I.
INTRODUCTION
The natural tendency of a particle or a system is to attain the position of minimum
potential energy or position of stable equilibrium. A particle or a system, when disturbed
from its position of minimum energy, tend to vibrate or oscillate about this mean position.
This oscillatory or periodic behaviour of a particle/system may damp out after a certain
time or can be sustained by application of a certain force. In this chapter we will deal with
analytical treatment of periodic motion under ideal, dammped and forced conditions.
• Waves and oscillation constitute one of the most important subjects of study in all
physics as virtually every system possesses this capability.
• Most systems will vibrate freely in large variety of ways.
• It is likely that small objects vibrate more rapidly and large objects slowly. A mosquito
wing vibrates hundred times per second whereas, after an earthquake the whole earth
vibrate once in a minute or two minutes.
• All these phenomena has one thing in common, namely periodicity. The pattern of
movement or displacement repeats itself over and over again.
• The pattern can be complicated like a heartbeat or it can be simple like the nearly
perfect sine curve of vibrations of a tuning fork.
II.
SIMPLE HARMONIC OSCILLATION
Let us start with an example. We will consider the ideal spring mass system with spring
force (restoring force) Fs = −k x, where, k is the spring constant and x is the displacement
of the spring from its equilibrium position. The equation of motion of such a system can be
written as
d2 x
= Fs ,
dt2
d2 x
m 2 + kx = 0,
dt
2
dx
k
+ x = 0,
2
dt
m
d2 x
+ ω02 x = 0 ,
dt2
m
1
(1)
where ω0 =
q
k
m
is the natural frequency of the oscillator. Eq. 1 is a second order linear,
homogeneous differential equation with constant coefficients. The solution of Eq. 1 is of the
form
x = ep t ,
(2)
where p can be determined from Eq. 1. Now putting Eq. 2 in Eq. 1, we get
p2 ep t + ω02 ep t = 0 ,
p2 + ω02 = 0 ,
p = ±i ω0 .
(3)
Now we see that x = ei ω0 t and x = e−i ω0 t are both solution of Eq. 1. So the most general
solution will be linear combination of both the solutions, i.e, the most general solution of
Eq. 1 can be written as
x(t) = C1 ei ω0 t + C2 e−i ω0 t ,
(4)
where C1 and C2 can be determined from given initial conditions. We, however, need to
know two initial conditions to determine the values of C1 and C2 . Let us examine few special
cases.
2
A.
Case I
Let us assume that the mass is pulled to one side by a distance a0 and released from rest
at t = 0. In this case, the two initial conditions are x(0) = a0 and
dx
(0)
dt
= 0. At any given
time t, the solution of the simple harmonic oscillator is given by Eq. 4. By differentiating
Eq. 4 with respect to t we can obtain
dx
,
dt
i.e,
dx
= C1 (i ω0 ) ei ω0 t + C2 (−i ω0 ) e−i ω0 t ,
dt
dx
= i ω0 C1 ei ω0 t − C2 e−i ω0 t .
dt
(5)
By putting t = 0 in Eq. 4 and Eq. 5 and using the initial conditions, we get
C 1 − C2 = 0 .
C 1 + C 2 = a0 ,
(6)
By solving these two equations in C1 and C2 , we get C1 = C2 = a0 /2. So the most general
solution can be written as
a0 i ω 0 t
−i ω0 t
x(t) =
e
+e
,
2
x(t) = a0 cos ω0 t
B.
(7)
Case II
Let us assume that the mass is hit and given a speed v0 at its equilibrium position at
t = 0. In this case, the initial conditions are x(0) = 0 and
dx
(0)
dt
= v0 . Now using Eq. 4 and
Eq. 5, we get
i ω0 (C1 − C2 ) = v0 .
C1 + C2 = 0 ,
(8)
By solving these two equations, we get
C1 =
v0
,
2 i ω0
C2 = −
v0
.
2 i ω0
(9)
Now the general solution can be obtained by replacing the values of C1 and C2 in Eq. 4.
That is
v0 i ω0 t
v0 −i ω0 t
e
−
e
,
2 i ω0
2 i ω0
v0
x(t) =
sin ω0 t
ω0
x(t) =
3
(10)
C.
Case III
Let us assume that the mass is given a speed v at a displacement a from the equilibrium
position at time t = 0. So the initial conditions are x(0) = a and
dx
(0)
dt
= v. Now using
these initial conditions we get
v
.
i ω0
(11)
1
v a−
.
2
i ω0
(12)
C1 − C2 =
C1 + C2 = a ,
Solving these two equations, we get
C1 =
1
v a+
,
2
i ω0
C2 =
So the general solution, in this case, takes the following form
1
v i ω0 t 1 v −i ω0 t
a+
e
+
a−
e
,
2
i ω0
2
i ω0
a
v
= (ei ω0 t + e−i ω0 t ) +
(ei ω0 t − e−i ω0 t ) ,
2
2 i ω0
v
sin ω0 t ,
= a cos ω0 t +
ω0
= a0 cos α cos ω0 t − a0 sin α sin ω0 t ,
x(t) =
= a0 cos (w0 t + α) ,
(13)
where
a0 sin α = −
a0 cos α = a ,
We can write a0 and α as follows
s
a0 =
III.
a2
v2
+ 2,
ω0
−1
α = tan
v
.
ω0
v −
.
a ω0
(14)
(15)
DAMPED HARMONIC OSCILLATOR
So far we have considered ideal oscillations where there is no friction or damping. In the
real world, of course, things always damp down.
• Frictional forces paly a crucial role in the ideal oscillatory motion of a particle.
• It tends to dampen the oscillatory behaviour of the system as the motion proceeds in
time.
4
y
Fext
Fd
v(t)
Fs
x(t)
m
x
O
A
• Oscillation decays gradually with time.
Let us now proceed to discuss damped harmonic oscillator. We again assume a spring mass
system with a spring force Fs = −k x. We introduce a simplest form of damping force that
is proportional to the velocity of the particle/system, i.e, Fd = −γ
dx
.
dt
Here γ represents
damping coefficient. Now the equation of motion becomes
d2 x
dx
,
=
−k
x
−
γ
dt2
dt
d2 x
dx
m 2 +γ
+ kx = 0,
dt
dt
dx
d2 x
+ 2β
+ ω02 x = 0 ,
2
dt
dt
m
(16)
where 2 β = γ/m and ω02 = k/m. This is a second order linear, homogeneous differential
equation with constant coefficient. The solution of Eq. 16 will take the following form.
x(t) = ep t ,
(17)
where p can be determined by substituting the value of x(t) in Eq. 16. That is
p2 ep t + 2 β p ep t + ω02 ep t = 0 ,
p2 + 2 β p + ω02 = 0 ,
q
p = −β ± β 2 − ω02 .
(18)
Hence the most general solution to Eq. 16 can be written as
x(t) = C1 ep1 t + C2 ep2 t ,
(19)
p
p
where p1 = −β + β 2 − ω02 and p2 = −β − β 2 − ω02 are the two solutions that we obtained
in Eq. 18. There are three regimes of interest. Those are
β 2 − ω02 < 0 ,
Underdamped oscillation ,
β 2 − ω02 > 0 ,
Overdamped oscillation ,
β 2 − ω02 = 0 ,
Critically damped oscillation .
5
(20)
Let us now proceed to discuss all the three regimes one by one.
A.
Underdamped Oscillation
In this regime β 2 − ω02 < 0. Hence p1 and p2 become complex. Let us write p1 and p2 as
follows:
p1 = −β + i ω ,
where ω =
p2 = −β − i ω ,
(21)
p
ω02 − β 2 > 0. Now the most general solution of Eq. 19 can be written as
x(t) = C1 ep1 t + C2 ep2 t ,
= C1 e(−β+i ω) t + C2 e(−β−i ω) t ,
= e−β t C1 ei ω t + C2 e−i ω t ,
(22)
where C1 and C2 can be determined from given initial conditions. Now let us discuss few
special cases.
1.
Case I
We assume that the mass is pulled to one side by a distance a and released from rest at
t = 0. So the initial conditions are x(0) = a and dx/dt(0) = 0. By differentiating Eq. 22
with respect to t, we obtain
dx
= C1 (−β + i ω) e(−β+i ω) t + C2 (−β − i ω) e(−β−i ω) t .
dt
(23)
Now putting t = 0 in Eq. 22 and using the initial condition x(0) = a, we obtain
C1 + C2 = a .
(24)
Similarly putting t = 0 in Eq. 23 and using the initial condition dx/dt(0) = 0, we obtain
C1 (−β + i ω) + C2 (−β − i ω) = 0 ,
−β(C1 + C2 ) + i ω (C1 − C2 ) = 0 ,
i ω (C1 − C2 ) = β (C1 + C2 ) ,
i ω (C1 − C2 ) = β a ,
βa
C1 − C2 =
.
iω
6
(25)
Solving Eq. 24 and Eq. 25, we obtain
β a
1+
,
C1 =
2
iω
a
β C2 =
1−
.
2
iω
(26)
Now the most general solution can be obtained by substituting the values of C1 and C2 in
Eq. 22. That is
−β t
x(t) = e
−i ω t
iωt
C1 e + C2 e
,
ha
β
a
β −i ω t i
−β t
iωt
= e
1+
e +
1−
e
,
2
iω
2
iω
h ei ω t + e−i ω t a β ei ω t − e−i ω t i
= e−β t a
+
,
2
ω
2i
h
i
aβ
= e−β t a cos ω t +
sin ω t .
ω
(27)
aβ
ω
= −a0 sin α, we get
h
i
−β t
x(t) = e
a0 cos α cos ω t − a0 sin α sin ω t ,
Now putting a = a0 cos α and
= a0 e−β t cos (ω t + α) ,
(28)
where
r β2 a0 = a2 1 + 2 ,
ω
α = tan−1
−
β
.
ω
(29)
When β = 0, we obtain x(t) = a0 cos ω0 t. There are two main features:
• The frequency of the damped harmonic motion is less than the frequency of undamped
motion, i.e, ω < ω0 . So the time period is increased.
• The amplitude A(t) = a0 e−β t is time dependent and decays exponentially with time
t.
Damping is a frictional force. So it generates heat and dissipates energy. When β is small,
we would still expect the system to oscillate, but with decreasing amplitude as its energy is
converted to heat. Over time it should come to rest at equilibrium.
2.
Case II
We assume that the mass is hit and is given a speed v0 at its equilibrium position at
t = 0. So the initial conditions are x(0) = 0 and dx/dt(0) = v0 . Using the initial condition
7
x(0) = 0 in Eq. 22, we get
C1 + C2 = 0 .
(30)
Similarly using dx/dt(0) = v0 in Eq. 23, we obtain
C1 (−β + i ω) + C2 (−β − i ω) = v0 ,
−β(C1 + C2 ) + i ω (C1 − C2 ) = v0 ,
i ω (C1 − C2 ) = v0 ,
v0
C1 − C2 =
.
iω
(31)
Solving Eq. 30 and Eq. 31, we obtain
C1 =
v0
,
2iω
C2 = −
v0
.
2iω
(32)
The most general solution can be obtained by substituting the values of C1 and C2 in Eq. 22.
That is
h v
v0 −i ω t i
0
x(t) = e−β t
ei ω t −
e
,
2iω
2iω
v0 −β t
e
sin ω t .
=
ω
3.
(33)
Case III
Let us assume that the mass is given a speed v at a displacement a from the equilibrium
position at time t = 0. So the initial conditions are x(0) = a and
dx
(0)
dt
= v. Now using
these initial conditions we get
C1 + C2 = a ,
(34)
and
C1 (−β + i ω) + C2 (−β − i ω) = v ,
−β(C1 + C2 ) + i ω (C1 − C2 ) = v ,
−β a + i ω (C1 − C2 ) = v ,
i ω (C1 − C2 ) = v + β a ,
v+βa
C1 − C2 =
.
iω
8
(35)
Now solving these two equations, we get
v + β a
1
a+
,
C1 =
2
iω
1
v + β a
C2 =
a−
.
2
iω
The most general solution takes the following form.
h1
v + β a iω t 1
v + β a −i ω t i
x(t) = e−β t
a+
e +
a−
e
,
2
iω
2
iω
h ei ω t + e−i ω t v + β a ei ω t − e−i ω t i
+
,
= e−β t a
2
ω
2i
h
i
v+βa
= e−β t a cos ω t +
sin ω t ,
ω
−β t
= a0 e
cos (ω t + α) ,
(36)
(37)
where
r
a0 =
B.
a2
(v + β a)2
+
,
ω2
−1
α = tan
v + β a
.
−
aω
(38)
Overdamped oscillation
In case of overdamped motion β 2 − ω02 > 0. Hence both the solutions p1 and p2 are real.
Let us write p1 and p2 as follows:
p1 = −β + δ ,
where δ =
p2 = −β − δ ,
(39)
p
β 2 − ω02 > 0. It is evident that δ < β. The most general solution can be
written as
h
i
x(t) = e−β t C1 eδ t + C2 e−δ t ,
= C1 e−(β−δ) t + C2 e−(β+δ) t .
(40)
It should be mentioned that
• both the exponent in Eq. 40 are negative because δ < β. Hence both the solutions go
asymptotically to the equilibrium x = 0.
• overdamped motion is non-oscillatory.
• overdamped motion is supperposition of a quickly decaying term and a slowly decaying
term.
• The slowly decaying term controls the rate at which it reaches equilibrium.
Let us now study few special cases.
9
1.
Case I
We assume that the mass is pulled to one side by a distance a and released from rest at
t = 0. So the initial conditions are x(0) = a and dx/dt(0) = 0. Differentiating Eq. 40, we
get
dx
= −C1 (β − δ) e−(β−δ) t − C2 (β + δ) e−(β+δ) t .
dt
(41)
Now using the initial conditions x(0) = a and dx/dt(0) = 0, we get
C1 + C2 = a ,
(42)
and
−C1 (β − δ) − C2 (β + δ) = 0 ,
−β(C1 + C2 ) + δ(C1 − C2 ) = 0 ,
−β a + δ(C1 − C2 ) = 0 ,
δ(C1 − C2 ) = β a ,
βa
.
C1 − C2 =
δ
Solving these two equations, we get
a
β
C1 =
1+
,
2
δ
(43)
C2 =
a
β
1−
.
2
δ
(44)
So the most general solution is
−β t
x(t) = e
=
=
=
=
h
δt
−δ t
i
C1 e + C 2 e
,
ha
β δ t a
β −δ t i
−β t
e
1+
e +
1−
e
,
2
δ
2
δ
h eδ t + e−δ t a β eδ t − e−δ t i
e−β t a
+
,
2
δ
2
h
i
aβ
e−β t a cosh δ t +
sinh δ t ,
δ
h
i
−β t
e
a0 cosh σ cosh δ t + a0 sinh σ sinh δ t ,
= a0 e−β t cosh (δ t + σ) ,
(45)
where
a0 cosh σ = a ,
a0 sinh σ =
10
aβ
.
δ
(46)
or
r β2 a0 = a2 1 − 2 ,
δ
2.
Case II
3.
Case III
C.
σ = tanh−1
β δ
.
(47)
Critically damped motion
In this case β 2 = ω02 , i.e, p1 = p2 = −β. Both the solutions are identical. Hence e−β t and
t e−β t are the two solutions. The most general solution will be
x(t) = e−β t C1 + C2 t .
(48)
As in the overdamped case, a critically damped motion also does not oscillate. Now let us
study few cases.
1.
Case I
Let us assume that x = 0 and dx/dt = v0 at t = 0. Differentiating Eq. 48, we get
dx
= −β C1 e−β t + C2 t β e−β t + C2 e−β t
dt
(49)
Now putting x = 0 at t = 0 in Eq. 48, we get C1 = 0. Similarly, using the initial condition
dx/dt = v0 at t = 0, we get
C2 = v0 .
(50)
x(t) = v0 t e−β t
(51)
So the most general solution is
It is a product of linearly increasing term and an exponentially decreasing term. For small
values of t, x(t) will linearly increase with t and reach its maximum and then it will decrease
exponentially. We are going to find out the maximum value of x(t). To find the maximum,
11
we need to differentiate Eq. 51 with respect to t and then equate it to zero. That is
dx
= v0 (e−β t − β t e−β t ) = 0 ,
dt
1 − β t = 0,
1
t= .
β
(52)
Now the maximum value of x(t) can be obtained by putting t = 1/β in Eq. 51. That is
xmax = v0
=
2.
Case II
3.
Case III
D.
1.
1 −1
e ,
β
v0
.
βe
(53)
Logarithmic decrement and Q value
Logarithmic decrement
Damping is measured in terms of the amplitude, i.e, how much the amplitude decreases
as the system/particle completes one oscillation. Let us assume that the amplitude of the
particle at time t1 is A(t1 ) and the amplitude at time (t1 + T ) after a time period T is
A(t1 + T ). Using the expression for the underdamped motion, we can have
A(t1 ) = a0 e−β t1 ,
A(t1 + T ) = a0 e−β(t1 +T ) .
(54)
So the ratio of these amplitudes will be
A(t1 )
= eβ T .
A(t1 + T )
(55)
Now the logarithmic decrement is defined as follows:
A(t1 )
=βT ,
A(t1 + T )
γ
λ=
T.
2m
λ = ln
(56)
Hence a measurement of time period T and the logarithmic decrement λ gives a very covenient method to determine the damping coeffiecient γ.
12
2.
Q value
Let us first calculate the average energy in a damped harmonic oscillator. The displacement x(t) of an underdamped oscillator is given by
x(t) = a0 e−β t cos(ω t + α) .
(57)
Now the potential energy at any instant of time t is
PE =
1
1
k x(t)2 = m ω02 a20 e−2 β t cos2 (ω t + α) .
2
2
(58)
Similarly, kinetic energy is
KE =
1 dx(t) 2
1
m v(t)2 = m
,
2
2
dt
h
i2
1
= m a20 e−2 β t ω sin(ω t + α) + β cos(ω t + α) ,
2
h
i
1
= m a20 e−2 β t ω 2 sin2 (ω t + α) + β 2 cos2 (ω t + α) + 2 ω β sin(ω t + α) cos(ω t + α) ,
2
h
i
1
= m a20 e−2 β t ω 2 sin2 (ω t + α) + β 2 cos2 (ω t + α) + ω β sin[2 (ω t + α)] .
(59)
2
Now average potential and kinetic energy are (average is taken over a complete cycle)
1
m ω02 a20 e−2 β t ,
4
h ω2 β 2 i 1
1
1
< KE >= m a20 e−2 β t
+
= m a20 e−2 β t (ω 2 + β 2 ) = m ω02 a20 e−2 β t , (60)
2
2
2
4
4
< PE >=
where for underdamped case, we have used ω 2 + β 2 = ω02 . So the average energy stored in
a damped harmonic oscillator is
< E >=< PE > + < KE >=
where E0 =
1
2
1
m ω02 a20 e−2 β t = E0 e−2 β t ,
2
(61)
m ω02 a20 is the energy of the undamped oscillator. The average energy lost in
a complete cycle is
∆E = < E > (t0 )− < E > (t0 + T ) ,
= E0 e−2 β t0 − E0 e−2 β (t0 +T ) ,
h
i
= E0 e−2 β t0 1 − e−2 β T ,
h
i
= E0 e−2 β t0 1 − (1 − 2 β T ) ,
= 2 β T E0 e−2 β t0 ,
13
(62)
where we have used ex = 1 + x and ignored the higher order terms in x. Now Q value is
defined as the ratio of total energy stored to the total energy lost in a complete cycle, or
more specifically
Q = 2π
E.
1
π
π
ω
<E>
= 2π
=
= =
.
∆E
2β T
βT
λ
2β
(63)
Relaxation time
The time at which the energy of the damped harmonic motion reduces to 1/e times its
energy at the equilibrium value is called the relaxation time. We know that the average
energy of a damped harmonic oscillator can be written as
< E >= E0 e−2 β t .
(64)
Let us assume that the relaxation time is τ so that
< E >= E0 e−2 β τ = E0 (1/e) ,
2β τ = 1 ,
1
.
τ=
2β
IV.
(65)
FORCED OSCILLATION
• In case of forced oscillation, the particle/system is subject to an external periodic force
which may or may not be time dependent.
• When a system is subject to an external force, the oscillation that results is called
forced oscillation and it oscillates with the frequency of the applied external force.
• The system takes a certain time to adjust to the frequency of the external force. This
is referred to as the transient period.
• In this time, free oscillation of the system die out and the system starts performing
harmonic motion of constant amplitude with frequency of the external force. This is
called steady state motion.
14
y
Fext
Fd
v(t)
Fs
x(t)
m
x
O
A
We are mainly interested in the steady state solution. Let us again consider a spring mass
which is subject to a
system with spring force Fs = −k x and damping force Fd = −γ dx
dt
harmonic external force Fext = F0 cos ω t. The equation of motion can be written as
d2 x
dx
+ k x = F0 cos ω t ,
+γ
2
dt
dt
d2 x
dx
+ ω02 x = f0 cos ω t ,
+ 2β
2
dt
dt
m
(66)
where f0 = F0 /m and ω is the frequency of the external force. This is a non-homogeneous
second order differential equation. General solution consists of mainly two parts, namely
particular integral and complimentary function. Complimentary function is the solution of
the homogeneous equation, i.e,
d2 x
dx
+ 2β
+ ω02 x = 0 .
2
dt
dt
(67)
We have already encountered this equation in damped harmonic oscillation. This oscillation
will decay and die down withtime. Hence we are not interested in this complimentary
function. We are mainly interested in the particular integral which is the only part that
survives after a long enough time. This is called the steady state solution.
Let us consider a complex parameter z whose real part is x, i.e, x = Re(z). We can
rewrite Eq. 66 in terms of z as follows:
Re
h d2 z
dt2
+ 2β
i
dz
+ ω02 z = f0 ei ω t .
dt
(68)
Now our main aim is to solve the following equation.
d2 z
dz
+ 2β
+ ω02 z = f0 ei ω t .
2
dt
dt
15
(69)
Let us assume that z = z0 ei ω t . Now substituting this in Eq. 69, we get
−ω 2 z0 ei ω t + 2 β (i ω) z0 ei ω t + ω02 z0 ei ω t = f0 ei ω t ,
h
i
z0 − ω 2 + 2 β i ω + ω02 = f0 ,
z0 =
(ω02
−
f0
2
ω )+
2iβ ω
,
h
i
f0 (ω02 − ω 2 ) − 2 i β ω
i.
z0 = h
2
2
2
2
2
(ω0 − ω ) + 4 β ω
(70)
So we see that z0 is a complex parameter. We can assume that z0 = a e−i φ . Now comparing
the real and imaginary parts, we obtain
a cos φ =
f0 (ω02 − ω 2 )
,
(ω02 − ω 2 )2 + 4 β 2 ω 2
a sin φ =
(ω02
2 f0 β ω
.
− ω 2 )2 + 4 β 2 ω 2
(71)
We can write a and φ as follows:
φ = tan−1
h 2β ω i
,
ω02 − ω 2
f0
a= p 2
.
(ω0 − ω 2 )2 + 4 β 2 ω 2
(72)
We can write z as
z = z0 ei ω t = a e−i φ ei ω t = a ei (ω t−φ) .
(73)
The steady state solution to Eq. 66 can be written as
x(t) = Re(z) = a cos (ω t − φ) .
(74)
• The mass oscillates with the same frequency ω as that of the external periodic force.
• There is a phase difference of φ between the forced oscillation and the external periodic
force.
• It has nothing to do with the spring force Fs and the damping force Fd .
• However, Fs and Fd have a great effect on the amplitude a and the phase φ.
There are three regimes of interest, i.e, low frequency regime, high frequency regime and
resonance. We will discuss each one by one.
16
A.
Low frequency regime
If the frequency of the forced oscillation ω is much smaller than the natural frequency ω0
then ω << ω0 . As ω tends to zero, the amplitude a will be
f0
F0
F0 m
F0
f0
→ 2 =
=
=
a= p 2
2
ω0
m ω0
m k
k
(ω0 − ω 2 )2 + 4 β 2 ω 2
(75)
Similarly the phase φ will be
h 2β ω i
→ 0.
ω02 − ω 2
−1
φ = tan
(76)
Hence the forced oscillation can be described by
x(t) =
F0
cos ω t .
k
(77)
We see that the motion is independent of the mass or the damping force. It depends only
on the spring constant k. Hence this regime is called ”stiffness-controlled” region.
B.
High frequency regime
If ω >> ω0 then the amplitude a will be
f0
f0
F0
a= p 2
→ 2 =
.
ω
m ω2
(ω0 − ω 2 )2 + 4 β 2 ω 2
(78)
Similarly the phase φ will be
φ = tan−1
h 2β ω i
2β
−1
→
tan
−
.
ω02 − ω 2
ω
(79)
As ω → ∞, we get a → 0 and φ → π. So the mass effectively at rest and can not follow the
external force because its frequency is very high compared to the natural frequency of the
spring mass system. Thus for ω mcuh larger than ω0 the motion can be described as
x(t) =
F0
F0
cos
(ω
t
−
π)
=
−
cos ω t .
m ω2
m ω2
(80)
We see that the motion is independent of the spring constant k or the damping force. It only
depends on the mass of the object. Hence this region is called ”mass controlled” region. It
is also observed that the mass is out of phase with the external phase as the phase difference
is φ = π.
17
C.
Mid frequency regime
At ω = ωres =
p
ω02 − 2 β 2 , the amplitude a will be maximum. This is called the resonance
frequency. Let us see this explicitly. The expression for a is
f0
.
a= p 2
2
(ω0 − ω )2 + 4 β 2 ω 2
(81)
Now a will be maximum if the denominator in Eq. 81 is minimum, i.e, for amplitude maxiumum, we need to have
i
d h 2
(ω0 − ω 2 )2 + 4 β 2 ω 2 = 0 ,
dω
2 (ω02 − ω 2 ) (−2 ω) + 8 β 2 ω = 0 ,
−(ω02 − ω 2 ) + 2 β 2 = 0 ,
ω 2 = ω02 − 2 β 2 ,
q
ω = ωres = ω02 − 2 β 2 .
(82)
For small values of the damping parameter, i.e, β << ω0 , we have ωres = ω0 . It should be
mentioned that the phase φ increases continuously from 0 to π and at ω = ω0 it becomes
φ = π/2. Hence for small value of β, the amplitude resonance and phase resonance coincide
with the natural undamped frequency ω0 of the oscillator. In this situation, the maximum
amplitude will be (putting ω = ω0 in Eq. 81)
amax =
f0
F0
=
.
2 β ω0
γ ω0
(83)
Neither the mass nor the spring force, but only the damping force γ controls the motion.
This region is called ”damping controlled region”. If the damping is too small, the amplitude
tends to be extremely large. As β → 0, the amplitude a → ∞. At ω = 0, the amplitude is
given by
a0 =
f0
.
ω02
(84)
So the ratio of amax to a0 is given by
amax
ω0
=
= Q.
a0
2β
(85)
• For a low driving frequency ω, i.e, ω << ω0 , the amplitude is nearly same, i.e, a =
F0 /k, for all the values of damping.
18
• As the frequency increases, the amplitude increases depending on the damping present
in the system.
• For zero damping, the amplitude is infinite at ω = ω0 .
• For low damping, the peak of the amplitude is higher but the maximum value is
shifted towards the left of ω0 because we know that amplitude resonance occurs at
p
ω = ωres = ω02 − 2 β 2 . So for zero damping, i.e, for β = 0, the amplitude resonance
occurs at ω = ω0 and as we increase β, it shifted towrads the left of ω0 depending on
the value of β.
• For higher damping, the peak of the amplitude resonance reduces further and is shifted
more to the left of ω0 .
• As the driving frequency increases beyond the resonant frequency, the amplitude tends
to reduce and goes to zero if ω → ∞ for all values of the damping.
• The amplitude curve falls more rapidly for low damping than for the high damping.
D.
Velocity resonance
The steady state solution for the forced oscillation is given by
x(t) = a cos(ω t − φ) .
(86)
So the instantaneous velocity is given by
v(t) =
dx(t)
= −a ω sin(ω t − φ) .
dt
19
(87)
0.1
0.15
0.2
0.3
2.
5
4
3
2
1
0
0.0
0.5
0.1
1.0
0.15
1.5
0.2
0.3
2.0
1.
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
20
1.5
2.0
Now velocity resonance will occur when aω will be maximum, i.e,
d
(aω) = 0 ,
dω
da
+ a = 0,
ω
dω
i
d h
f0
p
ω
+ a = 0,
dω
(ω02 − ω 2 )2 + 4 β 2 ω 2
i−1/2
d h 2
(ω0 − ω 2 )2 + 4 β 2 ω 2
ω f0
+ a = 0,
dω
1 h
i−3/2 d h
i
ω f0 −
(ω02 − ω 2 )2 + 4 β 2 ω 2
(ω02 − ω 2 )2 + 4 β 2 ω 2 + a = 0 ,
2
dω
h
i−3/2 h
i
1
− ω f0 (ω02 − ω 2 )2 + 4 β 2 ω 2
2 (ω02 − ω 2 ) (−2 ω) + 8 β 2 ω + a = 0 ,
2
h
i−3/2 h
i
f0
2
2 2
2 2
2
2
2
ω f0 (ω0 − ω ) + 4 β ω
− 2 ω (ω0 − ω ) + 4 β ω + p 2
= 0,
(ω0 − ω 2 )2 + 4 β 2 ω 2
(ω02 − ω 2 )(ω02 + ω 2 ) = 0 ,
ω = ω0 .
(88)
Hence the velocity resonance occurs at ω = ω0 .
E.
Power resonance
The instantaneous power absorbed by the oscillator from the driving force is
d
dx(t)
= F0 cos ω t a cos(ω t − φ) ,
dt
dt
= −a F0 ω cos ω t sin(ω t − φ) .
P = Fext
(89)
Now the average power absorbed by the oscillator is given by
< P > = −a F0 ω < cos ω t sin(ω t − φ) > ,
= −a F0 ω < cos ω t (sin ω t cos φ − cos ω t sin φ) > ,
h
i
= −a F0 ω cos φ < cos ω t sin ω t > − sin φ < cos2 ω t > ,
=
1
a F0 ω sin φ .
2
(90)
21
1.2
1.0
0.8
β=0.2
0.6
β=0.4
0.4
0.2
0.0
0
1
2
3
4
5
We know that
h 2β ω i
,
ω02 − ω 2
2β ω
tan φ = 2
,
ω0 − ω 2
sin φ
sin φ =
cos φ ,
cos φ
1
,
sin φ = tan φ
sec φ
1
sin φ = tan φ
,
1 + tan2 φ
2β ω
sin φ = p 2
.
(ω0 − ω 2 )2 + 4β 2 ω 2
φ = tan−1
(91)
Putting the value of sin φ in Eq. 90, we get
<P > =
1
a F0 ω sin φ ,
2
2β ω
1
f0
= p 2
F0 ω p 2
,
2 (ω0 − ω 2 )2 + 4β 2 ω 2
(ω0 − ω 2 )2 + 4β 2 ω 2
i
F02 γ h
ω2
=
.
2 m2 (ω02 − ω 2 )2 + 4β 2 ω 2
(92)
where we have used f0 = F0 /m and 2β = γ/m. Now < P > will be maximum if
i
d h
ω2
= 0,
dω (ω02 − ω 2 )2 + 4β 2 ω 2
h
i
h
i
ω2
2
2
2
2
2
ω
−
ω
2
((ω
−
ω
)(−2ω)
+
8
β
ω
= 0,
0
(ω02 − ω 2 )2 + 4β 2 ω 2
(ω02 − ω 2 )(ω02 + ω 2 ) = 0 ,
ω = ω0 .
(93)
22
Hence we see that just like velocity resonance, power resonance occurs at ω = ω0 . Now the
maximum average power absobed at the resonant frequency is (putting ω = ω0 in Eq. 92)
< P >max =
F02 γ 1
F02
.
=
2 m2 4 β 2
2γ
(94)
The sharpness of the power resonace curve depends inversely on the damping parameter.
F.
Bandwidth
Bandwidth is defined as the halfwidth or full width at half maximum. Half maximum will
occur at two different frequencies, one at lower frequency than ω0 and the other at higher
frequency than ω0 . Let us assume the lower frequency to be ω1 and the higher frequency to
be ω2 . Then the Bandwidth is difined as the difference between these two frequencies, i.e,
B = ω2 − ω1 .
(95)
Let us now find the values of ω1 and ω2 . We know that, at half maximum, the average power
absobed by the oscillator is
< P >max
,
2
i F2 γ 1
ω2
0
= 0 2
,
2 m2 (ω02 − ω 2 )2 + 4β 2 ω 2
2 m 8 β2
ω2
1
=
,
2
2
2
2
2
(ω0 − ω ) + 4β ω
8 β2
(ω02 − ω 2 )2 + 4β 2 ω 2 = 8 β 2 ω 2 ,
< P >=
F2 γ h
(ω02 − ω 2 )2 − 4β 2 ω 2 ,
(ω02 − ω 2 + 2 β ω)(ω02 − ω 2 − 2 β ω) = 0 .
(96)
Now we get two solutions. Let us look at the first one. i.e,
ω02 − ω 2 + 2 β ω = 0 ,
ω 2 − 2 β ω = ω02 ,
ω 2 − 2 β ω + β 2 = ω02 + β 2 ,
(ω − β)2 = ω02 + β 2 ,
q
ω − β = ω02 + β 2 ,
q
ω = β + ω02 + β 2 .
23
(97)
Since the value of ω is greater than ω0 , this must be our ω2 . Hence
ω2 = β +
q
ω02 + β 2 .
(98)
Similarly from Eq. 96, we get
ω02 − ω 2 − 2 β ω = 0 ,
ω 2 + 2 β ω = ω02 ,
ω 2 + 2 β ω + β 2 = ω02 + β 2 ,
(ω + β)2 = ω02 + β 2 ,
q
ω + β = ω02 + β 2 ,
q
ω = −β + ω02 + β 2 .
(99)
Hence
ω1 = −β +
q
ω02 + β 2 .
(100)
So from Eq. 95, we get
B = ω2 − ω1 = 2β .
(101)
This is another way to determine the damping parameter β.
V.
COUPLED OSCILLATOR
Study of coupled oscillations forms the natural link between simple harmonic motion
of a single particle and wave motion of a continuous infinite numbers of particles as in a
medium. Let us consider two identical spring mass system of spring constant k coupled by
a third spring of spring constant k0 in the middle. Let us denote the displacements of the
two masses from their equilibrium position by x1 and x2 , respectively. Now the equation of
motion of the masses are
d 2 x1
= −k x1 − k0 x1 + k0 x2 ,
dt2
d 2 x1
m 2 + (k + k0 ) x1 − k0 x2 = 0 ,
dt
m
24
(102)
and
d 2 x2
= −k x2 − k0 x2 + k0 x1 ,
dt2
d 2 x2
m 2 − k0 x1 + (k + k0 ) x2 = 0 .
dt
m
(103)
These are set of two second order linear homogeneous coupled differential equations with
constant coefficients.
A.
Eigenfrequencies and Eigenstates
The eigenstates are defined as those states of the oscillator in which the entire system
oscillates with a single frequency. Now let us assume that
x1 = A eiω t ,
x2 = B eiω t ,
(104)
d2 x2
= −ω 2 B eiω t .
dt2
(105)
so that
d 2 x1
= −ω 2 A eiω t ,
dt2
Putting these values in Eq. 102 and Eq. 103, we get
−m ω 2 A eiω t + (k + k0 ) A eiω t − k0 B eiω t = 0 ,
(−m ω 2 + k + k0 ) A − k0 B = 0 ,
(106)
and
−m ω 2 B eiω t − k0 A eiω t + (k + k0 ) B eiω t = 0 ,
−k0 A + (−m ω 2 + k + k0 ) B = 0 ,
25
(107)
Now from Eq. 106, we can write B in terms of A as
(−m ω 2 + k + k0 ) A − k0 B = 0 ,
(−m ω 2 + k + k0 )
B=
A.
k0
(108)
Now using this value of B in Eq. 107, we get
−k0 A + (−m ω 2 + k + k0 ) B = 0 ,
(−m ω 2 + k + k0 )
2
−k0 A + (−m ω + k + k0 )
A = 0,
k0
(−m ω 2 + k + k0 )2 = k02 ,
−m ω 2 + k + k0 = ±k0 ,
−m ω 2 = −k − k0 ± k0 ,
k + k0 ∓ k0
ω2 =
.
m
(109)
So we get two solutions depending on the plus and minus sign. Hence
k
,
m
r
k
,
ω1 =
m
ω12 =
(110)
and
k + 2 k0
,
m
r
k + 2 k0
,
ω2 =
m
ω22 =
(111)
These are the two eigenfrequencies of the system.
B.
Normal modes
Now let us see what will happen when we put ω = ω1 in Eq. 106, i.e,
(−m ω12 + k + k0 ) A − k0 B = 0 ,
(−k + k + k0 ) A − k0 B = 0 ,
A=B.
(112)
26
Similarly, putting ω = ω2 in Eq. 106, we get
(−m ω22 + k + k0 ) A − k0 B = 0 ,
(−k − 2 k0 + k + k0 ) A − k0 B = 0 ,
−k0 A − k0 B = 0 ,
A = −B .
(113)
So, if ω = ω1 , we get A = B. Hence from Eq. 104, we get
x1 (t) = A eiω1 t = x2 (t) .
(114)
This is true for all values of t. Hence
• The two masses always move synchronously.
• The middle spring remains unstretched all the time. It does not contribute to the
effective spring constant.
• Frequency remains equal to the uncoupled frequency ω1 .
Now if ω = ω2 , we get A = −B. Hence from Eq. 104, we get
x1 (t) = A eiω2 t = −x2 (t) .
(115)
This is true for all values of t. Hence
• The two masses always move antichronously (opposite in phase).
• The middle spring contributes maximally.
• It gives rise to higher frequency ω2 .
These two modes of vibration are called normal modes of the system.
C.
General solution
Now let us look at the most general solution of Eq. 102 and Eq. 103. Let us introduce
two new variables q1 (t) and q2 (t) as linear combination of the old variables x1 (t) and x2 (t),
i.e,
1
x2 = √ (q1 + q2 ) ,
2
1
x1 = √ (q1 − q2 ) ,
2
27
(116)
so that
1
q1 = √ (x1 + x2 ) ,
2
1
q2 = √ (x2 − x1 ) .
2
(117)
Now substituting the values of x1 and x2 from Eq. 116 in Eq. 102, we get
d2 x1
+ (k + k0 ) x1 − k0 x2 = 0 ,
dt2
i
1
1
d2 h 1
m 2 √ (q1 − q2 ) + (k + k0 ) √ (q1 − q2 ) − k0 √ (q1 + q2 ) = 0 ,
dt
2
2
2
2
2
d q1
d q2
m 2 − m 2 + k q1 − (k + 2 k0 ) q2 = 0 .
dt
dt
m
(118)
Similarly, substituting the values of x1 and x2 from Eq. 116 in Eq. 103, we get
d2 x 2
− k0 x1 + (k + k0 ) x2 = 0 ,
dt2
i
1
1
d2 h 1
m 2 √ (q1 + q2 ) − k0 √ (q1 − q2 ) + (k + k0 ) √ (q1 + q2 ) = 0 ,
dt
2
2
2
d2 q2
d2 q 1
m 2 + m 2 + k q1 + (k + 2 k0 ) q2 = 0 .
dt
dt
m
(119)
Now adding Eq. 118 and Eq. 119, we get
d2 q1
+ k q1 = 0 ,
dt2
k
d2 q 1
+ q1 = 0 ,
2
dt
m
2
d q1
+ ω12 q1 = 0 .
2
dt
m
(120)
Similarly, subtracting Eq. 118 from Eq. 119, we get
d2 q2
+ (k + 2 k0 ) q2 = 0 ,
dt2
d2 q2 k + 2 k0
+
q2 = 0 ,
dt2
m
d2 q2
+ ω22 q2 = 0 .
dt2
m
(121)
Now we see that the differential equations in x1 and x2 becomes decoupled. So the general
solution to Eq. 120 and Eq. 121 are
q1 (t) = a1 cos(ω1 t + α1 ) ,
q2 (t) = a2 cos(ω2 t + α2 ) .
28
(122)
So, in terms of x1 and x2 , we get
1
x1 (t) = √ (q1 − q2 ) =
2
1
x2 (t) = √ (q1 + q2 ) =
2
i
1 h
√ a1 cos(ω1 t + α1 ) − a2 cos(ω2 t + α2 ) ,
2
i
1 h
√ a1 cos(ω1 t + α1 ) + a2 cos(ω2 t + α2 ) ,
2
(123)
where a1 , a2 , α1 and α2 are arbitrary constants which can be determined from the given
initial conditions on x1 (t), x2 (t), dx1 /dt and dx2 /dt at t = 0.
D.
Case I: Normal mode 1
Pulling each of the masses to the right by the same amount A and leaving them from
rest will generate the first normal mode. So the initial conditions are
x1 (0) = x2 (0) = A ,
dx1
dx2
=
= 0.
dt t=0
dt t=0
(124)
Differentiating Eq. 123 with respect to t, we get
i
dx1
1 h
= √ − a1 ω1 sin(ω1 t + α1 ) + a2 ω2 sin(ω2 t + α2 ) ,
dt
2
i
1 h
dx2
= √ − a1 ω1 sin(ω1 t + α1 ) − a2 ω2 sin(ω2 t + α2 ) .
dt
2
(125)
Now putting the first two initial conditions x1 (0) = x2 (0) = A in Eq. 123, we get
i
1 h
√ a1 cos α1 − a2 cos α2 = A ,
2
i
1 h
√ a1 cos α1 + a2 cos α2 = A .
2
Similarly, putting the last two initial conditions
dx1
dt
=
t=0
dx2
dt
(126)
= 0 in Eq. 125, we get
t=0
i
1 h
√ − a1 ω1 sin α1 + a2 ω2 sin α2 = 0 ,
2
i
1 h
√ − a1 ω1 sin α1 − a2 ω2 sin α2 = 0 .
2
(127)
Adding and subtracting the equations in Eq. 126, we get
a1 cos α1 =
√
2A,
29
a2 cos α2 = 0 .
(128)
Similarly, adding and subtracting the equations in Eq. 127, we get
a1 sin α1 = 0 ,
a2 sin α2 = 0 ,
(129)
where we have used the fact that ω2 6= 0. Now from Eq. 143 and Eq. 144, we get
a21 cos2 α1 + a21 sin2 α1 = 2 A ,
a21 = 2 A ,
√
a1 = 2 A .
(130)
Similarly
a22 cos2 α2 + a22 sin2 α2 = 0 ,
a2 = 0 .
(131)
We know from Eq. 144 that a1 sin α1 = 0. Since a1 is non zero, it means sin α1 = 0. This
means, we can take α1 = 0. Again we find that a2 = 0, so α2 can take any arbitrary value.
Hence, we are going to take α2 = 0. Now putiing the values of a1 , a2 , α1 and α2 in Eq. 123,
we get
i
1 h√
x1 (t) = √
2 A cos ω1 t = A cos ω1 t ,
2
i
1 h√
x2 (t) = √
2 A cos ω1 t = A cos ω1 t .
2
(132)
The two masses move synchronously with same amplitude and same frequency. This is
called fundamental normal mode frequency.
E.
Case II: Normal mode 2
Pulling the first mass to the right and the second mass to the left by same amount A and
leaving them from rest will generate the second normal mode. So the initial conditions are
x1 (0) = −x2 (0) = A ,
dx1
dx2
=
= 0.
dt t=0
dt t=0
Now putting these initial conditions in Eq. 123 and Eq. 125, we get
i
1 h
√ a1 cos α1 − a2 cos α2 = A ,
2
i
1 h
√ a1 cos α1 + a2 cos α2 = −A .
2
30
(133)
(134)
i
1 h
√ − a1 ω1 sin α1 + a2 ω2 sin α2 = 0 ,
2
i
1 h
√ − a1 ω1 sin α1 − a2 ω2 sin α2 = 0 .
2
(135)
Adding and subtracting the equations in Eq. 134, we get
√
a2 cos α2 = − 2 A .
a1 cos α1 = 0 ,
(136)
Similarly, adding and subtracting the equations in Eq. 127, we get
a1 sin α1 = 0 ,
a2 sin α2 = 0 ,
(137)
Hence from these equations, we get
√
a2 = − 2 A ,
a1 = 0 ,
α1 = 0 ,
α2 = 0 .
(138)
Now putting these values of a1 , a2 , α1 and α2 in Eq. 123, we get
i
1 h√
x1 (t) = √
2 A cos ω2 t = A cos ω2 t ,
2
i
1 h √
x2 (t) = √ − 2 A cos ω2 t = −A cos ω2 t .
2
(139)
The two masses move antichronously with same amplitude and same frequency. This is
called the second harmonic normal mode frequency.
F.
Case III: Resonance
Pulling the first mass to the right by an amount A and left from rest, while the second
mass remains at rest in its initial position will generate resonance. So the initial conditions
are
x1 (0) = A ,
dx1
dx2
=
dt t=0
dt
x2 (0) = 0 ,
= 0.
(140)
t=0
Now putting these initial conditions in Eq. 123 and Eq. 125, we get
i
1 h
√ a1 cos α1 − a2 cos α2 = A ,
2
i
1 h
√ a1 cos α1 + a2 cos α2 = 0 .
2
31
(141)
i
1 h
√ − a1 ω1 sin α1 + a2 ω2 sin α2 = 0 ,
2
i
1 h
√ − a1 ω1 sin α1 − a2 ω2 sin α2 = 0 .
2
(142)
Adding and subtracting the equations in Eq. 141, we get
A
a1 cos α1 = √ ,
2
A
a2 cos α2 = − √ .
2
(143)
Similarly, adding and subtracting the equations in Eq. 127, we get
a1 sin α1 = 0 ,
a2 sin α2 = 0 ,
(144)
From these equations, we get
A
a1 = √ ,
2
A
a2 = − √ ,
2
α1 = α2 = 0 .
Now putting these values of a1 , a2 , α1 and α2 in Eq. 123, we get
i
Ah
x1 (t) =
cos ω1 t + cos ω2 t ,
2
i
Ah
cos ω1 t − cos ω2 t .
x2 (t) =
2
(145)
(146)
Now let us consider that
Ω=
ω2 + ω1
,
2
=
ω2 − ω1
,
2
(147)
so that
ω1 = Ω − ,
ω2 = Ω + .
Hence, in terms of Ω and , we can write Eq. 146 as
i
Ah
x1 (t) =
cos(Ω − )t + cos(Ω + )t = A cos t cos Ω t ,
2
i
Ah
x2 (t) =
cos(Ω − )t − cos(Ω + )t = A sin t sin Ω t ,
2
(148)
(149)
Now if ω1 ≡ ω2 then we get << Ω. Or the t oscillation is much slower than that of Ω t
oscillation. The t oscillation acts as an envelope for the Ω t oscillation. Few observations
are as follows:
• Oscillation of the first mass die down gradually while the second mass picks up oscillations.
32
1.0
x0 (t)
0.5
0.0
-0.5
-1.0
0
5
10
15
20
15
20
t
1.0
x1 (t)
0.5
0.0
-0.5
-1.0
0
5
10
t
• Ocillation of the second mass grow till the first mass comes to rest.
• Thereafter the roles are reversed, i.e, the oscillation of the second mass die down
gradually while the first mass pick up the oscillations.
• The periodic shuttling of the energy of the oscillator between the two masses is referred
to as resonance.
This can be understood in terms of driving force and resonance. At the start and until
t = π/2, the x1 (t) looks like cos Ω t with a slowly varying amplitude. However, x2 (t) looks
like sin Ω t with slowly varying amplitude. Again,
cos Ω t = sin(Ω t + π/2) .
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That means x1 (t) is 90◦ ahead of x2 (t). 90◦ phase difference between x1 (t) and x2 (t) means
that the first mass basically acts like a driving force for the second mass at resonance. Hence
energy is transferred from x1 (t) to x2 (t) in t ∈ (0, π/2) region.
But right after x1 (t) has zero amplitude at t = π/2, the cos t factor in x1 (t) switches
sign and x1 (t) looks like − cos Ω t. However, x2 (t) still looks like sin Ω t. Again, we know
that
− cos Ω t = sin(Ω t − π/2) .
(151)
Now the second mass acts like a driving force for the first mass. Hence the energy will now
transfer from x2 (t) to x1 (t). This will go on till we reach t = π. After this the role will be
reversed.
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