All 7 Chapters Covered
SOLUTIONS
,
, Contents
1 Free Oscillations of a Linear Oscillator 5
1.2 Review of the Principal Formulas .................................................. 5
1.3 Questions and Problems with Answers and Solutions.................. 6
1.3.1 Free Undamped Oscillations ............................................... 6
1.3.2 Damped Free Oscillations ................................................. 11
1.3.3 Non-oscillatory Motion of the System ............................. 15
2 Torsion Spring Oscillator with Dry Friction 21
2.2 Review of the Principal Formulas ................................................ 21
2.3 Questions and Problems with Answers and Solutions................ 22
2.3.1 Damping Caused by Dry Friction ................................... 22
2.3.2 Influence of Viscous Friction............................................ 26
3 Forced Oscillations in a Linear System 31
3.4 Review of the Principal Formulas ................................................ 31
3.5 Questions and Problems with Answers and Solutions................ 32
3.5.1 Steady-state Forced Oscillations........................................ 32
3.5.2 Transient Processes ........................................................... 44
4 Square-wave Excitation of a Linear Oscillator 59
4.8 Review of the Principal Formulas ................................................ 59
4.9 Questions and Problems with Answers and Solutions................ 60
4.9.1 Swinging of the Oscillator at Resonance ......................... 60
4.9.2 Non-resonant Forced Oscillations..................................... 69
5 Parametric Excitation of Oscillations 75
5.4 Questions and Problems with Answers and Solutions................ 75
5.4.1 Principal Parametric Resonance ....................................... 75
5.4.2 Manual Control of the Parameter ..................................... 90
5.4.3 Parametric Resonances of High Orders ........................... 91
3
K25515_SM_Cover.indd 5 16/12/
,4 CONTENTS
6 Sinusoidal Modulation of the Parameter 103
6.4 Questions and Problems with Answers and Solutions .............. 103
6.4.1 Principal Parametric Resonance...................................... 103
6.4.2 The Principal Interval of Parametric Resonance ............ 109
6.4.3 The Second Parametric Resonance.................................. 113
7 Free Oscillations of the Rigid Pendulum 115
7.5 Review of the Principal Formulas .............................................. 115
7.6 Questions and Problems with Answers and Solutions .............. 116
7.6.1 Small Oscillations of the Pendulum ............................... 116
7.6.2 Oscillations with Large Amplitudes .............................. 121
7.6.3 The Rotating Pendulum .................................................. 133
@LECTJULIESOLUTIONSSTUVIA
, Chapter 1
Free Oscillations of a
Linear Oscillator
1.2 Review of the Principal Formulas
The differential equation of a free linear torsion oscillator:
ϕ¨ + 2γϕ˙ +0ω2ϕ = 0. (1.1)
The frequency and the period of free oscillations without friction (at γ
≪ ω0): √
D 2π
ω0 = , T0 = . (1.2)
J ω0
An oscillatory solution (valid at γ < ω0):
ϕ(t) = A0 e−γt cos(ω 1t + δ0), (1.3)
where the constants A0 and δ0 are determined by the initial conditions ϕ(0), ϕ˙(0).
The frequency ω1 of damped oscillations
√
ω1 = ω 20 − γ2. (1.4)
An equivalent form of the general solution:
ϕ(t)
− =e γt(C cos ω1t + S sin ω1t), (1.5)
where the constants C and S are determined by the initial conditions.
They are related to A0 and δ0:
√
A0 = C2 + S2, tan δ0 = −S/C. (1.6)
5
K25515_SM_Cover.indd 7 16/12/
, 6 CHAPTER 1. FREE OSCILLATIONS OF A LINEAR
OSCILLATOR
In the case of weak damping (γ ≪ ω0)
ω1 ≈ ω0 − γ2 /(2ω0). (1.7)
The decay time (during which the amplitude is reduced by the factor e ≈
2.72):
τ = 1/γ. (1.8)
A non-oscillatory motion at γ = ω0:
ϕ(t) = (C1 t + C
2 )e
−γt. (1.9)
The quality factor Q of an oscillator:
τ ω0
Q=π = . (1.10)
T 2γ
0
The number of oscillations, during which the amplitude is halved:
ln 2 Q
N1/2 = Q = 0.22 Q = . (1.11)
π 4.53
The total mechanical energy of the oscillator consists of elastic potential
energy of the strained spring and kinetic energy of the flywheel:
E=E +E =1 2 1 2 (1.12)
pot kin Dϕ + Jϕ˙ .
2 2
The values of the potential energy and kinetic energy of the oscillator,
averaged over a cycle, equal one another, each of them constituting one half
the total energy:
⟨E ⟩ = ⟨E ⟩ = 1 E =1 DA2 =
1 Jω2A2. (1.13)
pot kin 0 0 0
2 4 4
1.3 Questions and Problems with Answers
and Solutions
1.3.1 Free Undamped Oscillations
1.3.1.1 The Initial Conditions and the Shape of the Plots. In the
absence of friction a linear oscillator executes simple harmonic motion,
which is charac- terized by purely sinusoidal time dependence of the
angular displacement and of the angular velocity.
(a) What initial conditions give rise to oscillations of cosine time
dependence, of sine time dependence? Suppose that you want to get
oscillations with the angu- lar amplitude of 90◦. What initial angular
displacement ϕ(0) = ϕ0 at zero initial angular velocity ϕ˙(0) = 0 ensures
the desired amplitude?
@LECTJULIESOLUTIONSSTUVIA
,K25515_SM_Cover.indd 9 16/12/
,1.3. PROBLEMS WITH ANSWERS AND SOLUTIONS 7
Free oscillations with a cosine time dependence of ϕ(t) = ϕ0 cos ω0t
occur if we set some arbitrary nonzero initial displacement ϕ(0) =
ϕ0, and the initial velocity zero: ϕ˙(0) = 0. To ensure an amplitude
of 90◦ in this case, the initial deflection ϕ0 must be ±90◦ (±π/2
rad). On the other hand, if we set a (nonzero) initial angular
velocity (ϕ˙(0) = Ω) and the initial displacement zero (ϕ(0) =
0), we obtain subsequent natural oscillations that are described
by a sine function: ϕ(t) = (Ω/ω0) sin ω0t.
(b) What initial angular velocity ϕ̇(0) = Ω ought you to impart to
the oscil- lator, at rest in the equilibrium position, in order to obtain the
same amplitude of 90◦? Remember, that the initial angular velocity Ω must
be expressed for input in units of the natural frequency ω0. Verify your
answer with a computer experiment, using the appropriate initial
conditions.
If the oscillator is excited by an initial velocity (ϕ˙(0) = Ω)
at the initial displacement zero (ϕ(0) = 0), the amplitude of
resulting os- cillations equals |Ω|/ω0. To ensure oscillations
whose amplitude is 90◦ (π/2 rad), we need to input Ω =
±(π/2)ω0 (±1.57 in units of the natural frequency ω0 ).
1.3.1.2 Maximal Deflection and Conservation of Energy. Imagine
exciting an oscillator initially at rest in the equilibrium position by a push
which produces an initial angular velocity Ω = 2ω0.
(a) Calculate the angle ϕmax of maximal deflection using the law of the
con- servation of energy.
To obtain this result from the law of energy conservation, we
should set equal the initial kinetic energy Ekin = 1 JΩ2 to
the potential 1
energy
E =
Dϕ2 2 . Thus we
pot 2 max at the maximal deflection ϕmax
find the same value √
f o r the maximal deflection from the equilibrium
position: ϕmax = Ω J/D = Ω/ω0 = 114.6◦.
(b) Verify your result experimentally. Note that the simulation
program per- forms the numerical integration of the differential equation
independently of con- servation laws, such as the conservation of
energy. That is, these laws are not used in the program. If we set the
initial angular velocity Ω = 2ω0, and the initial displacement zero, then the
amplitude (and the maximal deflection ϕmax from the equilibrium
position) equals Ω/ω0 = 2 radians, or 360◦ /π = 114.6◦.
1.3.1.3 The Phase Trajectory and the Initial Conditions. Compare the
motion of the representative point along the phase trajectory of a
conservative os- cillator with the time-dependent plots of the angle of
deflection and of the angular velocity.
(a) How is the phase trajectory changed if you change the initial conditions?
@LECTJULIESOLUTIONSSTUVIA
,K25515_SM_Cover.indd 11 16/12/
, 8 CHAPTER 1. FREE OSCILLATIONS OF A LINEAR
OSCILLATOR
When we change the initial conditions, we change the initial
mechan- ical state of the system. In the phase plane, changing
the initial con- ditions means changing the point from which
the phase trajectory originates. Since for the system with given
properties only one phase trajectory passes through a given
point of the phase plane, the further motion of the
representative point is defined uniquely.
(b) Does the direction of the motion of the representative point along the
phase trajectory depend on the initial conditions?
The representative point always moves clockwise along a closed
phase trajectory of a conservative linear oscillator,
independently of the ini- tial conditions.
(c) Is it possible that phase trajectories for different initial conditions
coincide? If so, formulate the requirements for the coincidence.
For a conservative system, phase trajectories corresponding to
differ- ent initial conditions coincide if these initial conditions
correspond to the same total energy. However, the
representative point starts its motion at different points of this
phase trajectory depending on the initial conditions.
1.3.1.4 Elliptical and Circular Shape of the Phase Trajectory.
(a) Prove analytically that the phase trajectory of a conservative linear
oscil- lator is an ellipse with its center at the origin of the phase plane.
What are the semiaxes of the ellipse?
To obtain the explicit equation of the phase trajectory, i.e., the
depen- dence of ϕ˙ on ϕ for a given motion, we can combine the
expressions for ϕ(t) and ϕ˙(t) in the general solution of the
differential equation of motion in order to eliminate time t.
Since sin2 α + cos2 α = 1, we
obtain:
ϕ2 ϕ˙2
+ = 1. (1.14)
A02 A02ω0 2
Thus the phase trajectory of a conservative linear oscillator is
an el- lipse whose center is located at the origin of the phase
plane and whose semiaxes are A0 and A0ω0 respectively.
(b) Show that the elliptical shape of the phase diagram of a conservative
linear oscillator follows immediately from the law of the conservation of
the energy.
For a conservative system, the relation between ϕ˙ and ϕ that
gives the equation of the phase trajectory, follows directly
from the law of energy conservation. This means that actually
we don’t need to solve the differential equation of motion in
order to get the equation of the
@LECTJULIESOLUTIONSSTUVIA
SOLUTIONS
,
, Contents
1 Free Oscillations of a Linear Oscillator 5
1.2 Review of the Principal Formulas .................................................. 5
1.3 Questions and Problems with Answers and Solutions.................. 6
1.3.1 Free Undamped Oscillations ............................................... 6
1.3.2 Damped Free Oscillations ................................................. 11
1.3.3 Non-oscillatory Motion of the System ............................. 15
2 Torsion Spring Oscillator with Dry Friction 21
2.2 Review of the Principal Formulas ................................................ 21
2.3 Questions and Problems with Answers and Solutions................ 22
2.3.1 Damping Caused by Dry Friction ................................... 22
2.3.2 Influence of Viscous Friction............................................ 26
3 Forced Oscillations in a Linear System 31
3.4 Review of the Principal Formulas ................................................ 31
3.5 Questions and Problems with Answers and Solutions................ 32
3.5.1 Steady-state Forced Oscillations........................................ 32
3.5.2 Transient Processes ........................................................... 44
4 Square-wave Excitation of a Linear Oscillator 59
4.8 Review of the Principal Formulas ................................................ 59
4.9 Questions and Problems with Answers and Solutions................ 60
4.9.1 Swinging of the Oscillator at Resonance ......................... 60
4.9.2 Non-resonant Forced Oscillations..................................... 69
5 Parametric Excitation of Oscillations 75
5.4 Questions and Problems with Answers and Solutions................ 75
5.4.1 Principal Parametric Resonance ....................................... 75
5.4.2 Manual Control of the Parameter ..................................... 90
5.4.3 Parametric Resonances of High Orders ........................... 91
3
K25515_SM_Cover.indd 5 16/12/
,4 CONTENTS
6 Sinusoidal Modulation of the Parameter 103
6.4 Questions and Problems with Answers and Solutions .............. 103
6.4.1 Principal Parametric Resonance...................................... 103
6.4.2 The Principal Interval of Parametric Resonance ............ 109
6.4.3 The Second Parametric Resonance.................................. 113
7 Free Oscillations of the Rigid Pendulum 115
7.5 Review of the Principal Formulas .............................................. 115
7.6 Questions and Problems with Answers and Solutions .............. 116
7.6.1 Small Oscillations of the Pendulum ............................... 116
7.6.2 Oscillations with Large Amplitudes .............................. 121
7.6.3 The Rotating Pendulum .................................................. 133
@LECTJULIESOLUTIONSSTUVIA
, Chapter 1
Free Oscillations of a
Linear Oscillator
1.2 Review of the Principal Formulas
The differential equation of a free linear torsion oscillator:
ϕ¨ + 2γϕ˙ +0ω2ϕ = 0. (1.1)
The frequency and the period of free oscillations without friction (at γ
≪ ω0): √
D 2π
ω0 = , T0 = . (1.2)
J ω0
An oscillatory solution (valid at γ < ω0):
ϕ(t) = A0 e−γt cos(ω 1t + δ0), (1.3)
where the constants A0 and δ0 are determined by the initial conditions ϕ(0), ϕ˙(0).
The frequency ω1 of damped oscillations
√
ω1 = ω 20 − γ2. (1.4)
An equivalent form of the general solution:
ϕ(t)
− =e γt(C cos ω1t + S sin ω1t), (1.5)
where the constants C and S are determined by the initial conditions.
They are related to A0 and δ0:
√
A0 = C2 + S2, tan δ0 = −S/C. (1.6)
5
K25515_SM_Cover.indd 7 16/12/
, 6 CHAPTER 1. FREE OSCILLATIONS OF A LINEAR
OSCILLATOR
In the case of weak damping (γ ≪ ω0)
ω1 ≈ ω0 − γ2 /(2ω0). (1.7)
The decay time (during which the amplitude is reduced by the factor e ≈
2.72):
τ = 1/γ. (1.8)
A non-oscillatory motion at γ = ω0:
ϕ(t) = (C1 t + C
2 )e
−γt. (1.9)
The quality factor Q of an oscillator:
τ ω0
Q=π = . (1.10)
T 2γ
0
The number of oscillations, during which the amplitude is halved:
ln 2 Q
N1/2 = Q = 0.22 Q = . (1.11)
π 4.53
The total mechanical energy of the oscillator consists of elastic potential
energy of the strained spring and kinetic energy of the flywheel:
E=E +E =1 2 1 2 (1.12)
pot kin Dϕ + Jϕ˙ .
2 2
The values of the potential energy and kinetic energy of the oscillator,
averaged over a cycle, equal one another, each of them constituting one half
the total energy:
⟨E ⟩ = ⟨E ⟩ = 1 E =1 DA2 =
1 Jω2A2. (1.13)
pot kin 0 0 0
2 4 4
1.3 Questions and Problems with Answers
and Solutions
1.3.1 Free Undamped Oscillations
1.3.1.1 The Initial Conditions and the Shape of the Plots. In the
absence of friction a linear oscillator executes simple harmonic motion,
which is charac- terized by purely sinusoidal time dependence of the
angular displacement and of the angular velocity.
(a) What initial conditions give rise to oscillations of cosine time
dependence, of sine time dependence? Suppose that you want to get
oscillations with the angu- lar amplitude of 90◦. What initial angular
displacement ϕ(0) = ϕ0 at zero initial angular velocity ϕ˙(0) = 0 ensures
the desired amplitude?
@LECTJULIESOLUTIONSSTUVIA
,K25515_SM_Cover.indd 9 16/12/
,1.3. PROBLEMS WITH ANSWERS AND SOLUTIONS 7
Free oscillations with a cosine time dependence of ϕ(t) = ϕ0 cos ω0t
occur if we set some arbitrary nonzero initial displacement ϕ(0) =
ϕ0, and the initial velocity zero: ϕ˙(0) = 0. To ensure an amplitude
of 90◦ in this case, the initial deflection ϕ0 must be ±90◦ (±π/2
rad). On the other hand, if we set a (nonzero) initial angular
velocity (ϕ˙(0) = Ω) and the initial displacement zero (ϕ(0) =
0), we obtain subsequent natural oscillations that are described
by a sine function: ϕ(t) = (Ω/ω0) sin ω0t.
(b) What initial angular velocity ϕ̇(0) = Ω ought you to impart to
the oscil- lator, at rest in the equilibrium position, in order to obtain the
same amplitude of 90◦? Remember, that the initial angular velocity Ω must
be expressed for input in units of the natural frequency ω0. Verify your
answer with a computer experiment, using the appropriate initial
conditions.
If the oscillator is excited by an initial velocity (ϕ˙(0) = Ω)
at the initial displacement zero (ϕ(0) = 0), the amplitude of
resulting os- cillations equals |Ω|/ω0. To ensure oscillations
whose amplitude is 90◦ (π/2 rad), we need to input Ω =
±(π/2)ω0 (±1.57 in units of the natural frequency ω0 ).
1.3.1.2 Maximal Deflection and Conservation of Energy. Imagine
exciting an oscillator initially at rest in the equilibrium position by a push
which produces an initial angular velocity Ω = 2ω0.
(a) Calculate the angle ϕmax of maximal deflection using the law of the
con- servation of energy.
To obtain this result from the law of energy conservation, we
should set equal the initial kinetic energy Ekin = 1 JΩ2 to
the potential 1
energy
E =
Dϕ2 2 . Thus we
pot 2 max at the maximal deflection ϕmax
find the same value √
f o r the maximal deflection from the equilibrium
position: ϕmax = Ω J/D = Ω/ω0 = 114.6◦.
(b) Verify your result experimentally. Note that the simulation
program per- forms the numerical integration of the differential equation
independently of con- servation laws, such as the conservation of
energy. That is, these laws are not used in the program. If we set the
initial angular velocity Ω = 2ω0, and the initial displacement zero, then the
amplitude (and the maximal deflection ϕmax from the equilibrium
position) equals Ω/ω0 = 2 radians, or 360◦ /π = 114.6◦.
1.3.1.3 The Phase Trajectory and the Initial Conditions. Compare the
motion of the representative point along the phase trajectory of a
conservative os- cillator with the time-dependent plots of the angle of
deflection and of the angular velocity.
(a) How is the phase trajectory changed if you change the initial conditions?
@LECTJULIESOLUTIONSSTUVIA
,K25515_SM_Cover.indd 11 16/12/
, 8 CHAPTER 1. FREE OSCILLATIONS OF A LINEAR
OSCILLATOR
When we change the initial conditions, we change the initial
mechan- ical state of the system. In the phase plane, changing
the initial con- ditions means changing the point from which
the phase trajectory originates. Since for the system with given
properties only one phase trajectory passes through a given
point of the phase plane, the further motion of the
representative point is defined uniquely.
(b) Does the direction of the motion of the representative point along the
phase trajectory depend on the initial conditions?
The representative point always moves clockwise along a closed
phase trajectory of a conservative linear oscillator,
independently of the ini- tial conditions.
(c) Is it possible that phase trajectories for different initial conditions
coincide? If so, formulate the requirements for the coincidence.
For a conservative system, phase trajectories corresponding to
differ- ent initial conditions coincide if these initial conditions
correspond to the same total energy. However, the
representative point starts its motion at different points of this
phase trajectory depending on the initial conditions.
1.3.1.4 Elliptical and Circular Shape of the Phase Trajectory.
(a) Prove analytically that the phase trajectory of a conservative linear
oscil- lator is an ellipse with its center at the origin of the phase plane.
What are the semiaxes of the ellipse?
To obtain the explicit equation of the phase trajectory, i.e., the
depen- dence of ϕ˙ on ϕ for a given motion, we can combine the
expressions for ϕ(t) and ϕ˙(t) in the general solution of the
differential equation of motion in order to eliminate time t.
Since sin2 α + cos2 α = 1, we
obtain:
ϕ2 ϕ˙2
+ = 1. (1.14)
A02 A02ω0 2
Thus the phase trajectory of a conservative linear oscillator is
an el- lipse whose center is located at the origin of the phase
plane and whose semiaxes are A0 and A0ω0 respectively.
(b) Show that the elliptical shape of the phase diagram of a conservative
linear oscillator follows immediately from the law of the conservation of
the energy.
For a conservative system, the relation between ϕ˙ and ϕ that
gives the equation of the phase trajectory, follows directly
from the law of energy conservation. This means that actually
we don’t need to solve the differential equation of motion in
order to get the equation of the
@LECTJULIESOLUTIONSSTUVIA
