EP225 Note No. 2
Oscillations: Mechanical Oscillations
2.1 Mass-spring system
Waves are in general created by something oscillating. For example, sound waves are
radiated by various oscillating elements such as speakers, vocal chord, strings in piano and
guitar, tuning fork, etc. Electromagnetic waves are also radiated by oscillating (or circulating) charges. In an antenna, electrons are forced to oscillate by an oscillating generator.
Through radiation of waves, oscillators lose energy to waves which carry energy (and momentum as well) through space.
Fig. 2.1. Mass-spring system.
Fig. 2.1. Mass-spring system.
1
Fig. 2.1. Mass-spring system.
Consider a mass M (kg) attached to the end of spring with spring constant k (N m 1 ).
The mass provides inertia and spring provides elasticity or restoring force. These properties
(inertia and elasticity) are basic ingredients of mechanical oscillations and also mechanical
waves. Oscillation in a mass-spring system can be started by various means, e.g., hitting the
mass, or displacing the mass from the equilibrium position and then release it. No matter
how oscillation is started, once started, the same oscillation frequency is observed. For ideal
spring-mass system without friction, oscillations prevail without damping.
To analyze oscillation of spring-mass system, let us consider equation of motion for the
mass,
dv
= kx
(1)
M
dt
where
dx
v=
(2)
dt
is the velocity of the mass, x(t) is the displacement of the mass from the equilibrium position,
and kx is the restoring force exerted on the mass by the spring. Note the negative sign
which indicates that the spring always tends to bring the mass back to the equilibrium
position (hence restoring force). Substitution of Eq. (2) into (1) yields the following second
order di¤erential equation,
k
d2 x(t)
+ x(t) = 0
(3)
2
dt
M
which has a general solution in the form
x(t) = A sin !t + B cos !t
where A and B are constants that can be determined from initial conditions, and
r
k
(rad s 1 )
!=
M
(4)
is the oscillation frequency. The quantity k=M has dimensions of
k
M
=
Nm
kg
1
=
kg m s
kg m
2
=s
2
p
and k=M indeed has dimensions of frequency, rad s 1 . (The angle radian is dimensionless.)
If the mass is displaced by a distance x0 and then released, the oscillation starts with
the initial value of x(t = 0) = x0 : Then, A = 0 and B = x0 ; and the solution for x(t) can be
uniquely determined,
x(t) = x0 cos !t
It is noted that giving an initial displacement to the mass involves change in the spring
length by the same amount, x = x0 ; and one has to do work with an amount of energy
1 2
kx (J)
2 0
2
given to the spring in the form of potential energy. After oscillation has started, the mass
acquires a velocity
dx(t)
v(t) =
= !x0 sin !t
dt
and corresponding kinetic energy
1
1
1
mass kinetic energy = M v 2 = M ! 2 x20 sin2 !t = kx20 sin2 !t
2
2
2
The instantaneous potential energy of the spring is
1
spring potential energy = kx20 cos2 !t
2
The sum of the two forms of energy is constant,
1
total energy = kx20
2
(Note that sin2 + cos2 = 1:) When the potential energy is maximum, the kinetic energy
becomes zero, and vice versa. The mass and spring toss energy between them as illustrated
in the …gure.
x(t)
2
1
0
0.05
0.10
0.15
0.20
0.25
0.30
-1
-2
Fig. 2.2. Displacement y = x(t) = 2 cos (!t) when
!=2
10 rad s 1 :
3
t
v(t)
100
50
0
0.05
0.10
0.15
0.20
0.25
0.30
t
-50
-100
Fig. 2.3. Velocity of the mass
v (t) = dx (t) =dt = 40 sin (20 t).
E 2.0
1.5
1.0
0.5
-0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32
t
Fig. 2.4. Kinetic energy (red), potential energy (black), and total energy (green).
Example 1: Derive the di¤erential equation
M
d2 x(t)
+ kx(t) = 0
dt2
(5)
from the condition of energy conservation,
1
1
M v 2 + kx2 = constant
2
2
Solution: Di¤erentiate
(6)
1
1
M v 2 + kx2 = constant
2
2
once to obtain
Mv
dv
dx
+ kx
=0
dt
dt
4
(7)
However,
v=
dx dv
d2 x
;
= 2
dt dt
dt
Therefore, Eq. (7) reduces to
d2 x(t)
k
+ x(t) = 0
2
dt
M
Example 2: (a) What mass should be attached to a spring with a spring constant k = 10
N m 1 for the system to oscillate at a frequency 5 Hz? (b) If oscillation is started with an
initial displacement of 3 cm, what is the maximum velocity of the mass?
Solution:
1. (a) The angular frequency is
!=2
From
!=
we …nd
M=
5 rad s
=2
r
1
k
M
k
10
=
= 0:01 kg
2
!
(10 )2
(b) The velocity of the mass is given by
v(t) =
dx (t)
=
dt
!x0 sin !t
Therefore, the amplitude of velocity oscillation is
!x0 = 2
0:03 m s
5
1
= 0.943 m s
1
Example 3: If the mass in a mass-spring system is subject to a friction force proportional
to the velocity, the equation of motion for the mass is modi…ed as
M
dv
=
dt
kx
fv
(8)
where f is a constant. Deriving a di¤erential equation to describe oscillation with the friction
force, show that the amplitude of oscillation decreases exponentially. For simplicity, assume
small friction force such that f
!M:
Solution: The di¤erential equation for the displacement can readily be found from Eq. (8)
by noting v = dx=dt,
d2 x
dx
M 2 +f
+ kx = 0
(9)
dt
dt
By assumption, the second term involving the …rst order derivative is small. Therefore, we
may assume for x(t);
x(t) = x0 e t cos !t;
!
5
where ! remains unchanged from the frictionless case, ! =
dx
=
dt
d2 x
=
dt2
'
p
x0 ( cos !t + ! sin !t)e
x0 [(! 2
2
) cos !t
x0 (! 2 cos !t
k=M : Noting
t
t
2 ! sin !t]e
2 ! sin !t)e
t
we …nd Eq. (9) reduces to
M x0 (! 2 cos !t
2 ! sin !t)e
t
f x0 ( cos !t + ! sin !t)e
t
+ kx0 e
t
cos !t = 0
This must hold at any time t: Therefore, the coe¢ cients of both cos and sin functions must
identically vanish. Recalling
!; we …nd
f
2M
The graph below shows damped oscillation when
=
e
y
0:005 t
= 0:0025!:
cos 2 t
1.0
0.8
0.6
0.4
0.2
0.0
10
20
30
40
50
-0.2
60
70
80
90
100
x
-0.4
-0.6
-0.8
-1.0
Example 4: Find (a) the e¤ective spring constant of two springs k1 and k2 connected in
tandem and then (b) the oscillation frequency when a mass M is attached.
6
Solution: Both springs experience the common force F ,
F = k1 l1 ;
where
F = k2 l2
l1 is the elongation of spring 1 and
l2 is that of spring 2. The total elongation is
l =
l1 + l2
1
1
=
+
F
k1 k2
k1 + k 2
=
F
k1 k 2
Therefore, the e¤ective spring constant is
ke¤ =
k 1 + k2
k 1 k2
and the oscillation frequency is
!=
r
ke¤
=
M
r
k1 k2 1
k1 + k2 M
Example 5: Find the oscillation frequency of the system shown.
Solution: The restoring forces provided by the springs are additive and the e¤ective spring
constant is k1 + k2 : The oscillation frequency is therefore given by
r
k 1 + k2
!=
M
Example 6: Find the oscillation frequency of the system shown. The cylinder (mass M ,
radius a) rolls on the ‡oor without slipping.
7
Solution: The moment of inertia of the cylinder about its axis is
1
IC = M a2
2
The translational velocity of the cylinder v is related to the rotational angular velocity d =dt
through
d
dx
=a
v(t) =
dt
dt
The sum of three energies (potential, kinetic, and rotational) is constant,
1 2 1
1
kx + M v 2 + IC
2
2
2
d
dt
2
= constant
or noting
1
IC
2
2
d
dt
1
= M v2
4
we have
1 2 3
kx + M v 2 = constant
2
4
Di¤erentiation with respect to time yields
3
dv
kxv + M v
=0
2
dt
which reduces to
d2 x 2 k
+
x=0
dt2
3M
Therefore, the oscillation frequency is given by
r
2 k
!=
3M
2.2 Ordinary Pendulum
Pendulum in a grand father’s clock is a familiar oscillation system. The restoring force
is provided by gravity and the mass M of the bob cancels between the inertia force M
dv=dt and the gravitational force M g: The oscillation frequency of a pendulum is practically
independent of the mass.
The motion of the bob is constrained along the arc determined by the length of pendulum
L and equation of motion for the mass may be written down as
d2
ML 2 =
dt
M g sin
where it is noted that the velocity of the bob along the arc is v(t) = L (t): The sin factor
is the component of the downward gravitational force in the direction as shown in Fig. 1.8
(p. 11). Cancelling the mass in both sides, we obtain
d2
g
+ sin = 0
2
dt
L
8
(10)
This is a nonlinear di¤erential equation for . If the oscillation angle
(rad), sin ' holds and Eq. (1) can be linearized as
is small, j j
d2
g
(t) = 0
+
dt2
L
1
(11)
which describes sinusoidal oscillation with a frequency
r
g
!=
(12)
L
p
If L = 1 m, the oscillation frequency is ! = 9:8 rad/sec = 3.13 rad/sec, or = 0:498 Hz.
The nonlinear equation Eq. (1) has been a subject of extensive studies in the past and its
solutions are now well understood. That the oscillation frequency should decrease with the
amplitude of oscillation can be seen from the following qualitative argument. The function
sin can be Taylor expanded as
1
3!
sin =
3
+
1
5!
5
:
Retaining up to the third order term and substituting into the original equation, we obtain
for 2 < 1 (not necessarily j j
1);
g
d2
+
2
dt
L
1
6
1
2
(13)
=0
This predicts an oscillation frequency which now depends on the amplitude,
s
2
g
0
!( 0 ) =
1
L
6
0;
A more detailed analysis only changes the numerical factor,
s
r
2
2
g
g
0
0
!( 0 ) =
1
'
1
L
8
L
16
(14)
It should be noted that the above result is still subject to the assumption of small oscillation
amplitude. For large amplitude approaching , the nonlinear equation has to be solved
without approximation. In the Web site, oscillations with small and large amplitudes are
shown in animation. Solutions for 0 = 0:1 (small amplitude) and 0 = =2 (large amplitude)
are shown below. The period when 0 = 0:1 is close to 2 but increases to about 6.67 when
0 = 1.
0
theta
0.10
0.05
0.00
-0.05
-0.10
2
4
6
8
10
= 0:1
12
9
14
16
18
20
22
24
26
28
30
t
0
= =2
y 00 + sin y = 0
y 0 (0) = 0
y (0) = =2
, Functions de…ned: y
y
theta
1.5
1.0
0.5
0.0
2
4
6
8
10
12
14
16
18
20
22
24
26
-0.5
28
30
t
-1.0
-1.5
Example 7:
Pendulum is still in wide use to measure the gravitational acceleration g
which varies from place to place. What is the % change in the oscillation frequency if g
changes by 1 %?
Solution: Let the change in g be
g): The oscillation frequency is
r
g+ g
! =
L
r s
g
g
=
1+
L
g
r
g
1 g
'
1+
;
L
2 g
g(
where use is made of the binomial expansion,
(1 + x)n ' 1 + nx;
jxj
1
If g=g = 0:01; the change in the oscillation frequency is approximately 0.5 %. Although
the change in the frequency is small, it manifests itself in the number of oscillations over
an extended period. For example, a pendulum having a length of L = 24:8 cm oscillates
at = 1 Hz or 3600 times per hour if g = 9:80 m/sec2 : 0.5 % change corresponds to 18
oscillations per hour which can be easily detected.
10
Example 8:
A pendulum on a boat or submarine is subject to unwanted (noise) acceleration as well as the gravitational acceleration. An ingenuous way is to use two pendulums
to get rid of unwanted acceleration. Explain how.
Solution: Let the unwanted force be f (t): The equations of motion for each pendulum are
d2 1
+g
dt2
d2 2
L 2 +g
dt
L
1
= f =M;
2
= f =M:
Subtraction yields
d2
L 2( 1
2 ) + g( 1
2 ) = 0;
dt
which means that the di¤erence between the two oscillation angles is not subject to the
unwanted force. This principle has made it possible to measure g in oceans.
2.3 Physical Pendulum
Any object freely pivoted becomes a pendulum. The oscillation frequency is determined
by the moment of inertia I (kg m2 ) about the pivot and the distance between the pivot and
the center of mass d;
r
M gd
(15)
!=
I
Although the mass M appears, it is cancelled because the moment of inertia I is proportional
to the total mass M: The case of ordinary pendulum can be recovered by noting that for a
lumped mass,
I = M L2
and thus
!=
r
M gL
=
M L2
r
g
L
Example 9: Calculate the oscillation frequency of a meter stick freely pivoted at its end.
Assume a small oscillation angle and g = 9:8 m s 2 .
Solution: The moment of inertia about the end of a stick having a length L is
Z
M L 2
x dx
I =
L 0
1
=
M L2
3
The distance between the pivot and center of mass is d = L=2: Therefore, the oscillation
frequency is given by
v
r
u
u M gL=2
3g
!=u
=
2
t ML
2L
3
11
If L = 1 m, we have
!=
Example 10:
hung.
r
3g
= 3:83 rad/sec or 0.61 Hz
2
Determine the oscillation frequency of a circular hoop of radius a freely
Solution: The moment of inertia about the pivot is
I = IC + M a2 = 2M a2
where
IC = M a2
is the moment about the center of the hoop.. The distance between the pivot and center of
mass is a: Therefore,
r
r
M ga
g
!=
=
2
2M a
2a
Example 11: The sweet point of baseball bat, tennis racket, etc. is the impact position
which gives a minimum reaction to the grip position. (a) For a physical pendulum with a
moment of inertia about the center of mass IC ; distance between the pivot and center of
mass d; and mass M; what is the length of a simple pendulum L which would have the
same oscillation frequency as the physical pendulum? (b) Show that the sweet point of the
physical pendulum is at distance L from the pivot.
Solution:
2. (a) The oscillation frequency of the physical pendulum is
r
M gd
!=
IC + M d2
where I = IC + M a2 is the moment of inertia about the pivot. From
r
r
M gd
g
!=
=
2
IC + M d
L
we …nd the length of a simple pendulum which has the same oscillation frequency
as the physical pendulum,
IC + M d2
L=
Md
(b) Suppose an impulse force F acts at the position L from the right. The pivot
experiences two forces, a translational leftward force equal to F; and clockwise
rotational force about the center of mass. If they cancel, the net force to act on
the pivot vanishes. The clockwise angular acceleration about the center of mass
is
(L d)F
=
IC
12
and the acceleration at the pivot is
a =
d
(L
=
d)dF
IC
IC +M d2
Md
=
d
IC
F
M
=
dF
to the right
Therefore, the rotational force exactly cancels the translational force.
Example 12: Sloshing Oscillation of Water. Water in a long lake or bay exhibits slow
sloshing oscillation. Its frequency is given by
r
12Hg
!=
a2
where a is the length and H is the water depth. Assume a sloshing amplitude h as shown.
The CM of the rectangle BCED is at
x = 0; y =
1
(H
2
h)
and that of the triangle ABC is at
1
x = a; y = H
6
2
h+ h=H
3
1
h
3
Then the CM of the whole water is at
xCM =
1
aH
ah
13
1
1a
a=
h
6
6H
1
a (H
aH
1 1 2
=
H +
H 2
yCM =
h)
1 2
h
6
1
(H h) + ah H
2
H 1 h2
=
+
2
6H
1
h
3
Deviation in y is
1 h2
H
= 6 2 x2
6H
a
The trajectory of (x; y) is a parabola and the CM executes harmonic oscillaltion with a
frequeny
r
12Hg
!=
a2
y=
14