Eq. (4.104). Neglecting everywhere _
X-terms compared to X
2 -terms, the transformed set of first-order equations becomes identical to that of the
constant-frequency case, (4.97), with the only difference that now X= X(s). Also,
performing the averaging operations as described in Sect. 4.7.3, it is easily shown
that the averaged equations become identical to those describing the
constant-frequency case. Equations (4.99)–(4.100) are thus applicable too for the
case of slow frequency-sweeps across resonance. But for sweeps, since X is a
function of time, no stationary solutions of the averaged equations exist. To obtain
a(s) and h(s), Eqs. (4.99)–(4.100) must be integrated numerically. This, however, is
a much more stable and fast undertaking than to integrate the full un-averaged
equations (4.104). The averaged equations (4.99)–(4.100) contain no
fast-oscillating components, and this gives leave for employing much larger
time-steps in the numerical procedure.
Only time-linear frequency-sweeps are considered. We assume the instantaneous
frequency of excitation to be prescribed externally by:
XðsÞ ¼
X 0 ;
s\s 0 ;
X 0 þ r s ðs À s 0 Þ; s 0 s s 1 ; r s ¼ ðX 1 À X 0 Þ=ðs 1 À s 0 Þ:
&
ð4:105Þ
As appears, the system is driven at constant frequency X 0 during an initial span
of time s 2 [0; s 0 [, sufficiently long for transients to decay and stationary conditions
to settle. Then, during s 2 [s 0 ; s 1 ], the frequency is either increased or decreased at
a constant rate r s towards the terminal frequency X 1 . An evaluation of the function
mðsÞ ¼
R ð1 À XðsÞÞds will show that if the sweep rate r s is small and if X 0 % 1,
then m(s) is indeed a slowly varying function of time, as assumed.
Fig. 4.18 depicts time histories for a sweep through X 2 [0.5; 1.5] at two
different sweep rates r s . Solid lines indicate results of numerically integrating the
un-averaged equations (4.104). Dashed lines indicate results of numerically integrating the averaged equations (4.99)–(4.100) with X(s) given by (4.105). At the
sweep rate r s = 0.002 the averaged response a(s) of the string accurately describes
the envelope of the fast-oscillating full response u(s) (Fig. 4.18(a), top). Also, the
averaged response of the point mass closely follows the un-averaged response
(Fig. 4.18(a), bottom). On raising the sweep rate to r s = 0.005 small discrepancies
between the averaged and the un-averaged responses become apparent (Fig. 4.18
(b)). Hence, for the particular parameter values chosen, values of sweep rates r s
0.002 will be considered sufficiently slow for averaged results to apply.
Fig. 4.19 compares frequency responses obtained at four different sweep rates
when the point mass is respectively fixed (Fig. 4.19(a)) and spring-loaded
(Fig. 4.19(b)). The stationary frequency responses of Fig. 4.17(a) (corresponding
to r s = 0) have been superimposed. With fixed point mass (Fig. 4.19(a)) the swept
response for the slowest sweep rate ðr s ¼ Æ2 Á 10
À4
Þ follows the stationary
response (r s = 0) rather closely. When the point mass is free to slide (Fig. 4.19(b)),
a jump-up to the stationary curve occurs for the up-sweep r s ¼ þ 2 Á 10
À4
ð
Þ ,
whereas a jump-down to the stationary curve occurs for the down-sweep
4.7 String with a Sliding Point Mass
253
X-terms compared to X
2 -terms, the transformed set of first-order equations becomes identical to that of the
constant-frequency case, (4.97), with the only difference that now X= X(s). Also,
performing the averaging operations as described in Sect. 4.7.3, it is easily shown
that the averaged equations become identical to those describing the
constant-frequency case. Equations (4.99)–(4.100) are thus applicable too for the
case of slow frequency-sweeps across resonance. But for sweeps, since X is a
function of time, no stationary solutions of the averaged equations exist. To obtain
a(s) and h(s), Eqs. (4.99)–(4.100) must be integrated numerically. This, however, is
a much more stable and fast undertaking than to integrate the full un-averaged
equations (4.104). The averaged equations (4.99)–(4.100) contain no
fast-oscillating components, and this gives leave for employing much larger
time-steps in the numerical procedure.
Only time-linear frequency-sweeps are considered. We assume the instantaneous
frequency of excitation to be prescribed externally by:
XðsÞ ¼
X 0 ;
s\s 0 ;
X 0 þ r s ðs À s 0 Þ; s 0 s s 1 ; r s ¼ ðX 1 À X 0 Þ=ðs 1 À s 0 Þ:
&
ð4:105Þ
As appears, the system is driven at constant frequency X 0 during an initial span
of time s 2 [0; s 0 [, sufficiently long for transients to decay and stationary conditions
to settle. Then, during s 2 [s 0 ; s 1 ], the frequency is either increased or decreased at
a constant rate r s towards the terminal frequency X 1 . An evaluation of the function
mðsÞ ¼
R ð1 À XðsÞÞds will show that if the sweep rate r s is small and if X 0 % 1,
then m(s) is indeed a slowly varying function of time, as assumed.
Fig. 4.18 depicts time histories for a sweep through X 2 [0.5; 1.5] at two
different sweep rates r s . Solid lines indicate results of numerically integrating the
un-averaged equations (4.104). Dashed lines indicate results of numerically integrating the averaged equations (4.99)–(4.100) with X(s) given by (4.105). At the
sweep rate r s = 0.002 the averaged response a(s) of the string accurately describes
the envelope of the fast-oscillating full response u(s) (Fig. 4.18(a), top). Also, the
averaged response of the point mass closely follows the un-averaged response
(Fig. 4.18(a), bottom). On raising the sweep rate to r s = 0.005 small discrepancies
between the averaged and the un-averaged responses become apparent (Fig. 4.18
(b)). Hence, for the particular parameter values chosen, values of sweep rates r s
0.002 will be considered sufficiently slow for averaged results to apply.
Fig. 4.19 compares frequency responses obtained at four different sweep rates
when the point mass is respectively fixed (Fig. 4.19(a)) and spring-loaded
(Fig. 4.19(b)). The stationary frequency responses of Fig. 4.17(a) (corresponding
to r s = 0) have been superimposed. With fixed point mass (Fig. 4.19(a)) the swept
response for the slowest sweep rate ðr s ¼ Æ2 Á 10
À4
Þ follows the stationary
response (r s = 0) rather closely. When the point mass is free to slide (Fig. 4.19(b)),
a jump-up to the stationary curve occurs for the up-sweep r s ¼ þ 2 Á 10
À4
ð
Þ ,
whereas a jump-down to the stationary curve occurs for the down-sweep
4.7 String with a Sliding Point Mass
253