Vibration 4 Exam paper May 2014, Solutions:

Q1
(a)

(3)

Free-body diagram

(b)
The equations of motion are:
̈ + ( − ) = sin
̈ + − ( − ) = ℎ sin
(3)

(c)
∴ � − 2 −

2 + − 2� �Θ� = � ℎ� (3)

For = 1, Θ = −2

and therefore the modal matrix can be written:
[] = � 1 1−12

−2
2

� = � 1 16.395 −0.408� (4)

where the natural frequencies are 3.8 Hz and 19.4 Hz, or 23.88 rad/s and 121.89 rad/s.
[][][] = �1 6.3951 −0.408� �0.5 00 0.2� � 1 16.395 −0.408� = �8.679 00 0.533� (3)
{} = [][] = �1 6.3951 −0.408� � 0.04� = �3.142.46� (3)

̈1 + 570.071 = 0.362 and ̈2 + 14858.102 = 4.615 (2)

(d)
1 = 0.362(570.07 − 246.74) = 1.1196 × 10−3 and 2 = 4.615(14858.10 − 246.74) = 3.1585 × 10−4

� = [] �12� = � 1 16.395 −0.408� �1.1196 × 10−33.1585 × 10−4�
∴ = 1.43 = 7.0310−3 = 0.4° (4)

Q2
12
1 k
2k
12
1 k
1m
2m

m1 = 9000 kg, k1 = 12x106 N/m,
11 = �12 × 1069000 = 36.5 /

The operating range is 320-400 rpm.

Therefore for the 2 DoF system, 1 < 33.5/, 2 > 41.9/ (3)

The frequency equation is
4 − (2222 + 222 )2 + 224 = 0 (4)

Tune the absorber to the primary system such that 11 = 22 = 36.5/ (2)

Let = 2 = 41.9/
41.94 − (2 × 36.52 + × 36.52) × 41.92 + 36.54 = 0 ∴ = 0.0766

Since = 2
1
, 2 = 689.7 (9)

If = 0.0766, check for 1

4 − (2 × 36.52 + 0.0766 × 36.52)2 + 36.54 = 0
4 − 2766.552 + 1774890.06 = 0
2 = 1383.27 ± 372.23
∴ 1 = 31.79/

which satisfies the requirement to be outside the operating range. (7)

Note:
If the solution lets = 1 = 33.5/, then = 0.0293,2 = 264, but then 2 =39.75/ which does not lie outside the operating range.

Q3
(a) Generalised coordinates are (z, θ, φ) where z in the vertical movement, θ is rotation about an x-
axis through the centre of the plate, φ is rotation about a y-axis through the centre of the plate. (2)

(b)
= 1
2
̇2 + 1
2
̇2 + 1
2
̇2 = ̇2 + 0.2(̇2 + ̇2) (2)

(c) Displacement of springs
+ 0.25 + 0.25
− 0.25 + 0.25
− 0.25 − 0.25
+ 0.25 − 0.25

The potential function is
= 250[( + 0.25 + 0.25)2 + ( − 0.25 + 0.25)2 + ( − 0.25 − 0.25)2+ ( + 0.25 − 0.25)2]
(8)
(d)

̇
� = 2̈

̇
� = 0.1̈

̇
� = 0.1̈

= 500( + 0.25 + 0.25) + 500( − 0.25 + 0.25) + 500( − 0.25 − 0.25)+ 500( + 0.25 − 0.25) = 2000

= 125( + 0.25 + 0.25) − 125( − 0.25 + 0.25) − 125( − 0.25 − 0.25)+ 125( + 0.25 − 0.25) = 125

= 125( + 0.25 + 0.25) + 125( − 0.25 + 0.25) − 125( − 0.25 − 0.25)
− 125( + 0.25 − 0.25) = 125

Substitute into Lagrange’s equation

̇
� −

+

= 0

2 0 00 0.1 00 0 0.1� �̈̈̈� + �2000 0 00 125 00 0 125� �� = �000�
(3X3 for each eqn)

(e) The equations of motion are uncoupled. (1)
The natural frequencies are
1 = 31.62/, 2 = 3 = 35.36/
(1)
In practice, slight asymmetries will result in the two natural frequencies being at close but separate
frequencies. (1)
The mode shapes are a vertical oscillation, a rotational oscillation about an x-axis through the centre,
a rotational oscillation about a y-axis through the centre. (1)

Q4
(a)
= 10 log10 Σ10 10�= 10 log10(1011 + 1011 + 1010.8 + 1011.2 + 1010.6 + 1011.5 + 1010.6 + 109.7)= 119.15
= 10 log10 22 = 20 (3)
= �10 10� = 20 × 10−6�10119.15 10� = 18.14
(2)
(b) Upper cut-off frequency is f1 Lower cut-off frequency is f2 Centre frequency is fc log1 + log2 = 2 log (1) and 2 = 21 (2)
From the first equation log12 = log2 ∴ 2 = 12 (3)
From equations (1) and (3) 212 = 2 ∴ 1 =
√2 (4)
(5)
Bandwidth ∆ = 2 − 1
From (1): ∆ = 1
From (4):
∆ =
√2

Spectrum level: = − 10 log∆
For the 1 kHz band ∆ = 1000
√2
= 707.11
= 106 − 10 log 707.11 = 77.5 /√
(3)
(c) A-weighting is the most widely used weighting network. It assigns a weight to each frequency
band in an octave band analysis that is related to the sensitivity of the human ear in that frequency
band. A-weighted noise measurements are therefore more useful for assessing noise nuisance and
safe noise exposure.

Centre Frequency (Hz) 63 125 250 500 1k 2k 4k 8k
Band SPL (dB(LIN)) 110 110 108 112 106 115 106 97
Protection -15 -17 -19 -21 -23 -25 -27 -29
A-weighting -24 -20 -16 -12 -8 -4 0 0
Total 71 73 73 79 75 86 79 68
(6)
= 10 log10 Σ10 10�= 10 log10(107.1 + 107.3 + 107.3 + 107.9 + 107.5 + 108.6 + 107.9 + 106.8)= 88.11 ()
(4)
With the ear protectors, the exposure is below the 90 dB limit for an 8 hour working day and the
worker’s hearing should not be at risk. (2)

Q5
(a) The rigid-body modes should be well separated from the structural modes (2)

(b) The rod or support strings should react the smallest possible unrecorded forces (2)

(c) Low stiffness connections do not transmit HF signals well (2)

(d) (i) 45° and (ii) 90° (2)

(e)

(3)

(f)
̈ + ̇ + = ()
(−2) + () + () = (−2 + + ) =

() = 1(−2)+() (5)

(g)

The equation is re-expressed in the form of 1/(a+ib).

Re-expressing in the original form gives:

(4)

(h) From the two expressions above, let the common denominator be d

2() = �( − 2

)2 + (−

)2

2() = �( − 2)2 + ()2
2

)()(
1)( 2 cimk ωω
ωα
+−
=
22 )())((
1
1
1
1
ibaibaiba
iba
ibaiba
iba
iba
iba
iba
iba
iba
iba −+−

=
−+

=

+
−=

+
[ ] 222
2
)()(
)()(
cmk
mk
ωω
ωωα
+−

=ℜ [ ] 222 )()()( cmk
ci
ωω
ωωα
+−

=ℑ
2() = �
2

2() = �1

Therefore:
|()| = 1
�( − 2)2 + ()2

⟨() = tan−1(
− 2
)
(5) 