Dynamics Tutorial 14-Natural vibrations-one degree of freedom-27p.pdf
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Transcript of Dynamics Tutorial 14-Natural vibrations-one degree of freedom-27p.pdf
© 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)