Mechanical Vibrations
MDPN471
Fall 2014
Problem Set 04
Problem 1. A weight of 50 N is suspended from a spring of stiffness 4000 N/m and is subjected to
a harmonic force of amplitude 60 N and frequency 6 Hz. Find (a) the extension of the spring
due to the suspended weight, (b) the static displacement of the spring due to the maximum
applied force, and (c) the amplitude of forced motion of the weight.
Problem 2. Consider a spring-mass system, with k = 4000 N/m and m = 10 kg, subject to a
harmonic force F(t) = 400 cos ωt N. Find and plot the total response of the system under the
initial conditions: xo = 0.1 m, vo = 0 for the frequency ω equals: a) 10, b) 20, c) 20.1, d) 30.
Problem 3. A mass m is suspended from a spring of stiffness 4000 N/m and is subjected to a
harmonic force having amplitude of 100 N and a frequency of 5 Hz. The amplitude of the
forced motion of the mass is observed to be 20 mm. Find the value of m.
Problem 4. A spring-mass system with m = 10 kg and k = 5000 N/m is subjected to a harmonic
force of amplitude 250 N and frequency ω. If the maximum amplitude of the mass is observed
to be 100 mm, find the value of ω.
Problem 5. A spring-mass system, resting on an inclined plane, is subjected to a harmonic force as
shown in Figure 5. Find the response of the system by assuming zero initial conditions.
Problem 6. A spring-mass system is set to vibrate from zero initial conditions under a harmonic
force. The response is found to exhibit the phenomenon of beats with the period of beating
equal to 0.5 s and the period of oscillation equal to 0.05 s. Find the natural frequency of the
system and the frequency of the harmonic force.
Problem 7. The spring actuator shown in Figure 7 operates by using the air pressure from a
pneumatic controller (p) as input and providing an output displacement to a valve (x)
proportional to the input air pressure. The diaphragm, made of a fabric-base rubber, has an area
A and deflects under the input air pressure against a spring of stiffness k. Find the response of
the valve under a harmonically fluctuating input air pressure p(t) = po sin ωt for the following
data: po=70 kPa, ω=8 rad/s, A=650 cm2, k=70 kN/m, mass of spring=7.5 kg, and mass of valve
and valve rod=10 kg.
SDOF Forced Undamped : mx(t ) =
t , x p (t ) d st cos(ωt ) / (1 − r 2 )
+ kx (t ) F0 cos ω=
For zero ICs : x (t ) =
(ω − ωn ) t 
2d st  (ω + ωn ) t
sin
sin

2 
1− r 
2
2

− X cos (ωt − φ )
When r < 1: x p (t ) =
X cos (ωt − φ ),When r > 1: x p (t ) =
When=
r 1,=
(t ) x0 cos ωn t + ( x0 / ωn ) sin ωn t + 0.5d stωn t sin ωn t
z 0 : x=
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Mechanical Vibrations
MDPN471
Fall 2014
Figure 5
Figure 7
SDOF Free Undamped : mx(t ) + kx (t ) =
0
=
x (t ) A sin(ωn t + j )
ωn2 x 02 + v 02 / ωn
A
=
φ = tan −1 (ωn x 0 / v 0 )
=
x (t ) A1 cos ωn t + A 2 sin ωn t
A=
A12 + A 22 , A1 =
a1 + a2 ,
=
x (t ) a1e j ωn t + a2e − j ωn t
SDOF Free Damped
c
Critically
2 km
damped: x (t ) =  x 0 + (v 0 + ωn x 0 ) t  e −ω t
= e −ζωn t
n
Over damped:
A1,2 =
e −ζωn t (A1e −ωn t
ζ 2 −1
=
a1 0.5(A1 − A 2 j ),
=
a2 0.5(A1 + A 2 j )
j = tan −1 ( A1 / A 2 )
=
Underdamped
: x(t) Ae −ζωn t sin( ωd t + ϕ )
0, ζ =
mx(t ) + cx (t ) + kx (t ) =
x(t)
A=
(a1 − a2 ) j
2
+ A 2e ωn t
v 0 + ( ζζ
+
− 1 ) ωn x 0
2
2ωn ζ 2 − 1
ζ 2 −1
x cos 1 − ζ 2 ω t

n
 0

 v 0 + ζωn x 0

sin 1 − ζ 2 ωn t 
+
2
1 − ζ ωn


1
ωn 1 − ζ 2
A = (v 0 + ζωn x 0 ) 2 + (x 0ωd ) 2 , ωd =
)
:
ωd
 x 0ωd

x
1
=
=
ϕ tan −1 
ln 1
 ,d =
n x n +1
 v 0 + ζωn x 0 
2px
1− x 2
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