# Stabilization of Driven Pendulum with Periodic Linear Forces

Authors: Babar Ahmad
Journal: Journal of Nonlinear Dynamics
 (2013)

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## Abstract

Using Kapitza method of averaging for arbitrary periodic forces, the pendulum driven by different forms of periodic piecewise linear forces is stabilized. These periodic piecewise linear forces are selected in the range[−1,1] to establish an exact comparison with harmonic forces. In this contest, the rectangular force was found to be the best, but this force is more effective when it has a time-dependent structure. This time-dependent structure is found by defining a parametric control on some other periodic piecewise linear forces.


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## Body

## 1. Introduction

A pendulum with fixed suspension has only one stable point, while a pendulum whose suspension has fast oscillation can have more stable points (can oscillate). Such phenomena were first studied by Stephenson in 1908 [1–3]. In 1951, Kapitza presented this problem in a different way [4], so-called Kapitza pendulum. In 1960,Landau et al. studied the stability of such a system driven by harmonic force [5]. Then, its rapid growing applications started such as trapping of particles by laser [6–8], control of robotic devices [9, 10], effect on price equilibrium [11], and control by lasers in cybernetics [12].Next in place of harmonic force Ahmad and Borisenok (2009) used periodic kicking forces, modifying Kapitza method for arbitrary periodic forces [13]. Also, Ahmad used symmetric forces and stabilized the system with comparatively low frequency of fast oscillation [14].


## 2. Kapitza Method for Arbitrary Periodic Forces

A classical particle of massm is moving in time-independent potential field U and a fast oscillating control field. For simplicity, consider one-dimensional motion. Then, the force due to time-independent potential U(x) is
(1)f1(x)=-dUdx,
and a periodic fast oscillating force with zero mean in Fourier series is
(2)f2(x,t)=∑k=1∞‍[ak(x)cos(kωt)+bk(x)sin(kωt)].
This fast oscillation has frequency ω≡2π/T≫2π/TU≡ω0. Here, ω0 is the frequency of motion due to f1. The mean value of a function is denoted by bar and is defined as
(3)f-=1T∫0Tf(x,t)dt.
Also, the Fourier coefficient a0 is
(4)a0(x)=2T∫0Tf2(x,t)dt.
From (3) and (4), it follows that(5)f-≅a0.
In (2), ak and bk are the Fourier coefficients given as
(6)ak(x)=2T∫0Tf2(x,t)coskωtdt,bk(x)=2T∫0Tf2(x,t)sinkωtdt.
Due to (1) and (2), the equation of motion is
(7)mx¨=f1(x)+f2(x,t).
Here at a time two motions are observed: one along a smooth path due to f1 and the other small oscillations due to f2. So the path can be written as x(t)=X(t)+ξ(t) (see Figure 1). Here, ξ(t) represents small oscillations.Figure 1
Path of the particle.By averaging procedure, the effective potential energy function is [13]
(8)Ueff=U+14mω2∑k=1∞‍(ak2+bk2)k2.The pendulum driven by a periodic force is stabilized by minimizing (8). These forces are chosen in the range [-1,1] to establish an exact comparison with harmonic forces. Next, an ϵ-parametric control is developed with one of the driving forces and has better results.


## 3. The Pendulum Driven by Harmonic Force

Consider a pendulum whose point of support oscillates horizontally (see Figure3), under the influence of harmonic force (Kapitza pendulum). The harmonic force is (see Figure 2)
(9)f(t)=sinωtif0≤t<T,
and the fast oscillating force is
(10)f2(ϕ,t)=mω2cosϕf(t)
with the meaning  f¯2=0, which follows from the Fourier coefficient a0=0, for (9). The other Fourier coefficients are given by using (6):
(11)ak=0,bk=mω2cosϕ.
Then, the forces acting on the particle are
(12)f1=-d(-mglcosϕ)dϕ,f2=mω2cosϕsin(ωt),
and the effective potential energy is obtained by using (8):
(13)Ueff=mgl(-cosϕ+ω24glcos2ϕ).
The stable equilibrium is found by minimizing (13). Ueff has the extrema at ϕ=0,π,±arccos2gl/ω2.(i)
The downward pointϕ=0 is stable if ω2<2gl.(ii)
Vertically upward pointϕ=π is not stable.(iii)
The point given bycosϕ=2gl/ω2 is stable if ω2>2gl [5].These stable points are illustrated in Figure 4.Figure 2
Harmonic force.Figure 3
Horizontal modulation with harmonic force.Figure 4
Stable points with horizontal oscillations.


## 4. The Pendulum Driven by Periodic Piecewise Linear Forces

The goal is to stabilize the pendulum with low frequency as compared to harmonic force. Next, this harmonic force is replaced with some periodic piecewise linear forces within the range of harmonic force. These periodic piecewise linear forces areT-periodical: R(t+T,n)≡R(t,n). For horizontal modulation, the force acting on the particle is
(14)f2(ϕ,t)=mω2cosϕ·R(t,n).


### 4.1. Triangular Type Force

First of all introduce the triangular type force (see Figure5) given by
(15)Rs(t)={4Ttif0≤t<T4,4T(-t+T2)ifT4≤t<3T4,4T(t-T)if3T4≤t<T.
For (15), the Fourier coefficient a0=0 indicates R-s=0. The Fourier expansion of (14) is
(16)f2(ϕ,t)=mω2cosϕ8π2×∑j=0∞‍(-1)j(2j+1)2sin(2π(2j+1)tT).
With
(17)ak=0,bk=mω2cosϕ8π2(-1)j(2j+1)2,
the effective potential energy is
(18)Ueff=U+mω2cos2ϕ·14(8π2)2∑j=0∞‍1(2j+1)6=U+π260mω2cos2ϕ
which has extrema at ϕ=0,π,±arccos30gl/ω2π2. Minimization of Ueff shows that(i)
the downward pointϕ=0 is stable if ω2<3.0396gl,(ii)
vertically upward pointϕ=π is not  stable,(iii)
the point given bycosϕ=3.0396gl/ω2 is stable if ω2>3.0396gl.From (iii), it is observed that, at nontrivial position, the oscillator is stabilized with higher frequency as compared to harmonic force [13].Figure 5
Triangular type force.Hence, this force is less effective than sin- or cos-type force. So, this force is replaced by some other periodic piecewise linear forces.


### 4.2. Hat Type Force

Now if sine function is traced by linear pulses (see Figure6), defined by (19)(19)Lc(t)={12if0≤t<16T,1if16T≤t<13T,12if13T≤t<12T,-12if12T≤t<23T,-1if23T≤t<56T,-12if56T≤t<T,
Next if the sine force is traced by a linear force forming a hat (see Figure 6), defined by (19). For horizontal modulation, the oscillating force is
(20)f2(ϕ,t)=mω2cosϕ·Lc(t).
Then, by Fourier expansion in place of (19),
(21)ak=0bk=mω2cosϕ1kπ(1-coskπ+2coskπ3).Using the above coefficients, the oscillating force acting on the particle is
(22)f2(t)=mω2cosϕ×∑k=1∞‍1kπ(1-coskπ+2coskπ3)sinkωt,
and the effective potential energy is
(23)Ueff=U+mω2cos2ϕ·14π2∑k=1∞‍1k2(1-coskπ+2coskπ3)2Ueff=U+0.2302mω2π2cos2ϕ
which has extrema at ϕ=0,π,±arccos2.172gl/ω2. Here,(i)
the positionϕ=0 is stable if ω2<2.172gl,(ii)
the inverse positionϕ=π is not stable,(iii)
the positionϕ=arccos2.172gl/ω2 is stable if ω2>2.172gl.Again a less effective result is obtained. So this periodic piecewise linear force is replaced by another one.Figure 6
Hat type force.


### 4.3. Trapezium Type Force

If the sine force is traced by a linear shape forming a trapezium (see Figure7), given by (24)(24)Tm(t)={8tTif0≤t<T8,1ifT8≤t<3T8,8T(T2-t)if3T8≤t<5T8,-1if5T8≤t<7T8,8(t-T)Tif7T8≤t<T,with T-m=0. For horizontal modulation, the force acting on the particle is
(25)f2(ϕ,t)=mω2cosϕ·TM(t,n).
Then, by Fourier expansion in the place of (24),
(26)ak=0,bk=mω2cosϕ16π21k2sinkπ4.
Using the above coefficients, the oscillating force is
(27)f2(ϕ,t)=mω2cosϕ16π2×∑k=0∞‍1k2sinkπ4sinkωt,
and the effective potential energy is
(28)Ueff=U+mω2cos2ϕ·14(16π2)2×∑k=0∞‍1k6sin2kπ4=U+0.3393mω2cos2ϕ
which has extrema at ϕ=0,π,±arccos1.4736gl/ω2. Here,(i)
the pointϕ=0 is stable if ω2<1.4736gl,(ii)
the pointϕ=π is not stable,(iii)
the point given bycosϕ=1.4736gl/ω2 is stable if ω2>1.4736gl.From (iii), it is observed that, at nontrivial position, the oscillator is stabilized with lower frequency as compared to harmonic force. So this type of force is much effective than sin- or cos-type force. Next, modify this trapezium shape force to have a better result.Figure 7
Trapezium type force.


### 4.4. Quadratic Type Force

If slopes are removed in the beginning and at the end from it and define a quadratic type force:Qc(t)=Qc(t+T) (see Figure 8), given by (29)(29)Qc(t)={1if0≤t<3T8,8T(T2-t)if3T8≤t<5T8,-1if5T8≤t<T,with the same property Q-c=0. For horizontal modulation, the oscillating force is
(30)f2(ϕ,t)=mω2cosϕ·Qc(t,n).
The Fourier expansion of (30) is
(31)Qc(t)=mω2cosϕ×∑k=0∞‍(2kπ+8π2k2sinkπ4)sinkωt,and the effective potential energy is
(32)Ueff=U+mω2cos2ϕ·14∑k=0∞‍1k2(2kπ+8π2k2sinkπ4)2=U+0.3856mω2cos2ϕ
which has extrema at ϕ=0,π,±arccos1.2967gl/ω2. Here,(i)
the positionϕ=0 is stable if ω2<1.2967gl,(ii)
the positionϕ=π is not stable,(iii)
the positionϕ=arccos1.2967gl/ω2 is stable if ω2>1.2967gl.Again the frequency of oscillation is lower at nontrivial position.Figure 8
Quadratic type force.


### 4.5. Rectangular Type Force

Now if we introduce rectangular type force:Rl(t)=Rl(t+T) (see Figure 9), given by (33)(33)Rl(t)={10≤t≤T2-1T2≤t≤Twith the same property R-l=0. For horizontal modulation, the force acting on the particle is
(34)f(t)=mω2cosϕ·Rl(t,n),
and its Fourier expansion is
(35)Rl(t)=mω2cosϕ4π×∑k=0∞‍1(2k-1)sin(2k-1)ωt;
the effective potential energy is
(36)Ueff=U+mω2cos2ϕ·14(16π2)2∑k=0∞‍1(2k-1)4=U+0.4112mω2cos2ϕ
which has extrema at ϕ=0,π,±arccos1.2159gl/ω2. Here,(i)
the pointϕ=0 is stable if ω2<1.2159gl,(ii)
the pointϕ=π is not stable,(iii)
the pointϕ=arccos1.2159gl/ω2 is stable if ω2>1.2159gl [13, 14].From (iii), it is observed that, at nontrivial position with the help of this type of external force, the frequency of oscillation has become much lower. At nontrivial position, the above results are summarized in Table 1. From these results, it is also observed that, as the area under the curve increases, the frequency of oscillation decreases, at nontrivial position. The triangular type force has minimum area and so has maximum frequency, while rectangular type force has maximum area and has minimum frequency.Table 1
Stability comparison of periodic piecewise linear forces with harmonic force under horizontal fast oscillation.
Force type
Trivial position
Stability condition
Nontrivial position
Stability condition
Sin
0
ω
2
<
2
g
l
±arccos2gl/ω2
ω
2
>
2
g
l
Triangular
0
ω
2
<
3.03
96
g
l
±arccos3.0396gl/ω2
ω
2
>
3.0396
g
l
Linear (sine)
0
ω
2
<
2.1
72
g
l
±arccos2.172gl/ω2
ω
2
>
2.172
g
l
Trapezium
0
ω
2
<
1.47
36
g
l
±arccos1.4736gl/ω2
ω
2
>
1.4736
g
l
Quadratic
0
ω
2
<
1.29
67
g
l
±arccos1.2967gl/ω2
ω
2
>
1.2967
g
l
Rectangular
0
ω
2
<
1.21
59
g
l
±arccos1.2159gl/ω2
ω
2
>
1.215
9
g
lFigure 9
Rectangular type force.


## 4.1. Triangular Type Force

First of all introduce the triangular type force (see Figure5) given by
(15)Rs(t)={4Ttif0≤t<T4,4T(-t+T2)ifT4≤t<3T4,4T(t-T)if3T4≤t<T.
For (15), the Fourier coefficient a0=0 indicates R-s=0. The Fourier expansion of (14) is
(16)f2(ϕ,t)=mω2cosϕ8π2×∑j=0∞‍(-1)j(2j+1)2sin(2π(2j+1)tT).
With
(17)ak=0,bk=mω2cosϕ8π2(-1)j(2j+1)2,
the effective potential energy is
(18)Ueff=U+mω2cos2ϕ·14(8π2)2∑j=0∞‍1(2j+1)6=U+π260mω2cos2ϕ
which has extrema at ϕ=0,π,±arccos30gl/ω2π2. Minimization of Ueff shows that(i)
the downward pointϕ=0 is stable if ω2<3.0396gl,(ii)
vertically upward pointϕ=π is not  stable,(iii)
the point given bycosϕ=3.0396gl/ω2 is stable if ω2>3.0396gl.From (iii), it is observed that, at nontrivial position, the oscillator is stabilized with higher frequency as compared to harmonic force [13].Figure 5
Triangular type force.Hence, this force is less effective than sin- or cos-type force. So, this force is replaced by some other periodic piecewise linear forces.


## 4.2. Hat Type Force

Now if sine function is traced by linear pulses (see Figure6), defined by (19)(19)Lc(t)={12if0≤t<16T,1if16T≤t<13T,12if13T≤t<12T,-12if12T≤t<23T,-1if23T≤t<56T,-12if56T≤t<T,
Next if the sine force is traced by a linear force forming a hat (see Figure 6), defined by (19). For horizontal modulation, the oscillating force is
(20)f2(ϕ,t)=mω2cosϕ·Lc(t).
Then, by Fourier expansion in place of (19),
(21)ak=0bk=mω2cosϕ1kπ(1-coskπ+2coskπ3).Using the above coefficients, the oscillating force acting on the particle is
(22)f2(t)=mω2cosϕ×∑k=1∞‍1kπ(1-coskπ+2coskπ3)sinkωt,
and the effective potential energy is
(23)Ueff=U+mω2cos2ϕ·14π2∑k=1∞‍1k2(1-coskπ+2coskπ3)2Ueff=U+0.2302mω2π2cos2ϕ
which has extrema at ϕ=0,π,±arccos2.172gl/ω2. Here,(i)
the positionϕ=0 is stable if ω2<2.172gl,(ii)
the inverse positionϕ=π is not stable,(iii)
the positionϕ=arccos2.172gl/ω2 is stable if ω2>2.172gl.Again a less effective result is obtained. So this periodic piecewise linear force is replaced by another one.Figure 6
Hat type force.


## 4.3. Trapezium Type Force

If the sine force is traced by a linear shape forming a trapezium (see Figure7), given by (24)(24)Tm(t)={8tTif0≤t<T8,1ifT8≤t<3T8,8T(T2-t)if3T8≤t<5T8,-1if5T8≤t<7T8,8(t-T)Tif7T8≤t<T,with T-m=0. For horizontal modulation, the force acting on the particle is
(25)f2(ϕ,t)=mω2cosϕ·TM(t,n).
Then, by Fourier expansion in the place of (24),
(26)ak=0,bk=mω2cosϕ16π21k2sinkπ4.
Using the above coefficients, the oscillating force is
(27)f2(ϕ,t)=mω2cosϕ16π2×∑k=0∞‍1k2sinkπ4sinkωt,
and the effective potential energy is
(28)Ueff=U+mω2cos2ϕ·14(16π2)2×∑k=0∞‍1k6sin2kπ4=U+0.3393mω2cos2ϕ
which has extrema at ϕ=0,π,±arccos1.4736gl/ω2. Here,(i)
the pointϕ=0 is stable if ω2<1.4736gl,(ii)
the pointϕ=π is not stable,(iii)
the point given bycosϕ=1.4736gl/ω2 is stable if ω2>1.4736gl.From (iii), it is observed that, at nontrivial position, the oscillator is stabilized with lower frequency as compared to harmonic force. So this type of force is much effective than sin- or cos-type force. Next, modify this trapezium shape force to have a better result.Figure 7
Trapezium type force.


## 4.4. Quadratic Type Force

If slopes are removed in the beginning and at the end from it and define a quadratic type force:Qc(t)=Qc(t+T) (see Figure 8), given by (29)(29)Qc(t)={1if0≤t<3T8,8T(T2-t)if3T8≤t<5T8,-1if5T8≤t<T,with the same property Q-c=0. For horizontal modulation, the oscillating force is
(30)f2(ϕ,t)=mω2cosϕ·Qc(t,n).
The Fourier expansion of (30) is
(31)Qc(t)=mω2cosϕ×∑k=0∞‍(2kπ+8π2k2sinkπ4)sinkωt,and the effective potential energy is
(32)Ueff=U+mω2cos2ϕ·14∑k=0∞‍1k2(2kπ+8π2k2sinkπ4)2=U+0.3856mω2cos2ϕ
which has extrema at ϕ=0,π,±arccos1.2967gl/ω2. Here,(i)
the positionϕ=0 is stable if ω2<1.2967gl,(ii)
the positionϕ=π is not stable,(iii)
the positionϕ=arccos1.2967gl/ω2 is stable if ω2>1.2967gl.Again the frequency of oscillation is lower at nontrivial position.Figure 8
Quadratic type force.


## 4.5. Rectangular Type Force

Now if we introduce rectangular type force:Rl(t)=Rl(t+T) (see Figure 9), given by (33)(33)Rl(t)={10≤t≤T2-1T2≤t≤Twith the same property R-l=0. For horizontal modulation, the force acting on the particle is
(34)f(t)=mω2cosϕ·Rl(t,n),
and its Fourier expansion is
(35)Rl(t)=mω2cosϕ4π×∑k=0∞‍1(2k-1)sin(2k-1)ωt;
the effective potential energy is
(36)Ueff=U+mω2cos2ϕ·14(16π2)2∑k=0∞‍1(2k-1)4=U+0.4112mω2cos2ϕ
which has extrema at ϕ=0,π,±arccos1.2159gl/ω2. Here,(i)
the pointϕ=0 is stable if ω2<1.2159gl,(ii)
the pointϕ=π is not stable,(iii)
the pointϕ=arccos1.2159gl/ω2 is stable if ω2>1.2159gl [13, 14].From (iii), it is observed that, at nontrivial position with the help of this type of external force, the frequency of oscillation has become much lower. At nontrivial position, the above results are summarized in Table 1. From these results, it is also observed that, as the area under the curve increases, the frequency of oscillation decreases, at nontrivial position. The triangular type force has minimum area and so has maximum frequency, while rectangular type force has maximum area and has minimum frequency.Table 1
Stability comparison of periodic piecewise linear forces with harmonic force under horizontal fast oscillation.
Force type
Trivial position
Stability condition
Nontrivial position
Stability condition
Sin
0
ω
2
<
2
g
l
±arccos2gl/ω2
ω
2
>
2
g
l
Triangular
0
ω
2
<
3.03
96
g
l
±arccos3.0396gl/ω2
ω
2
>
3.0396
g
l
Linear (sine)
0
ω
2
<
2.1
72
g
l
±arccos2.172gl/ω2
ω
2
>
2.172
g
l
Trapezium
0
ω
2
<
1.47
36
g
l
±arccos1.4736gl/ω2
ω
2
>
1.4736
g
l
Quadratic
0
ω
2
<
1.29
67
g
l
±arccos1.2967gl/ω2
ω
2
>
1.2967
g
l
Rectangular
0
ω
2
<
1.21
59
g
l
±arccos1.2159gl/ω2
ω
2
>
1.215
9
g
lFigure 9
Rectangular type force.


## 5. Vertical Modulation

For vertical modulation with harmonic force (see Figure10), the fast oscillating force is
(37)f2=mω2sinϕ·sinωt.Figure 10
Kapitza pendulum with vertical oscillation.Here, the positionϕ=0 is always stable, and the inverse point ϕ=π is stable if ω2>2gl (see Figure 11) [5].Figure 11
Stable points for vertical oscillation.Using external periodic piecewise linear forces, (37) takes the form
(38)f=mω2sinϕ·R(t,n),
where  R(t,n) is the external periodic piecewise linear forces. The stability results at ϕ=π are summarized in Table 2.Table 2
Stability comparison of periodic piecewise linear forces with harmonic force under vertical fast oscillation.
Force type
Position
Stability condition
Position
Stability condition
Sin
0
Always
π
ω
2
>
2
g
l
Triangular
0
Always
π
ω
2
>
3.0396
g
l
Linear (sine)
0
Always
π
ω
2
>
2.172
g
l
Trapezium
0
Always
π
ω
2
>
1.4736
g
l
Quadratic
0
Always
π
ω
2
>
1.2967
g
l
Rectangular
0
Always
π
ω
2
>
1.2159
g
l


## 6. Parametric Control

All the above results can be considered as nonparametric control. Next, anϵ-parametric control is defined for one of the periodic piecewise linear forces. At nontrivial position, the frequency of oscillation is calculated. This ϵ-parametric force with 0<ϵ<1 is given by (similar to external force (29))
(39)Rϵ(t)={1if0≤t<1-ϵ2T,1ϵ(-2Tt+1)if1-ϵ2T≤t<1+ϵ2T,-1if1+ϵ2T≤t<T,
and illustrated in Figure 12.Figure 12
ϵ-Parametric quadratic type force.The Fourier coefficienta0=0 indicates R-ϵ=0. For horizontal modulation, the oscillating force acting on the particle is
(40)f2(ϕ,t)=mω2cosϕ·Rϵ(t,n).
With (39), the Fourier coefficients are
(41)ak=0bk=mω2cosϕ(2(kπ)+2ϵk2π2sinϵkπ),
and the oscillating force in fourier expansion is
(42)f2(t)=mω2cosϕ×∑k=1∞‍(2(kπ)+2ϵk2π2sinϵkπ)sinkωt.
The effective potential energy is
(43)Ueff=U+mω2cos2ϕ·14π2∑k=1∞‍4k4(1+1ϵkπsinϵkπ)2=-mglcosϕ+mω2cos2ϕ·b,
where
(44)b=1π2∑k=1∞‍1k4(1+1ϵkπsinϵkπ)2
which has extremum at ϕ=0,π,±arccos0.5gl/ω2b.The stability of the system is discussed under the force with different values ofϵ. See Figure 13. First of all consider ϵ=0.9; the infinite sum is
(45)b=0.1320,
and the effective potential energy is
(46)Ueff=-mglcosϕ+0.132mω2cos2ϕ·
The nontrivial position ±arccos(3.7879gl/ω2) is stable under the condition ω2>3.7879gl. This value is larger than the above considered examples, such a poor result. Next for ϵ=0.8, the infinite sum is 0.1607, and the nontrivial position ±arccos(3.1114gl/ω2) is stable if ω2>3.1114gl, such a better result. Also, it is found that, as ϵ decreases, the infinite sum b increases and the system is stabilized with a relatively low frequency. For different values of ϵ, the results of infinite sum b and the nontrivial position ±arccos0.5gl/ω2b with stable condition are given in Table 3.Table 3
Stability conditions withϵ-parametric force.
ϵ 0<ϵ<1
Sumb
Nontrivial position
Stability condition
0.9
0.1320
±arccos(3.7879gl/ω2)
ω
2
>
3.7879
g
l
0.8
0.1607
±arccos(3.1114gl/ω2)
ω
2
>
3.1114
g
l
0.75
0.1775
±arccos(2.8169gl/ω2)
ω
2
>
2.8169
g
l
0.7
0.1956
±arccos(2.5562gl/ω2)
ω
2
>
2.5562
g
l
0.6
0.2357
±arccos(2.1213gl/ω2)
ω
2
>
2.1213
g
l
0.5
0.2793
±arccos(1.7902gl/ω2)
ω
2
>
1.7902
g
l
0.4
0.3239
±arccos(1.5437gl/ω2)
ω
2
>
1.5437
g
l
0.3
0.3664
±arccos(1.3647gl/ω2)
ω
2
>
1.3647
g
l
0.25
0.3856
±arccos(1.2967gl/ω2)
ω
2
>
1.2967
g
l
0.2
0.4029
±arccos(1.241gl/ω2)
ω
2
>
1.241
g
l
0.1
0.4287
±arccos(1.1663gl/ω2)
ω
2
>
1.1663
g
lFigure 13
Quadratic type force with differentϵ(0.9-0.1).Also asϵ→0, the term b≅0.4386, and the position ±arccos(1.14gl/ω2) is stable under the condition ω2>1.14gl  which is lower than with rectangular type force. Hence, with parametric control, the rectangular type force is approached, and the system is stabilized with a relatively low frequency. The minimization of dimensionless effective potential energy function with horizontal modulation is shown in Figure 14 and with vertical modulation is shown in Figure 15.Horizontal oscillation.
(a)
U
eff is minimum at ϕ=0 if ω2<1.14gl
(b)
U
eff is minimum at cosϕ=1.2159gl/ω2 if ω2>1.14glVertical oscillation.
(a)
U
eff is always minimum at ϕ=0
(b)
U
eff is minimum at ϕ=π if ω2>1.14glHere, the same effect is also observed; asϵ decreases, the area under the curve increases, and the value of b increases; consequently, the frequency of oscillation becomes low at nontrivial position. In this connection, an interesting result is obtained; when ϵ→0, the quadratic type force approaches the rectangular type force, so at nontrivial position the frequency of oscillation should be almost the same, but, with parametric force, the frequency of oscillation is low.Observe from Table 3, the rectangular force fall between ϵ=0.2 and ϵ=0.1, more clearly, the parametric force with ϵ=0.17…, gives the frequency of oscillation almost equals with rectangular type force, hence comparatively less area shows low frequency at non-trivial position.


## 7. Conclusions

Using Kapitza method of averaging for an arbitrary periodic force, the modulated pendulum with periodic piecewise linear force is stabilized with frequencyω  that is sufficiently lesser than that in the case of harmonic modulation. In this contest, rectangular force was found to be the best. But this force is more effective when it has a time-dependent structure. This time-dependent structure is found by defining a parametric control on some other periodic piecewise linear forces. Hence, a more suitable form of rectangular force is found.The parametric control can be applied to control the nontrivial stable position, for horizontally or vertically modulated pendulum.



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*Source: 824701-2013-10-03.xml*