[CHAP. 1
VIBRATIONS AND W AVES
12
The resultant wave (solid line) is obtained by adding the two waves graphically. We begin in
Fig. 1-9(a) with zero phase angle between the two waves, i.e. the two waves are completely in phase
with each other. The resultant amplitude is equal to twice the amplitude of the given waves.
Fig. 1-9(b) shows the addition of two identical waves with 18 0 phase difference between them.
Similarly, Fig. 1-9(c) to Fig. 1-9(j) are the resultants of the additions of two identical waves with
progressively greater values of phase angle between the two identical waves.
When the two identical waves are completely out of plfase, i.e. the phase angle between the two
waves is 180 0 , the resultant wave is zero. In other words, the two waves cancel each other.
SIMPLE HARMONIC MOTIONS
1.8. A simple harmonic motion is given as x(t) ::::: 10 sin (lOt - 30 0 ) where x is measured
in meters, t in seconds, and the phase angle in degrees. Find (a) the frequency and
period of the motion, (b) the maximum displacement, velocity and acceleration, (c)
the displacement, velocity and acceleration at t::::: 0 and t::::: 1 seconds.
(a) = 10 sin (lOt I = w/2rr = 1.6 eye/sec,
x(t)
Then
w
= 10 rad/sec,
(b) Displacement is x(t) = 10 sin (lOt - 30°).
Velocity is dx/dt::::: wAo cos (wt - e).
d2x/dt 2
Acceleration is
-10 2(10) ::::: -1000 m/sec 2 •
(c) At t ;::: 0:
x(O)
x(O)
At t
30 0 )
and
= Ao sin (wt - Ii) p = 1/1 = 0.63 sec. Thus the maximum displacement is 10 m.
Thus the maximum velocity is 10(10)::::: 100 m/sec.
-w2Ao sin (wt - 8),
:::::
and so the maximum acceleration is
= 10 sin (-30 = 10(-0.5) -5m
= wAo cos (-30 ::::: 10(10)(0.866) = 86.6 m/sec
0
;:::
)
0
)
x(O)
:::::
-w2Ao sin (-30 0 )
x (1)
;:::
10 sin (10 - 30 0 ) -
x(1)
:::::
10(10) cos 180
x(l)
:::::
-(10)2 (10) sin 180 0
= 1:
0
-(10)2 (10)(-0.5)
:::::
10 sin (570 0
-
30 0 )
:::::
-
500 m/sec 2
10 sin 180 0
0
= -100 m/sec
-
0
FREE VIBRATION
fl::g:-; Determine the differential equation of motion and natural frequency of vibration of
" - - / the simple single-degree-of-freedom spring-mass system shown in Fig. 1-10.
Apply Newton's law of motion, :IF;::: mao For vertical
motion, the forces acting on the mass are the spring force
k(8 st + x) and the weight mg of the mass. Therefore the dif­
ferential equation of motion is
mx = -k(ost + x) + my
where Bst is the static deflection due to the weight of the mass
acting on the spring. Then my::::: Ostk, and the equation of
motion becomes
+ kx = 0
mx
which is the differential equation for SHM.
of solution for this equation are
x(t) = A sin Vklm t + B cos Vk/m t
x(t) ::::: C sin (Vk/m t
x(t) The general forms
+
= D cos (Vk/m t -
¢)
Ii)
where A, B, C, D, ¢ and e are arbitrary constants depending on
initial conditions x(O) and x(O). Two constants must appear in
each of the general solutions because this is a second order
differential equation.
Fig.I-IO
CHAP. 1)
~
,
B
13
VIBRATIONS AND WAVES
For an initial displacement
and hence
= Xo
x(O)
= Xo
x(t)
and zero initial velocity x(O)
0,
we have A
0,
Xo cos yklm t
=:
Physically, this solution represents an undamped free vibration, one cycle of which occurs when
varies through 360 degrees. Therefore the period P and the natural frequency In are
y kIm t
P
where
"'n
=:
y k/m rad/sec
=
2".
In
and
--sec
Vklm
=
= -Vklm
2 - cyc/sec
lIP
11' is the circular natural frequency of the system. --+-~----~r----T----~---.r--~~---~t
Fig.l-ll.
Free vibration without damping
/;';:O~A g~neralized
singl~:degr~e-of-freedom
spring-masS system with damping is shown
FIg. 1-12. InvestIgate Its general motion.
" ' - . - - / lD
= ma,
m x + ex + kx = 0
Employing Newton's law of motion
mx
=:
-c:i; - kx
or
~
F
where k is the spring constant, tn the mass, and c the damping coeffici :nt.
We cannot assume solutions of the sine or cosine functions because
of the term
We assume x
ert ; then
== re rt , == r 2 ert . Sub­
stituting these values into the differential equation of motion, we obtain
ex.
=
or
x
x
mr2
+ cr + k
0
=:
The two values of r satisfying the above equation are
r1,r Z
==
-c :± yc 2 -
where "'n =: y k/m, and
tion of motion is
4mk
r = c/2m"'n
=
(-t:± y~2-l)wn
is called the damping factor.
x(t) =: Aer,t
Fig. 1·12
Thus the solution to the equa­
+ Be rzt
where A and B are arbitrary constants determined by the two initial conditions imposed on the
system.
Since the values of r depend on the magnitude of
vibration with damping:
r,
we have the following three cases of free
If ~ is greater than unity, the values of r are real and distinct; the amplitude of x
Case 1:
is decr~asing but will never change sign. Therefore oscillatory motion is not possible for the system
regardless of initial conditions. This is o~)erdamped, where
x(f) == Ae-rl t
+ Be- r t
2
Case 2: If r is equal to unity, the values of r are real and negative, and are equal to -Wn­
The motion of the system is again not oscillatory, and its amplitude will eventually reduce to zero.
This is critically-damped, where
x(t)
=
(C
+ Dt)e- w • t
[CHAP. 1
VIBRATIONS AND WAVES
14
If
Case 3:
And if we define
r is
less than unity, the values of r are complex conjugates:
wd::::
VI - ~2
Wn
as the damped natural frequency in rad/sec, we have
and
Expanding,
Letting E
+ i(E
x(t)
+F
= G and i(E - F)
F) sin Wdt)
H, we finally obtain
e-Cw"t(G coswdt +H Sinwdt)
x(t)
As shown before, we may combine a cosine and sine function of the same frequency into a
single sine or cosine function as
x{t)
1e-,",,,t sin (Wdt + 0)
1e- Cw"t cos (Wd t - if;)
x(t)
where 1:::: yG2 + H2,
(J
= tan-l(G/H),
if;
tan-1(H/G).
The motion is oscillatory with angular frequency Wd' The amplitude of motion will decrease
expunentially with time because of the term e-'",,,I, which is known as the decaying factor. This
is underdamped vibration. Refer to Fig. 1-13.
Hence it may be concluded that the motion of a dynamic system with damping and having
free vibration depends on the amount of damping present in the system. The resulting motion will
be periodic only if the amount of damping present is less than critical, and the system oscillates
with angular frequency slightly less than the free natural frequency of the system.
4:e
critically·damped
jOVerdamped
underdamped
Fig.I·13. Free vibration with damping
~
o
ED VIBRATION
1.11. nvestigate the general motion of a simple spring.mass
system with damping excited by a sinusoidal force
Fo cos wt as shown in Fig. 1-14.
Employing Newton's law of motion,
m
x = sum of forces in the x direction
-k(x + .sst)
+ mg - eX + Fo cos wt
But k.s st mg, the Weight of the mass; hence the equation of
motion takes its most general form
mx+c:i:+k:e:::: Focoswt
The general solution for this second order differential equation
with constant coefficients is
x
::::
Xc
+ xp
Fig.I-14
CHAP. 1]
15
VIBRATIONS AND WAVES
where Xc is called the complementary solution, or the solution of the homogeneous equation,
m + cx + kx = O. xI' is the particular solution for the given equation.
The complementary solution, known as free vibration, has been solved previously in Problem
1·10. The particular solution, obtained from the nonhomogeneous part Fo cos wt of the differential
equation of motion, is
Xp(t)
A sin wt + B coswt x
and so
=
=
xp(t)
wA cos wt -
xp(t)
wB sin wt w2B cos wt sind
-w 2A
Substituting these expressions into the equation of motion, we obtain +
(kA - mAw 2 - cwB) sin wt
mBw 2 + cOlA) cos wt
(kB
Fo cos wt Equating the coefficients, (k -
from which
A
mw~)A
=
(k -mw2)2
,COlA. + (k -
0,
cwB
Fowc
(k - mOl 2)2 + (cw)2 sin wt
Then
.
= Fo
Fo(k -mw 2 )
B
+ (cw)2 '
mw 2 )B
+
Fo(k - mw 2 )
(k - mOl2)" + ()"
COl cos wt
W
W
We may combine these two sinusoidal functions of the same frequency either by rotating vectors
or by trigonometric identities to obtain
Fo
cos (Olt - ¢)
-;::.======
Y(k - mw
+ (cOl)2
2 )2
Folk
-;::.======
Y(1 - + (2ir)2 cos (Olt -
or
where r = wlw n ,
1. 2)2
Wn
and
yklm,
tan- 1
¢
1=
COl
k - 7nOl2
¢)
= tan- 1 1 2- i r?
r'
ofl\ r = 0
I V " locus of maxima
" I I\
I
I
/
/
I
/
I
I
\
I
I,
\
\
\
13 \
0
< \3 < \2 < it
\
\
Fig. 1·15
Hence it may be concluded that the particular solution xp(t), which is known as the steady
state response or forced vibration, is of the same frequency as that of the excitation force regard­
less of initial conditions. The amplitude of forced vibration depends on the amplitude and
16
VIBRATIONS AND WAVES
[CHAP. 1
frequency of the excitation force, and the parameters of the systems. At resonance, Le. when the
forcing frequency is equal to the natural frequency, or w/w n = I, the amplitude of forced vibration
is limited only by the damping factor I and hence the amount of damping present. Therefore
resonance should be avoided at all times. Finally, the steady state response of the system is not
in phase with the excitation force; its variation by the phase angle ¢ is due to the presence of
damping in the system. Without damping, the steady state response is either in phase or 180 0 out
of phase with the excitation force. See Fig. 1-15 to Fig. 1-19.
x
Fig.I-I6.
Forced vibration without damping (2/" == 3f)
Fig. I-Ii.
Forced vibration without damping (in
= 6.28/)
Forced vibration with damping (f n
6.28f)
x
a
Fig. I-IS.
'x
Fig. 1-19.
Free and forced vibration (f n == 6.28f)
ENERGY OF VIBRATION
~
Determine the power requirements for vibration testing and analysis.
In vibration testing, we have forced vibration. The work done is the product of the excitation
and displacement, while power required is the rate of doing work. Let F
Fo cos wt and
x = A cos (wt - ¢); then the work done is
=
w =
f
F dx
=
f
Fo cos wt[-A sin (wt - ¢) d(wt)]
CHAP. 1] VIBRATIONS AND WAVES
17
and work done per cycle of motion is
(2".
-FOA)" cos wt sin (wt - rp) d(wt)
=
W
o
. as the angle wt goes through a cycle of 2... Since sin (Olt - rp)
done per cycle of motion becomes
1
= sin Olt cos rp
2
=
W
FoA sin rp
".
o
cos 2 wt d(wt) - FoA cosrp
FA'
[Olt, sin 2wtJ2'"
o sm rp 2 -:- --4- 0
FoA cosrp [
f2'" coswt sinwt d(Olt)
o
sin 2 tJ2:r
-r
0
Z
cos 2OltJ27l'
1
'"
FA'
sin 2wtJ 0
o sm rp [Olt,
2 -:- --4' - FoA cosrp [ - - - - ­
.
4
2
0
-_
..AFo sin rp Fo sin wt and x
If F
cos Olt sin rp, the work
\
A sin (wt - rp), then the work done is
f F dx
W
=
f F ~: dt
=
f
Fx dt
The expression for work done in one cycle of motion is then
fo
2"./",
W
=
Since cos (wt - rp)
W
f
Fo sin wt[OlA cos (wt - rp) dtl
= cos Olt cos rp + sin wt sin rp, 2
= "./", OlAFo sin wt cos wt cos rp dt +
1
o
As shown earlier, the above expression can be reduced to
sin20lt
[ AFoOlcosrp-2--
W
[ AFow cos rp
=
+
(L
211"/'"
FoAw sin wt cos (Olt - rp) dt
o
f2lTIW wAFo sin rp sin 2 Olt dt
0
(t
t) Jo
.
sin 2 w
AFoOlsInrp 2-~
co:;Olt)
+
AFow sin rp
""'211"/",
G-Si~:wt)
J:lTIW
..AFo sin rp
Thus the power required is proportional to the amplitude Fo of the excitation force as well as
to the amplitude A of the displacement. When there is no damping in the system, the work done
by the driving force is zero because rp = 0° or 180°. At resonance, energy is needed to build up
the amplitude of vibration; and for this case, rp
90°.
The steady state response of a simple dynamic system to a sinusoidal excitation
10 sin O.bt newtons is 0.1 sin (O.h-t 30°) meters. Determine the work done by
the excitation force in (a) one minute and (b) one second.
(a)
From Problem 1.12, the work done per cycle by the excitation force is given by
f2- F dx
W
o
=
Ii> Fx dt
=
r.AFo sm rp
0
=
where Fo = 10 newtons is the amplitude of the excitation force, A
0.1 m is the amplitude
of the steady state response, and rp 30° is the phase angle. Hence work done by the
excitation force is W
3.14(0.1)(10)(0.5) = 1.57 joules/eye.
The angular frequency is
1//
20 sec. In one minute, the excitation force will complete
0.1.. rad/sec and the period P
three cycles. Therefore work done by the excitation force in one minute is 4.71 joules.
=
=
(b) Work done per cycle is W
W
=
i
=
=
f20 Fx dt.
Then work done in one second is
o
1
o
(10 sin O.l;;-t)(O.Olr.-) cos (O.lr.-t - 30°) dt
=
0.05 jqule
18
VIERA TIONS AND WAVES
[CHAP. 1
Prove that the mean kinetic and potential energies of nondissipative vibrating systems
are equal.
For free vibration without damping, the motion can be assumed harmonic and is given by
=
x(t)
Kinetic energy KE
A sin "'Ilt
,
= tmx2 = tm(",~A2 cos 2 "'nt) = ~kA2 cos 2 "'nt,
where
"'~
= kim.
Potential energy PE = tkx2 = tkA2 sin 2 "'nt.
~ fP
(KE)mean
(tkA2 cos 2 "'Ilt) dt
o
If
=
(PE)mcan
p
sin 2 '" " t) dt
p o (J.kA2
~
1.15. A uniform string fixed at both ends is displaced a distance h at the center and released
from rest as shown in Fig. 1-20. Find the energy of transverse vibration of the
string.
~I
-
h
Fig. 1·20
The free transverse vibration of a uniform string can be expressed as
_= ~
~
y(x, t)
1,2....
t
where
Ai
is the amplitude of motion and
A
' i;;-x
i smycos
(i;;-a
-y;t + )
(Ii
is the phase angle.
(Ii
(See Problem 1.17.)
Then
KE
L
PE
or 5 [02YJ2
~
o
uX
KE
2
+
dx
oc
PE
i=
~
1~2~
...
where S is the tension in the string, PL is the mass per unit length of the string, and a
the speed of wave propagation.
. . .
. .
From the ImtJal condItIons
A~
= 64h 21i 2lT 2.
•
y(x,O)
=0
and y(x,O)
= {2hXIL'
2h(1- xIL),
0 "" x "" L/2
L/2 "" x"" L
= "";SlpL
is
.
we obtam
The expression for the energy of transverse vibration of the string becomes
i
Let the total energy associated
~ith
=
1,3, ...
the fundamental mode of vibration be E I, i.e.
El
=
16pLa 2 h 21L;;-2
Then the energies associated with the first harmonic, second harmonic, third harmonic, ... are
respectively
Thus the main part of the energy of vibration is associated with the normal modes of low order.
The quality of a tone is governed by the proportion of energy in each of the modes of vibration.
Though the fundamental frequency may be the same, the -energy distribution in the harmonics
characterizes each musical instrument.