Selected Topics in Physics
a lecture course for 1st year students
by W.B. von Schlippe
Spring Semester 2007
Lecture 2: Oscillations
• simple harmonic oscillations;
• coupled oscillations; beats;
• damped oscillations;
• forced oscillations.
• pendulum
1
2 Oscillations
2.1 Simple Harmonic Oscillations
2
A mass m is attached to a spring as shown in the figure, and slides
without friction on a horizontal surface. Its movement is confined to
one dimension which we choose to be the
x
axis.
The force exerted on the mass is
F = − kx
where
The equation of motion is
k
is the spring constant.
mx = −kx
or with ω2=k/m
x + ω2x = 0
We solve this differential equation by putting
hence
x = eλt
2
2
λt
λ
+
ω
e
(
) =0
3
and since
eλt ≠ 0
we get
λ = ±iω
i.e. we have two solutions.
The general solution of the DEq is a linear superposition of these solutions:
x = c1eiωt + c2 e − iωt
and with Euler’s formula
we get
e ± iα = cos α ± i sin α
x = ( c1 + c2 ) cos ωt + i ( c1 − c2 ) sin ωt
and with
we have finally
1
1
c1 = ( a − ib ) ; c2 = ( a + ib )
2
2
x = a cos ωt + b sin ωt
where a and b are two integration constants to be defined by the
initial conditions.
4
An alternative form of the solution is
x = A cos (ωt + α )
A is the amplitude and α is the initial phase.
The expression
ωt + α
is the (instantaneous) phase of oscillation.
x
x(t)=cos(2t+0.75)
0
0
2
4
6
8
10
t
Shown in the figure is the s.h.o.
x = A cos ( 2t + 0.75 )
5
2.2 Coupled Oscillations
Consider two masses, each connected to a wall through a spring
of spring constant k, coupled to each other by a spring of spring
constant κ and sliding without friction on a horizontal surface.
The displacements of the masses
from their equilibrium positions are
x1 and x2, respectively.
The equations of motion are
mx1 = −kx1 − κ ( x1 − x2 )
(2.1)
mx2 = −kx2 + κ ( x1 − x2 )
(2.2)
To solve these simultaneous DEqs we take their sum and difference:
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(2.1)+(2.2):
m ( x1 + x2 ) = − k ( x1 + x2 )
(2.3)
(2.1)-(2.2):
m ( x1 − x2 ) = − ( k + 2κ )( x1 − x2 )
(2.4)
and then define two new variables:
y = x1 + x2
hence
and
z = x1 − x2
(2.5)
my = −ky
(2.6)
mz = − ( k + 2κ ) z
(2.7)
i.e. the equations (2.1) and (2.2) are decoupled into two s.h.o. equations.
Define:
ω 2 = k m and ω12 = ( k + 2κ ) m
(2.8)
7
then our eqs (2.6) and (2.7) become
y + ω2 y = 0
z + ω12 z = 0
whose solutions are
y = A cos (ωt + α )
z = B cos (ω1t + β )
and going back to the coordinates
x1 and x2
by Eq. (2.5), we get
1
x1 = ⎡⎣ A cos (ωt + α ) + B cos (ω1t + β ) ⎤⎦
2
1
x2 = ⎡⎣ A cos (ωt + α ) − B cos (ω1t + β ) ⎤⎦
2
(2.9)
(2.10)
This completes the derivation of the general solution of the equations
of motion.
8
To discuss particular cases we must specify the initial conditions.
For instance, let us assume that initially, i.e. at time t=0, both masses
are at rest, with the right-hand mass in its equilibrium position and
the left-hand mass displaced to
x1=a:
IC: t = 0, x1 = a, x2 = 0, x1 = x2 = 0
Substituting into (2.9) and (2.10) we get
A cos α + B cos β = 2a
A cos α − B cos β = 0
and
Aω sin α + Bω1 sin β = 0
Aω sin α − Bω1 sin β = 0
( 2.11)
( 2.12 )
( 2.13)
( 2.14 )
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which we can satisfy with α=β=0 (but not with A=0 or B=0: Exercise!)
and hence B=A, thus
1
x1 = A ⎡⎣cos (ωt ) + cos (ω1t ) ⎤⎦
2
1
x2 = A ⎡⎣cos (ωt ) − cos (ω1t ) ⎤⎦
2
or, using a well known trigonometric identity, we get
⎛ ω + ω1 ⎞
⎛ ω1 − ω ⎞
x1 = A cos ⎜
t ⎟ cos ⎜
t⎟
⎝ 2
⎠
⎝ 2
⎠
⎛ ω + ω1 ⎞ ⎛ ω1 − ω ⎞
x2 = A sin ⎜
t ⎟ sin ⎜
t⎟
⎝ 2
⎠ ⎝ 2
⎠
(2.15)
(2.16)
10
An interesting case arises when the coupling is weak, i.e. when the spring
constant κ is much less than
Recall:
k.
ω = k m and ω1 =
( k + 2κ )
hence
ω + ω1
2ω and ω1 − ω
ω
m
κ
k
ω
and hence Eqs. (2.15) and (2.16) become
x1 = A cos (δ t ) cos (ωt )
(2.17)
x2 = A sin (δ t ) sin (ωt )
(2.18)
Here the rapid oscillation with circular frequency ω is modulated by
the slow oscillation of frequency δ=ωκ/2k.
It is usual to consider the expression
slowly varying amplitude.
A cos (δ t )
to be a
11
The functions (2.17) and (2.18) are shown in the next figure.
The initial energy of oscillator 1, which was residing in the springs, is
transferred to oscillator 2. At the first node of oscillator 1 all energy
has been transferred to oscillator 2, then the process is reversed.
This phenomenon is called beats.
12
2.3 Damped Oscillations
Realistically, there is always some amount of damping resistance present
in every oscillation. In a system like the one of 2.1, the resistance is
the friction between the mass and the supporting surface.
Intuitively we say that the resistance depends on the speed of the mass
but not on its position. At zero velocity the resistance must vanish, therefore
it is reasonable to set
R ( x ) = R′ ( 0 ) x + O ( x 2 )
and for this to be a resistance we must have
R′ ( 0 ) < 0
hence, neglecting the terms of order x2, we get the equation of motion
of the damped harmonic oscillator in the form of
x + 2 β x + ω02 x = 0
with
2 β = − R′ ( 0 ) m and ω02 = k m
(2.17)
13
Equation (2.17) is a linear differential equation with constant coefficients.
To solve it, we make the substitution (Ansatz):
hence
but
(λ
2
x ( t ) = eλt
+ 2 βλ + ω02 ) eλt = 0
eλt ≠ 0
and therefore we get the characteristic equation
λ 2 + 2βλ + ω02 = 0
(2.18)
λ1,2 = − β ± β 2 − ω02
(2.19)
whose two roots are
The general solution of DEq. (2.17) is therefore
x ( t ) = c1eλ1t + c2 eλ2t
(2.20)
14
Now we must distinguish three cases:
(i )
( ii )
( iii )
undercritical damping: β 2 − ω02 < 0
critical damping: β 2 − ω02 = 0
overcritical damping: β 2 − ω02 > 0
( i ) define: ω 2 = ω02 − β 2
hence
λ1,2 = − β ± iω
x ( t ) = c1e( − β +iω )t + c2 e( − β −iω )t = e − β t ( c1eiωt + c2 e− iωt )
Let c = c1 and c2 = c* ( c* is the complex conjugate of c )
hence
or putting
we get
x ( t ) = e − β t ( ceiωt + c*e − iωt ) = 2e − β t Re ( ceiωt )
c=
1 iα
Ae
2
with real
A and α
x ( t ) = Ae − β t cos (ωt + α )
where A and α are integration constants.
15
Example of undercritically damped oscillation:
A = 1,
ω = 3,
β = 0.2,
α = −0.8
e −0.2t
x
0
0
2
4
6
8
10
t
16
( ii )
critical damping: β 2 − ω02 = 0
the problem here is that we do not have two linearly independent solutions:
x2 = x1
So leave
x1
at the value that comes out of the calculation:
x1 = e − β t
But to construct the second, linearly independent solution, use the
following:
define: Λ = β 2 − ω02 > 0
and put
(Λ → 0
at the end )
1 Λt −Λt
e − e ) = te− β t
(
Λ→ 0 2Λ
x2 = e − β t lim
and then the general solution takes on the form
x ( t ) = ( c1 + c2t ) e − β t
17
x = (1 + t ) e −0.08t
Example of critically damped system:
6
x
4
2
0
0
20
40
60
80
100
t
Critical damping is of interest if one wants to make the system settle
down as fast as possible after a disturbance, such as in the wheel
suspensions of vehicles and in many measuring instruments, e.g.
in ballistic galvanometers.
18
( iii )
overcritical damping: β 2 − ω02 > 0
define: Λ 2 = β 2 − ω02 > 0
hence
( Λ < β !)
x1 = e( − β +Λ ) t ;
x2 = e( − β −Λ ) t
and hence the general solution:
x ( t ) = e − β t ( c1eΛ t + c2 e −Λ t )
and for sufficiently large times
decaying displacement:
x ( t ) Λt
t
we get an exponentially
−( β −Λ ) t
∼
c
e
1
1
19
Example of overcritically damped system:
at t = 0 : x = 0 and x = v > 0
IC:
c1 =
hence
thus
v
, c2 = −c1
2Λ
v − β t Λt −Λt
v −βt
x (t ) =
e ( e − e ) = e sinh Λt
2Λ
Λ
1.0
0.8
x
x=5e
-0.5t
sinh(0.2t)
0.6
0.4
0.2
0.0
0
2
4
6
t
8
10
20
2.4 Forced Oscillations
Oscillating system with sinusoidal forcing term:
x + 2 β x + ω02 x = f cos Ω t
(2.21)
It is convenient to write down a second, similar equation:
y + 2 β y + ω02 y = f sin Ω t
(2.22)
If we multiply Eq. (2.22) by i, then add it to Eq. (2.21), define the new variable
z = x + iy
and use Euler’s formula
cos Ω t + i sin Ω t = eiΩ t
then we get the following DEq.:
z + 2β z + ω02 z = feiΩ t
To solve this DEq we make the
hence
(2.23)
Ansatz: z = AeiΩ t
A ( −Ω 2 + 2i βΩ + ω02 ) eiΩ t = feiΩ t
21
and after cancellation of the exponential factors we get
A=
and with
A = A eiα
f
ω02 − Ω 2 + 2i βΩ
we get the modulus of the amplitude and the phase:
A=
f
(ω
2
0
tan α =
−Ω
)
2 2
+ 4β Ω
2
Im A
2βΩ
= 2
Re A Ω − ω02
(2.24)
2
(2.25)
22
The general solution of DEq (2.23) is
z = AeiΩ t + e − β t ( a cos ω t + b sin ω t )
where
For
zt = e − β t ( a cos ω t + b sin ω t ) is the transient oscillation
βt
1
we get the steady state solution:
z = AeiΩ t = A ei( Ω t +α )
and hence the steady state solution of the original DEq (2.21):
x = Re z = A cos ( Ω t + α )
where the amplitude and phase are given by Eqs (2.24) and (2.25).
The amplitude is shown in the next figure as function of the forcing
frequency Ω.
23
Amplitude of forced oscillation as function of the forcing frequency Ω
β=0.1
4
|A|
β=0.15
2
β=0.3
β=1
0
0
2
Ω
4
We see that the amplitude has a peak which gets more pronounced
at smaller resistance. Such a behaviour is called a resonance.
The position of the maximum of |A| is the resonance frequency;
it is given by
Ω R = ω02 − 2β 2
(2.26)
24
and the height of the resonance peak is given by
AR =
f
2β ω − β
2
0
2
(2.27)
By Eq. (2.26) reality of the resonance frequency implies
β ≤ ω0
2
and therefore also |A|R is seen to be real for all allowed values of β.
There is no resonance if β≥ω0/√2
For β→0, i.e. in the absence of damping, the amplitude tends to infinity
as we approach resonance. In practice the system breaks up before
resonance is reached.
Of interest is also the behaviour of the phase shift α as a function of
the driving frequency.
25
Phase shift α as function of the driving frequency Ω for ω0=1
Note that α is passing through π/2 at Ω=ω0
β=0.1
0.15
β=0.3
0.5
26
2.5 Pendulum
A pendulum consists of a mass m attached
to one end of a weightless rigid rod which is
suspended without friction from a rigid support
Two forces act on m: its weight and the
string tension (not shown in the drawing),
which is equal in magnitude and opposite
in direction to the component of the weight
in the direction of the string; therefore we
have
mr = W + T
with
and
(
(2.28)
)
T = − W ⋅ rˆ rˆ
W = mgxˆ where xˆ is the unit vector in x direction
Now take the cross product of (2.28) with the vector
r
27
On the left-hand side we have
(
)
d
r ×r =
r × r = l 2ϕ nˆ
dt
and on the right-hand side we have
(
)
r × W + T = r × W = − mgl sin ϕ nˆ
since the vectors
r and T
are collinear and therefore their
cross product is equal to nought.
l
n̂
is the length of the pendulum rod and
is a unit vector perpendicular to
r and W
Thus the equation of motion takes on the form of
or with
ω2 = g l
ϕ = − ( g l ) sin ϕ
ϕ + ω 2 sin ϕ = 0
(2.29)
28
Equation (2.29) is a nonlinear second order differential equation. The standard
way of solving it is slightly artificial: we begin by multiplying the equation by
ϕ
and we note that
hence
and hence
ϕ ϕ + ω 2ϕ sin ϕ = 0
d ⎛ϕ2 ⎞
ϕϕ = ⎜ ⎟
and
dt ⎝ 2 ⎠
d
ϕ sin ϕ = − cos ϕ
dt
⎞
d ⎛ϕ2
2
⎜ − ω cos ϕ ⎟ = 0
dt ⎝ 2
⎠
ϕ2
2
− ω 2 cos ϕ = C
where C is the integration constant. We can fix C by putting φ to the
maximum angle φ0, which we call the amplitude. For φ=φ0 the angular
velocity is zero, and hence
C = −ω 2 cos ϕ0
29
and hence we get
or
and hence
ϕ2
2
= ω 2 ( cos ϕ − cos ϕ0 )
ϕ = ± 2ω 2 ( cos ϕ − cos ϕ0 )
dϕ
dt = ±
2ω 2 ( cos ϕ − cos ϕ0 )
Now the period of oscillation T is the time of a full swing from –φ0 to φ0
and back. Therefore, making also use of the symmetry of the system, we get
T=
4
ω
ϕ0
∫
0
dϕ
2 ( cos ϕ − cos ϕ0 )
This integral belongs to the class of elliptic integrals. They cannot be
expressed in terms of elementary functions. The standard way of
handling such an integral is the following.
30
First we apply the identity
cos ϕ = 1 − 2sin
hence
T=
4
ω
0
ϕ
2
d (ϕ 2 )
ϕ0
∫
2
sin 2 (ϕ0 2 ) − sin 2 (ϕ 2 )
and then we make the substitution
sin
ϕ
2
= sin
ϕ0
2
sin u
and we note that u=0 for φ=0 and u=π/2 for φ=φ0, hence
T=
where
4
ω
π 2
∫
m = sin 2 (ϕ0 2 )
0
du
1 − m sin u
2
(2.30)
is the parameter of the elliptic integral.
31
For amplitudes small enough for m to be negligibly small we get
T=
2π
ω
= 2π
l
g
which is the well known result that we get more easily if we approximate
sin φ by φ in Eq. (2.29). But the work done to get Eq. (2.30) pays when
we want to apply our result to large amplitudes. Then we can proceed as
follows: we note that msin2u < 1 for φ0 < π, and hence we expand:
2
1
1⋅ 3
2
2
= 1 + m sin u +
m sin u ) + …
(
2
2
2⋅4
1 − m sin u
1
and we get a first correction if we neglect terms of order m2:
with
⎛ 1 2 ϕ0 ⎞
T = T0 ⎜ 1 + sin
⎟
4
2
⎝
⎠
T0 = 2π l g
32
For an amplitude of φ0 = 6 degrees we have
1 2 ϕ0
≅ 0.0007
sin
4
2
and we see that neglecting this correction is justified. But for large
amplitudes the power series is inconvenient since one needs to carry
a large number of terms. Then it is simpler to do the integration numerically
to any required accuracy. The results are shown in the next figure:
pendulum
In the figure the ratio
T / T0 is shown as a
function of the amplitude
φ0. One can see that
even at an amplitude of
90 degrees the period
of oscillation is increased
by less than 20%.
1.2
T/T0
1.15
1.1
1.05
1
0
30
60
amplitude (deg)
90
33
Problems for Lecture 1:
1.) Show that the speed of a satellite in a circular orbit about the Earth
is given by
v = RE g r
where RE =6378 km is the radius of the Earth (to be assumed spherical),
g=9.8 ms-2 is the acceleration due to gravity and r is the radius of the orbit.
Find the speed of the International Space Station, taking its altitude to be
330 km (the actual orbit of the ISS is not perfectly circular but nearly so).
2.) Find the radius of the geostationary orbit about the Earth.
(Answer: 4.22x107 m)
34