1 ejωt and x2 = X2 ejωt

1x&& = -ω2 X1 ejωt and = -ω2x&& 2 X2 ejωt

If these are substituted back into the equations of motion we have:

0)x(xkxkxm 1221111 =−−+&& { }

{ }( ) ( ) 0kXkkωmX

0XXkXkωXm

0eXeXkeXkeωXm

22212

11

122112

11

ωt j1

ωt j22

ωt j11

tω j211

=−+++−

=−−+−

=−−+−

and 0)x(xkxm 12222 =−+&&

{ }{ }

( ) ( ) 0kωmXkX

0XXkXωm

0eXeXkeXωm

22

2221

12222

2

tω j1

tω j22

tω j2

22

=+−+−

=−+−

=−+−

This gives us a pair of simultaneous equations and so A and B are solved. In matrix form we have

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−−++−

2

1

22

22

2212

1

XX

kωmkkkkωm

X1 and X2 could be solved by equating the determinant to zero.

0kωmk

kkkωm

22

22

2212

1 =+−−

−++−

This is called the characteristic equation and yields the values of ω. ( )( ) ( )( )( )

( ) 0kkmkmkkmωωmm

0kkωmkkkωmkkωmωmm

0kkkωmkkωm

2122212124

21

22

22

22221

2212

21

421

2222

2212

1

=+++−

=−+−+−−

=−−−+−++−

( )

0mmkk

mk

mk

mkωω

0mmkk

mmmkmkkmωω

21

21

1

2

1

1

2

224

21

21

21

22212124

=+⎟⎟⎠

⎞⎜⎜⎝

⎛++−

=+++

−

This gives a solution for ω by solving ω2 with the quadreatic equation.

2mmkk4

mk

mk

mk

mk

mk

mk

ω 21

212

1

2

1

1

2

2

1

2

1

1

2

2

2

−⎟⎟⎠

⎞⎜⎜⎝

⎛++±⎟⎟

⎠

⎞⎜⎜⎝

⎛++

=

21

212

1

2

1

1

2

2

1

2

1

1

2

22

mmkk

mk

mk

mk

41

mk

mk

mk

21ω −⎟⎟

⎠

⎞⎜⎜⎝

⎛++±⎟⎟

⎠

⎞⎜⎜⎝

⎛++=

This gives two values for ω and these are the natural frequencies. The lower frequency is called the fundamental frequency. Note that √k/m is the natural frequency of a single DOF system so :

2n1

2n2

2

1

22n1

2n2

1

22n1

2n2

2 ωωmkωω

41

mkωω

21ω −⎟⎟

⎠

⎞⎜⎜⎝

⎛++±⎟⎟

⎠

⎞⎜⎜⎝

⎛++=

MATRIX SOLUTIONS The equations for the system were stated at the beginning as :-

0)x(xkxkxm 1221111 =−−+&& and 0)x(xkxm 12222 =−+&& If we change these to make x the common factor we have

0xkxkxkxm 12221111 =+−+&&

( ) 0xkxkkxm 2212111 =−++&& and

0xkxkxm 122222 =−+&& In matrix form this is:

0 0

0

2

1

22

221

2

1

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−++⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡xx

kkkkk

xx

mm

&&

&&

The mass matrix is The stiffness matrix is ⎥⎦

⎤⎢⎣

⎡=

2

1

00

mm

M ⎥⎦

⎤⎢⎣

⎡−

−+=

22

221

kkkkk

K

The matrix from may be written as 0 xx =+KM && .......................(A) The displacement equations are x1 = A sin (ωt + φ) and x2 = B sin (ωt + φ) Let the amplitude vector be X The displacement vectors are x = X sin (ωt + φ) ................(B) The second derivative of this is ............(C) Xx )sin(2 φωω +−= t&&

Substitute (B) and (C) into (A). { } { } 0)sin( )sin( 2 =+++− XKXM φωφωω tt

{ } { } 0 )sin(2 =++− φωω tXKM { } 0 2 =+− XKMω ....................(D) If there is a non trivial solution then [ ] 0 det 2 =+− XKMω ...................(E)

© D.J.DUNN freestudy.co.uk 3

Equation (E) gives us the exact same solution for ω as found earlier.

[ ] 0 det 2 =+− XKMω

0BA

kωmkkkkωmdet

22

22

2212

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−−++−

Mode shapes are displacement patterns representing the relative positions of both masses. It is normal to take A= 1 and solve B so the mode shape is A/B. There is one mode shape for each degree of freedom so in this case there are two. These are found by substituting ω and finding A/B for each frequency. WORKED EXAMPLE No.1

Given ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

4002

00

2

1

mm

M ⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡−

−+=

5005005001000

22

221

kkkkk

K

Find the two natural frequencies and mode shapes.

0BA

kωmkkkkωmdet

22

22

2212

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−−++−

05004500

50010002det 2

2=⎥

⎦

⎤⎢⎣

⎡

+−−−+−ω

ω

( )( ) ( )( ) 0500500500410002 22 =−−−+−+− ωω 0250000500000400010008 224 =−+−− ωωω 8ω4 – 5000ω2 +250000 = 0 Solving the quadratic for ω2 we get ω2 = 570.2 and 54.8 ω = 23.9 rad/s and 7.4 rad/s now substitute these into

0XX

kωmkkkkωm

2

1

22

22

2212

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−−++−

First substitute ω = 54.8

0XX

005)8.54(400500510004.8)52(

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−

−+−0

XX

8.280005005890.4

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−

890.4X1 – 500X2 = 0 and -500 X1 + 280.8 X2 = 0 Let X1 = 1 and X2 = 890.4/500 = 1.781 or 500/280.8 = 1.781. The mode shape is the ratio

1.781/1 Next substitute ω = 23.9

0 0XX

005)2.570(40050051000)2.0752(

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−

−+−XX

8.1780005005140.4-

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−−

-140.4X1 – 500X2 = 0 and -500X1 – 1780.8 X2 = 0 Let X1= 1 and X2 = -140.4/500 = -0.281 or -500/1780.8 = -0.281. The mode shape is the ratio -0.281/1

© D.J.DUNN freestudy.co.uk 4

In the example the interpretation of the mode shapes is as follows. At the lower frequency they move in phase reaching maximum amplitude with a ratio 1.781/1. At the higher frequency they move in opposite phase with and amplitude -0.281/1. You might puzzle why the frequency is the same for both masses at each mode. Remember that both masses are affected by the other and they are not independent systems. If each were independent the frequency would simply be ω = √k/m for each (15.8 and 11.2 rad/s in this case)

The shape modes can be determined from the following formulae.

Mode 1 2

212

2

2

21211

1

2

kωmkkkωm

XX

+−=

++−=

k

Mode 2 2

222

2

2

21221

1

2

kωmkkkωm

XX

+−=

++−=

k

EIGEN VALUES Equation (D) earlier was { } 0 2 =+− XKMω Divide through by M [M-1 K – ω2]X = 0 If we let λ = ω2

[M-1 K – λ] X = 0 [M-1 K] X = λ X Hence λ is the eigenvalue of [M-1 K] and this may be found easily with computer packages such as Mathcad. The method can be used to solve problems with three or more degrees of freedom. There is a natural frequency for each degree. The method can be used for any two degree of freedom system in which the stiffness and mass matrices are clearly identified.

© D.J.DUNN freestudy.co.uk 5

WORKED EXAMPLE No.2 Given m1 = 2 kg , k1 = 30 000 N/m, m2 = 30 kg and k2 = 40 000 N/m determine the two natural

frequencies using the mass and stiffness matrices and check with the derived formula. Determine the mode shapes.

SOLUTION

The mass matrix is ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

30002

00

2

1

mm

M

The stiffness matrix is ⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡−

−+=

40000400004000070000

22

221

kkkkk

K

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡00

40000400004000070000

30002

2

1

2

1

xx

xx&&

&&

To avoid trivial solutions we must have [ ] 0.det 2 =+− KMω

040000400004000070000

30002

.det 2 =⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡−ω

0400003040000

40000700002det 2

2=⎥

⎦

⎤⎢⎣

⎡

+−−−+−ω

ω

( )( ) ( )( ) 040000400004000030700002 22 =−−−+−+− ωω 010x6.110x8.210x1.210x8060 9926234 =−+−− ωωω 60ω4 – 2.18x106ω2 +1.2x109 = 0 Solving the quadratic for ω2 we get ω2 = 35.77 x 103 and 560 ω = 189.1 rad/s and 23.6 rad/s

21

212

1

21

2

2

1

21

2

22

mmkk

mkk

mk

41

mkk

mk

21ω −⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++±⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++=

rad/s23.63and189.1ω 558.7 and 35774ω

176081816760

12000000002

7000030

4000041

270000

3040000

21ω

2

22

==

±=−⎟⎠⎞

⎜⎝⎛ +±⎟

⎠⎞

⎜⎝⎛ +=

Next substitute ω = 23.63

0BA

00400)7.558(30004000040070000)7.5582(

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−

−+−0

BA

23239004000040068883

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−

68883X1 – 40000X2 = 0 and -40000X1 +23239 X2 = 0 Let X1 = 1 and X2 = 68883/40000 = 1.722 or 40000/23239 = 1.722 The mode shape is the ratio 1.722/1 Next substitute ω = 189.1

0BA

00400)57743(300040000400700005774)32(

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−

−+−0

BA

103322000400004001548-

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−−

-1548 X1 - 40000 X2 = 0 and -40000 X1 -1033220 X2 = 0 Let X1 = 1 and X2 = -1548/40000 = - 0.0387 or -40000/1033220 = - 0.0387

© D.J.DUNN freestudy.co.uk 6

SELF ASSESSMENT EXERCISE No.1 Calculate the natural frequencies and mode shapes for the systems below.

(37 rad/s, 76.3 rad/s, 1.523 and -2.189)

(839 rad/s, 94 rad/s, -0.018 and 3.456)

(51.1 rad/s, 19.5 rad/s, 1.618 and – 0.618) THREE DEGREES OF FREEDOM For a three degree of freedom system the mass and stiffness matrices are as follows.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

000000

mm

mM K=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+−

−+

33

3322

221

0

0

kkkkkk

kkk

The pattern for higher DOF may be discerned. The solution is difficult without a computer package. © D.J.DUNN freestudy.co.uk 7

FORCED VIBRATIONS Since we are not examining cases with no damping, the resonant frequencies will produce very large amplitudes and these will occur at a single frequency. There will be two resonant frequencies for a two DOF system. CASE 1 – FORCE APPLIED TO m2There are more cases of forced vibrations as to be examined than for single DOF systems. In this tutorial we will only examine the application of a sinusoidal force. The following is a guide to the general approach and we start with the case when the force is applied to m2 as shown. The free body diagram shown yields:- m1 + k1x&& 1 x1 – k2(x2 - x1) = 0 and m2 + k2x&& 2 (x2 - x1) = Fosin(ωt) Change to the complex operator x = X ejωt

Note that is dx&& 2x/dt2 = - ω2 X ejωt

X is the amplitude of the motion. -m1 ω2X1 ejωt + k1 X1 ejωt – k2(X2 - X1) ejωt = 0 -m1 ω2X1 + k1 X1 – k2(X2 - X1) = 0 -m1 ω2X1 + k1 X1 – k2X2 + k2X1 = 0 ………….(1) -m2 ω2X2 ejωt + k2 (X2 - X1) ejωt = Fo ejωt

-m2 ω2X2 + k2 (X2 - X1) = Fo

-m2 ω2X2 + k2 X2 - k2X1 = Fo………………….(2)

From (1) we get ⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+=

2

2121

12 kωmkkXX ....................(3) Noting that k1/m1 = ω1

2

⎥⎦

⎤⎢⎣

⎡−+=⎥

⎦

⎤⎢⎣

⎡−+= 2

1

2

1

2

2

12

1

1

1

2

2

1

1

2

ωω

kk1

kkω

km

kk1

kk

XX ................(4)

Substitute (3) into (2) o122

2121

122

2121

12

2 FXkk

ωmkkXkk

ωmkkXωm- =−⎥⎦

⎤⎢⎣

⎡ −++⎥

⎦

⎤⎢⎣

⎡ −+

{ } o1222

22

2121

1 FXkkωm-k

ωmkkX =−+⎥⎦

⎤⎢⎣

⎡ −+ { } o222

22

2121

1 Fkkωm-k

ωmkkX =⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+⎥⎦

⎤⎢⎣

⎡ −+

{ } o122

222

12

1

1

1

21 FXkωm-k

kkω

km

kk1X =−⎥

⎦

⎤⎢⎣

⎡−+ o121

2

2

22

1

1

1

21 FXkkω

km-1ω

km

kk1X =−

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−+

o212n2

2

2n1

2

1

21 Fkk

ωω-1

ωω

kk1X =

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−+

212n2

2

2n1

2

1

2

o1

kkωω-1

ωω

kk1

FX−

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−+

= ⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+=

2

2121

12 kωmkkXX

ωn1 and ωn2 are the natural frequencies of the single DOF mass-spring system given by the formula derived earlier.

© D.J.DUNN freestudy.co.uk 8

WORKED EXAMPLE No.3 A system as shown in the previous diagram has the following data. m1 = 10 kg, m2 = 2 kg, k1 = 10 000 N/m, k2 = 6000 N/m and Fo = 100 N Plot X2 and X1 and the ratio X2/X1 against frequency. Calculate the natural frequencies for the

two DOF system and the individual one DOF systems. Draw conclusions about the behaviour of the system.

SOLUTION ωn1 =√(k1/m1) = 31.62 rad/s ωn2 =√(k2/m2) = 54.77 rad/s

2n1

2n2

2

1

22n1

2n2

1

22n1

2n2

2 ωωmkωω

41

mkωω

21ω −⎟⎟

⎠

⎞⎜⎜⎝

⎛++±⎟⎟

⎠

⎞⎜⎜⎝

⎛++=

( ) ( ) 22222222 62.1377.4560062.1377.454160062.1377.45

21ω −++±++=

300000052900002300ω2 −±= 1513.32300 22900002300ω2 ±=±= ω2 = 3813.3 or 786.7 ω = 61.75 or 28 rad/s If we tried to evaluate and plot without the aid of a computer we would find it difficult to pin

point the resonant frequencies because without damping they are just spikes as shown.

The plots for

212n2

2

2n1

2

1

2

o1

kkωω1

ωω

kk1

FX−

⎭⎬⎫

⎩⎨⎧−⎥

⎦

⎤⎢⎣

⎡−+

= and ⎥⎦

⎤⎢⎣

⎡−+= 2

n1

2

1

2

2

112 ω

ωkk1

kkXX are shown

below.

We can see that there are two resonant frequencies that occur at the natural frequencies of the 2

DOF system. Now plot the amplitude ratio.

© D.J.DUNN freestudy.co.uk 9

⎥⎦

⎤⎢⎣

⎡−+= 2

n1

2

1

2

2

1

1

2

ωω

kk1

kk

XX

Evaluate and plot

We can see that at ω = 40 rad/s the ratio is 0 and this means that m2 is not moving. In this case only m1 is oscillating and both springs are changing length and effectively combine to form one spring of stiffness k1 + k2.

The forcing frequency must then correspond to 1

21

mkkω +

= and if we check this out

rad/s 4010

1600ω ==

CASE 2 – FORCE APPLIED TO m1 The free body diagram shown yields:- m1 + k1x&& 1 x1 – k2(x2 - x1) = Fosin(ωt) and m2 + k2x&& 2 (x2 - x1) = 0 Change to the complex operator x = X ejωt

Note that is dx&& 2x/dt2 = ω2 X ejωt

X is the amplitude of the motion. -m1 ω2X1 ejωt + k1 X1 ejωt – k2(X2 - X1) ejωt = Fo ejωt

-m1 ω2X1 + k1 X1 – k2(X2 - X1) = Fo -m1 ω2X1 + k1 X1 – k2X2 + k2X1 = Fo ………….(1) -m2 ω2X2 ejωt + k2 (X2 - X1) ejωt = 0 -m2 ω2X2 + k2 (X2 - X1) = 0-m2 ω2X2 + k2 X2 - k2X1 = 0………………….(2)

From (2) we get ⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡ −= 2

n2

2

22

222

21 ωω1X

kωmkXX ....................(3) Noting that k2/m2 = ωn2

2

⎥⎦

⎤⎢⎣

⎡−= 2

n2

2

2

1

ωω1

XX ................(4)

Substitute (3) into (1)

o2n2

2

22222n2

2

212n2

2

22

1 Fωω1XkXk

ωω1Xk

ωω1Xωm- =⎥

⎦

⎤⎢⎣

⎡−+−⎥

⎦

⎤⎢⎣

⎡−+⎥

⎦

⎤⎢⎣

⎡−

{ } o22212

12n2

2

2 FXkkkωm-ωω1X =−++⎥

⎦

⎤⎢⎣

⎡−

{ } o22

1212n2

2

2 Fkωm-kkωω1X =

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+⎥⎦

⎤⎢⎣

⎡−

o212

1

1

1

22n2

2

2 Fkkωkm-

kk1

ωω1X =

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎭⎬⎫

⎩⎨⎧+⎥

⎦

⎤⎢⎣

⎡−

o212n1

2

1

22n2

2

2 Fkkωω-

kk1

ωω1X =

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎭⎬⎫

⎩⎨⎧+⎥

⎦

⎤⎢⎣

⎡−

212n1

2

1

22n2

2o

2

kkωω-

kk1

ωω1

FX−

⎭⎬⎫

⎩⎨⎧+⎥

⎦

⎤⎢⎣

⎡−

= ⎥⎦

⎤⎢⎣

⎡−= 2

n2

2

21 ωω1XX

ω1 and ω2 are the natural frequencies of the single DOF mass-spring system.

© D.J.DUNN freestudy.co.uk 10

WORKED EXAMPLE No.4 A system as shown in the previous diagram has the following data. m1 = 10 kg, m2 = 2 kg, k1 = 10 000 N/m, k2 = 6000 N/m and Fo = 100 N. Plot X2, X1 and the ratio X2/X1 against frequency. Calculate the natural frequencies for the two

DOF system and the individual one DOF systems. Draw conclusions about the behaviour of the system.

SOLUTION This is the sdame as the last example but with the force applied to m1. As before : ωn1 =√(k1/m1) = 31.62 rad/s ωn2 =√(k2/m2) = 54.77 rad/s and for the 2 DOF system ω = 61.75 and 28 rad/s If we tried to evaluate and plot without the aid of a computer we would find it difficult to pin

point the resonant frequencies because without damping they are just spikes as shown.

Evaluate

2121

2

1

22n2

2o

2

kkω-kk1

ωω1

FX−

⎭⎬⎫

⎩⎨⎧+⎥

⎦

⎤⎢⎣

⎡−

=

nω

and ⎥⎦

⎤⎢⎣

⎡−= 2

n2

2

21 ωω1XX and plot and we get the

following results.

We can see that there are two resonant frequencies that occur at the natural frequencies of the 2

DOF system. Now plot the amplitude ratio.

© D.J.DUNN freestudy.co.uk 11

⎥⎦

⎤⎢⎣

⎡−= 2

n2

2

2

1

ωω1

XX

Evaluate and plot

We can see that at ω = 54.77 rad/s the ratio is 0 and this means that m1 is not moving. In this case only m1 is oscillating and the frequency must be ωn2 =√k2/m2= 54.77 rad/s

VIBRATION ABSORBER When a mass and spring is added to a vibrating system with a natural frequency ωn1 and tuned so that the motion of the main mass is reduced (to zero), the system is called a vibration absorber. In this case m2 and k2 are selected so that ωn2 = forcing frequency.

In this case ⎥⎦

⎤⎢⎣

⎡−= 2

n2

2

2

1

ωω1

XX = 1 - 1 = 0

WORKED EXAMPLE No.5 A machine of mass 20 kg is mounted on flexible supports and has a significant level of

vertical vibration at the resonant frequency of 80 Hz. A second mass is to be attached to the top on a sprung platform to absorb this vibration. If the mass is to be 1.2 kg, calculate the spring rate of the mountings and the absorber spring rate required.

SOLUTION

The initial system has a natural frequency 1

1n m

k732.12280ω ===π

20k732.12

280ω 1===π

k1 = 3242.28 N/m

When the absorber mass is added it must absorb the vibrations at 12.732 rad/s so

2

2

mk732.12 =

1.2k2=

k2 = 1.2(12.732)2 = 194.54 N/m

If we plot ⎥⎦

⎤⎢⎣

⎡−= 2

n2

2

2

1

ωω1

XX we can see that this is zero at 12.7 rad/s

© D.J.DUNN freestudy.co.uk 12

SELF ASSESSMENT EXERCISE No.2

© D.J.DUNN freestudy.co.uk 13

1. For the system shown, calculate the resonant frequencies when Fo = 4 N. (21.6 and 46.2 rad/s) Calculate the forcing frequency that results in no motion of the 0.5 kg mass. (31.6 rad/s) Calculate the absolute amplitudes of both masses at 5 rad/s. (8.34 mm and 8.55 mm) 2. For the system shown, calculate the resonant frequencies when Fo = 12 N. (17.3 and 31.6 rad/s) Calculate the forcing frequency that results in no motion of the 0.5 kg mass. (28.2 rad/s) Calculate the absolute amplitudes of both masses at 1 rad/s. (10.043 mm and 40.1 mm)