1.6 Other Simple Vibrating Systems
1.6.5 Combinations of Springs and Masses
Several combinations of masses and springs are shown in Fig. 1.9, along with their resonance frequencies. Note the effect of combining springs in series and parallel combinations. Two springs with spring constants K1 and K2 will have a combined spring constant Kp = K1 + K2 when connected in parallel but only K8 = K1K2/(K1 + K2) in series. When K1 = K2, the parallel and series values become 2K1 and Kl/2, respectively. The combinations in Fig. 1.9 all have a single degree of freedom. In Section 1.12, we discuss two-mass systems with two degrees of freedom; that is, the two masses move independently.
1. 6. 6 Longitudinal and Transverse Oscillations of a Mass-Spring System
Consider the vibrating system shown in Fig. 1.10. Each spring has a spring constant K, a relaxed length ao, and a stretched length a. Thus, each spring exerts a tension K (a - ao) on the mass when it is in its equilibrium position ( x = 0). When the mass is displaced a distance x, the net restoring force
(a) (b) (c) (d) (e)
FIGURE 1.9. Mass-s ring combinations that vibrate at single frequencies:
(a) fo = (1/271') K/2m; (b) fo = (1/27r)J2K/m; (c) fo = (1/271')~;
(d) fo = (1/27r)JK/4m; (e) fo = (1/27r)JKjffi.
1.6. Other Simple Vibrating Systems 17
(a) (b)
FIGURE 1.10. Longitudinal (a) and transverse (b) oscillations of a mass-spring system.
is the difference between the two tensions:
Fx = K(a- x- ao)- K(a + x- ao) = -2Kx.
The natural frequency for longitudinal vibration is thus given by
fz = ]_ !2K_
2?T v -;;:;:
(1.52)
(1.53) Now, consider transverse vibrations of the same systems, as shown in Fig. l.lO(b). When the mass is displaced a distance y from its equilibrium position, the restoring force is due to they component of the tension:
Fy = -2K( Ja2 + y2 - a0 ) sin(}= -2K( Ja2 + y2 - a0 ) y Ja2 + y2
= -2Ky (1 - Ja2 ao + y2 ) . (1.54)
For small deflection y, the force can be written as
(1.55) When the springs are stretched to several times their relaxed length (a ~ ao), the force is approximately -2Ky, and the natural frequency is practically the same as the frequency for longitudinal vibrations given in Eq. (1.53):
(1.56) When the springs are stretched only a small amount from their re- laxed length (a ~ ao), however, the first term in Eq. (1.55) becomes very small, so the vibration frequency is considerably smaller than that given in Eqs. (1.53) and (1.56). Furthermore, the contribution from the cubic
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18 1. Free and Forced Vibrations of Simple Systems
term in Eq. (1.55) takes on increased importance, making the vibration nonsinusoidal for all but the smallest amplitude.
1. 7 Forced Oscillations
When a simple oscillator is driven by an external force f(t), as shown in Fig. 1.11, the equation of motion Eq. (1.32) then becomes
mx + RX + Kx = f(t). (1.57) The driving force f(t) may have harmonic time dependence, it may be impulsive, or it may even be a random function of time. For the case of a sinusoidal driving force f(t) = F cos wt turned on at some time, the solution to Eq. (1.57) consists of two parts: a transient term containing two arbitrary constants, and a steady-state term that depends only on F and w.
To obtain the steady-state solution, it is advantageous to write the equation of motion in complex form:
mf + R1: + Kx = Fejwt. (1.58) Since this equation is linear in x and the right-hand side is a harmonic function with angular frequency w, in the steady state the left-hand side should be harmonic with the same frequency. Thus, we replace x by Aeiwt and obtain
(1.59) The complex displacement is
Fejwt
x= ~--~---~
K-w2m + jwR F/m (1.60)
w5 - w2 + jw2o: '
where F = Fejwt, w5 = Kjm, and o: = R/2m.
f(t)
FIGURE 1.11. A damped harmonic oscillator with driving force f(t).
1. 7. Forced Oscillations 19 Differentiation of x gives the complex velocity v:
_ FeJwt Fwlm
v - - ---....,:.-..,.---..,.,.
- R + j ( wm - K I w) - 2wa + j ( w2 - w5) ã (1.61)
The mechanical impedance Z is defined as F I v:
Z = P lv = R + j(wm- Klw) = R + jXm, (1.62) where Xm = wm- Klw is the mechanical reactance. The actual steady- state displacement is given by the real part of Eq. (1.60):
x = Re _P_ = FZ sin(wt + ¢).
JWZ w (1.63)
A quantity Xs = Fl K = F I mw5 can be defined as the static displace- ment of the oscillator produced by a constant force of magnitude F. At very low frequency, the displacement amplitude will approach F I K, and the os- cillator is said to be stiffness dominated. When w = Wct, the amplitude becomes
xo = F l2amwo = Qxs. (1.64) In other words, Q becomes a sort of amplification factor, which is the ratio of the displacement amplitude at resonance ( wo = w) to the static displacement.
There is a direct relation between the damping coefficient a, the decay time T of Eq. (1.39), and the width of the resonance peak in Fig. 1.12. If we take the absolute value of both sides of Eq. (1.61), then we see that the denominator, which largely determines the shape of the resonance, is [(w2 - w5)2 + 4w2a2]112 . Provided that we are only concerned with frequencies w quite close to w0 , we can write this approximately as 2w0[(w -w0 ) 2 +a2]112 .
The magnitude of the denominator thus increases by a factor 2112 relative to its value at w = wo when lw - wo I ~ a. The response decreases by the same factor, which represents a 3 dB decline from the peak value at the resonance. This 3 dB half-width of the resonance curve, measured in radians per second, is thus equal to the damping coefficient a, and also, by Eq. (1.39), to the reciprocal of the decay timeT in seconds. The 3 dB full-width ~w of the curve is 2a =Rim, and its relative value 2~wlwo is equal to Q-1 .
At high frequency (w :ằ wo), the displacement falls toward zero. The frequency response of a simple oscillator for different values of a (or Q) is shown in Fig 1.12(a). The magnitude of x is less than Xs for frequencies above WoV2- 82 (where 8 = 1IQ = 2alw0 ), which, for small values of a, is about v'2wo. If a > wolv'2, x < Xs at all frequencies.
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20 1. Free and Forced Vibrations of Simple Systems
xlx,
stiffness dominated
resistance dominated
(a)
(b) mass dominated
FIGURE 1.12. Frequency dependence of the magnitude x and phase (¢x- ¢F) of the displacement of a linear harmonic oscillator.
The phase angle between the displacement and the driving force is the phase angle of the denominator in Eq. (1.60):
2a:w
<Px - ¢F = tan-1 2 w -w0 2 . (1.65) At low frequency (w ~ 0), <Px-¢F = 0. When w = wo, <Px-¢F = 90°, and at high frequency (w ằ wo), <Px-ÂF ~ 180°, as shown in Fig. 1.12(b ).
There are other convenient ways to represent the frequency response of a simple oscillator. One way is to show how the real and imaginary parts of the mechanical impedance Z( = F jv) or the mechanical admittance (mo- bility) Y = 1/Z(= vjF) vary with frequency. At resonance, the real part of the admittance has its maximum value, while that of the impedance re- mains equal to R at all frequencies. The imaginary parts of both quantities are zero at resonance. Figure 1.13 shows the real and imaginary parts of the
1.8. Transient Response of an Oscillator 21 z 'IZI
' ' B
y
0
'
R ...
(a)
' ' ' \ \ (c)
a' ' ' ... I
(b)
FIGURE 1.13. Real and imaginary parts of the mechanical impedance and admit- tance for a harmonic oscillator of the same type as in Fig. 1.12: (a) mechanical impedance; (b) mechanical admittance or mobility; (c) Nyquist plot showing the imaginary part of admittance versus the real part, with frequency as a parameter.
mechanical impedance and admittance for an oscillator of the same type as in Fig. 1.12. The graph of imaginary part versus the real part in Fig. 1.13(c) is sometimes called a Nyquist plot.