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Department of PhysicsHong Kong Baptist University
Simple Harmonic Motion
Physics Enhancement Programme for Gifted Students
The Hong Kong Academy for Gifted Educationand
Department of Physics, HKBU
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Department of PhysicsHong Kong Baptist University
Simple harmonic motion
In mechanical physics, simple harmonic motion is a type of periodic
motion where the restoring force is directly proportional to the displacement. No
matter what the direction of the displacement, the force always acts in a
direction to restore the system to its equilibrium position. It can serve as
a mathematical model of a variety of motions, such as the oscillation of a spring,
motion of a simple pendulum as well as molecular vibration.
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Department of PhysicsHong Kong Baptist University
Mathematics of simple harmonic motion
Simple harmonic motion is a type of periodic motion which can use mathematical
model to express it.
)cos()( txtx m
Displacement(position) x(t)
Amplitude xm
Phase
Angular frequency
Frequency f
Period Tf
T1
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Department of PhysicsHong Kong Baptist University
Mathematics of simple harmonic motion
Since the motion returns to its initial value after one period T,
)])(cos[)cos( Ttxtx mm
)(2 Ttt
2T
fT
22
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Department of PhysicsHong Kong Baptist University
Mathematics of simple harmonic motion
Velocity)]cos([)( tx
dt
d
dt
dxtv m
)sin()( txtv m
Velocity amplitude v xm m
Acceleration
Acceleration amplitude
)]sin([)( txdt
d
dt
dvta m
)()cos()( 22 txtxta m
a xm m2
02
2
2
xdt
xd
Equation of motion
This equation of motion will be very useful in
identifying simple harmonic motion and its frequency.
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Department of PhysicsHong Kong Baptist University
Simple harmonic motion of spring
Simple harmonic motion of a mass on a spring is subject to the linear elastic
restoring force given by Hooke's Law. The motion is sinusoidal in time and
demonstrates a single resonant frequency.
For one-dimensional simple harmonic motion, the equation of motion, which is a
second-order linear ordinary differential equation with constant coefficients, could
be obtained by means of Newton's second law and Hooke's law.
where m is the inertial mass of the oscillating body, x is
its displacement from the equilibrium, and k is the spring constant.
kxdt
xdmmaF
kxF
2
2
02
2
xm
k
dt
xd
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Period
Angular frequency
Simple harmonic motion of spring
Comparing with the equation of motion for simple harmonic motion,
Simple harmonic motion is the motion executed by a particle of mass m subject to a
force that is proportional to the displacement of the particle but opposite in sign.
m
k2
m
k
k
mT 2
2T
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Department of PhysicsHong Kong Baptist University
A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 Nm-1. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0.What are the angular frequency, the frequency, and the period of the resulting oscillation?What is the amplitude of the oscillation?What is the maximum speed of the oscillating block?What is the magnitude of the maximum acceleration of the block?What is the phase constant for the motion?What is the displacement function x(t)?
Examples
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Department of PhysicsHong Kong Baptist University
1rads 78.968.0
65 m
k Hz 56.1
2
f s 643.0
1
fT
cm 11mx
1ms 08.1)11.0)(78.9( mm xv
222 ms 5.10)11.0()78.9( mm xa
11.0cos)0( mxx
0sin)0( mxv
0sin 0
)78.9cos(11.0)cos()( ttxtx m
(a)
(d)
(e) At t = 0,
(f)
(b)
(c)
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Department of PhysicsHong Kong Baptist University
At t = 0, the displacement of x(0) of the block in a linear oscillator is 8.50 cm. Its velocity v(0) then is 0.920 ms1, and its acceleration a(0) is +47.0 ms2.What are the angular frequency ?What is the phase constant and amplitude xm?
Examples
)cos()( txtx m)sin()( txtv m
)cos()( 2 txta m
085.0cos)0( mxx
920.0sin)0( mxv
0.47cos)0( 2 mxa
2
)0(
)0(
x
a1rads 5.23
0850.0
0.47
)0(
)0(
x
a
At t = 0, (1)
(2)
(3)
(3) (1):
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Department of PhysicsHong Kong Baptist University
tan
cos
sin
)0(
)0(
x
v
4603.0)085.0)(51.23(
920.0
)0(
)0(tan
x
v
o7.24 ooo 1557.24180
cos
)0(xxm cm 4.9m 094.0
7.24cos
085.0o
mx
cm 4.9m 094.0155cos
085.0o
mx
(2) (1):
or
(1): If = 24.7o,
If = 155o,
Since xm is positive, = 155o and xm = 9.4 cm.
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Energy in Simple Harmonic Motion
Potential energy
)cos()( txtx m
)(cos2
1
2
1)( 222 tkxkxtU m
Kinetic energy)sin()( txtv m
)(sin2
1
2
1)( 2222 txmmvtK m
mk /
).(sin2
1)( 22 tkxtK m
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Department of PhysicsHong Kong Baptist University
Energy in Simple Harmonic Motion
Mechanical energy
)(sin2
1)(cos
2
1 2222 tkxtkxKUE mm
)](sin)([cos2
1 222 ttkxm
1)](sin)([cos 22 tt
2
2
1mkxKUE
The mechanical energy is conserved !
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Department of PhysicsHong Kong Baptist University
Suppose the damper of a tall building has mass m = 2.72 105 kg and is designed to oscillate at frequency f = 10 Hz and with amplitude xm = 20 cm.(a) What is the total mechanical energy E of the damper?(b) What is the speed of the damper when it passes through the equilibrium point?
Examples
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Department of PhysicsHong Kong Baptist University
22 )2( fmmk
N 10073.1)20)(1072.2( 925
22
2
1
2
1kxmvUKE
29 )2.0)(10073.1(2
10
MJ 521J 10147.2 7 .
22
2
1
2
1kxmvUKE
0)1072.2(2
110147.2 257 v
1ms 6.12 v
(a)
The energy:
(b) Using the conservation of energy,
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Department of PhysicsHong Kong Baptist University
.
I ,
I ,
d
dt I
2
20
.
2 I
.
T 2
, T
I 2
.
An Angular Simple Harmonic Oscillator
When the suspension wire is twisted through an angle , the torsional pendulum produces a restoring torque given by
is called the torsion constant.Using Newton’s law for angular motion,
Comparing with the equation of motion for simple harmonic motion,
Since
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Department of PhysicsHong Kong Baptist University
A thin rod whose length L is 12.4 cm and whose mass m is 135 g is suspended at its midpoint from a long wire. Its period Ta of angular SHM is measured to be 2.53 s. An irregularly shaped object, which we call X, is then hung from the same wire, and its period Tb is found to be 4.76 s. What is the rotational inertia of object X about its suspension axis?
Example
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2
12
1MLIa
2)124.0)(135.0(12
1
24 kgm 107298.1
a
a
IT 2
b
b
IT 2
b
a
b
a
I
I
T
T
a
a
bb I
T
TI
2
244
2
kgm 1012.6)1073.1(53.2
76.4
Rotational inertia of the rod about the center
Since and
Thus
Therefore,
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Department of PhysicsHong Kong Baptist University
,sin 2 mLLmg
.0sin2
2
L
g
dt
d
.02
2
L
g
dt
d
.2
L
g
T 2
, T
L
g 2 .
The Simple Pendulum
The restoring torque about the point of suspension is = mg sin L.Using Newton’s law for angular motion, = I,
When the pendulum swings through a small angle, sin . Therefore
Comparing with the equation of motion for simple harmonic motion,
Since
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Department of PhysicsHong Kong Baptist University
mg h Isin ,
.0sin2
2
I
mgh
dt
d
.02
2
I
mgh
dt
d
.2
I
mgh
T 2
, T
I
mgh 2 .
The Physical Pendulum
The restoring torque about the point of suspension is = mg sin h.Using Newton’s law for angular motion, = I,
When the pendulum swings through a small angle, sin . Therefore
Comparing with the equation of motion for simple harmonic motion,
Since
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If the mass is concentrated at the center of mass C, such as in the simple pendulum, then
.2222
g
L
mgL
mL
mgh
IT
We recover the result for the simple pendulum.
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Department of PhysicsHong Kong Baptist University
A meter stick, suspended from one end, swings as a physical pendulum.(a) What is its period of oscillation T?(b) A simple pendulum oscillates with the same period as the stick. What is the length L0 of the simple pendulum?
Examples
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Department of PhysicsHong Kong Baptist University
2
3
1ML
mgh
IT 2
2/
3/2
2
mgL
mL
g
L
3
22
g
LT 02
g
L
g
L
3
222 0
cm 7.663
20 LL
(a) Rotational inertia of a rod about one end
Period
(b) For a simple pendulum of length L0,
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Department of PhysicsHong Kong Baptist University
A physical pendulum has a radius of gyration k. When it is suspended at distances l and l’ from the center of mass, the periods of oscillation are the same. (a) Find the relation between l and l’. (b) This has been used to determine g accurately. Find an expression for g.
Examples
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Department of PhysicsHong Kong Baptist University
22 MlMkI
gl
lk
Mgl
MlMkT
2222
22
'
'2'
22
gl
lkT
'
'2222
l
lk
l
lk
2222 ''' lllkllkl
2' kll
(a) When it is suspended at a distance l from the center of mass,
.
Similarly,
Equating T and T’,
g
ll
gl
lkT
'22
22
2
2 '4
T
llg
(b) Substituting into the expression of T,
,
.
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Department of PhysicsHong Kong Baptist University
A diver steps on the diving board and makes it move downwards. As the board rebounds back through the horizontal, she leaps upward and lands on the free end just as the board has completed 2.5 oscillations during the leap. (With such timing, the diver lands when the free end is moving downward with greatest speed. The landing then drives the free end down substantially, and the rebound catapults the diver high into the air.) Modeling the spring board as the rod-spring system (Fig. 15-12(d)), what is the required spring constant k? Given m = 20 kg, diver’s leaping time tfl = 0.62 s.
Examples
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Department of PhysicsHong Kong Baptist University
sin2kLkxL
2kL
I
2
222
3
1
dt
dmLkL
03
2
2
m
k
dt
d
m
k32
22 2
33
T
mmk
5.2
62.0
5.2
fltT
1
2
Nm 42805.2/62.0
2
3
20
k
When the board is displaced by an angle ,
The restoring torque:
Using Newton’s law for angular motion,
Comparing with the equation of motion for simple harmonic motion,
The period should be
Therefore
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Department of PhysicsHong Kong Baptist University
,bvFd
.makxbv
.02
2
kxdt
dxb
dt
xdm
),'cos()( 2/ textx mbt
m
.4
'2
2
m
b
m
k
mk /
kmb
Damped Simple Harmonic Motion
The liquid exerts a damping force proportional to the velocity. Then,
Using Newton’s second law,
Solution:
where
If b = 0, ’ reduces to
of the undamped oscillator.
, then ’ .
b = damping constant.
If
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mbt
mextx 2/)(
.2
1)( /2 mbt
mekxtE
The mechanical energy decreases exponentially with time.
The amplitude, , gradually decreases with time.
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Department of PhysicsHong Kong Baptist University
For the damped oscillator with m = 250 g, k = 85 Nm1, and b = 70 gs1.
(a) What is the period of the motion?
(b) How long does it take for the amplitude of the damped oscillations to drop to half
its initial value?
(c) How long foes it take for the mechanical energy to drop to half its initial value?
Example
s 34.085
25.022
k
mT
m
mbt
m xex2
12/
2
12/ mbte
2ln2
1ln
2
m
bt
s 95.407.0
)2)(ln25.0)(2(2ln2
b
mt
(a)
(b) When the amplitude drops by half,
Taking logarithm,
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2/2
2
1
2
1
2
1m
mbt
m kxekx
2
1/ mbte
2ln2
1ln
m
bt
s 48.207.0
)2)(ln25.0(2ln
b
mt
(c) When the energy drops by half,
Taking logarithm,
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Department of PhysicsHong Kong Baptist University
mk /
x t x tm d
( ) ). cos(
,cos2
2
tFkxdt
dxb
dt
xdm d
.cos)sin()cos()( 2 tFtbtmkxdddddm
BABABA sinsincoscos)cos(
,cos)cos()( 2222 tFtxbmkddmdd
.)(
cos2222
dd
d
bmk
b
,)( 222222
dd
m
bm
Fx
.
When a simple harmonic oscillator is driven by a periodic external force, we have forced
oscillations or driven oscillations.
Its behavior is determined by two angular frequencies:
(2) the angular frequency d of the external driving force.
The motion of the forced oscillator is given by
Substituting into the equation of motion,
Using the identity
where
and
Forced Oscillations and Resonance
(1) the natural angular frequency
Hence
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d .
(1) It oscillates at the angular frequency d of the external driving force.(2) Its amplitude xm is greatest when
This is called resonance.
See Youtube “Tacoma Bridge Disaster”.
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Department of PhysicsHong Kong Baptist University
k
x
1
x
2
kk0
m m
),(12011
xxkkxxm
).(21022
xxkkxxm
tiAex Re1
tiBex Re2
,0)(0
2
0 BkAmkk
.0)(0
2
0 AkBmkk
.0
2
0
2
0
0
k
mkk
mkk
k
B
A
,0
02
00
0
2
0
B
A
mkkk
kmkk
Two Coupled Oscillators and Normal Coordinates
Using Newton’s second law,
Possible solution, and
It is convenient to adopt the third trial solution. Then
For non-trivial solutions, we have
More generally, we can use the matrix form:
(A and B are complex.)
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.02
00
0
2
0
mkkk
kmkk
,0)( 2
0
22
0 kmkk
,2
m
k .
202
m
kk
,1
m
k ,BA
,2
0
2
m
kk .BA
and for non-trivial solutions,
Either way, we arrive at a secular equation,
or
If
If
Hence we obtain two solutions. In each solution, the two particles oscillate
with the same frequency. They are called normal modes. Their frequencies
are called normal frequencies. Any other solutions are combinations of the
normal modes.
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m
k
1 tAxx cos
21
m
kk0
1
2 tAx cos
1 tAx cos
2
Symmetric mode: and
Antiymmetric mode: and
.
In general, the mode that has the highest symmetry will have
the lowest frequency, while the antisymmetric mode has the
highest frequency.
Axx )0()0(21
0)0()0(21
xx
Axx )0()0(21
0)0()0(21
xx
The symmetric mode can be excited by pulling the two particles from their equilibrium
positions by equal amounts in the same direction so that
and
The antisymmetric mode can be excited by pulling apart the two particles
equally in opposite directions and then released, so that
and
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Department of PhysicsHong Kong Baptist University
)sin(sin 121 TmgFt
L
x11sin
L
xx 122sin
L
xx 1212
2)sin(
tFmL 1 )2( 1211 xx
L
mgx
L
mgxm
22 sin mgmL )( 122 xxL
mgxm
2
1
2
1
//
//3
x
x
LgLg
LgLg
x
x
Examples
Find the frequencies of small oscillations of a double pendulum.
Tangential component of forces acting on the upper particle:
For small oscillations, T mg,
Using Newton’s second law,
L
L
m
m1
2
x1
x2
L
L
m
m1
2
x1
x2
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tieB
A
x
x
Re
2
1
0//
//3
2
1
2
2
x
x
LgLg
LgLg
0//
//32
2
LgLg
LgLg
Lg /)22(2 AB 2
Lg /)22(2 AB 2
Let
Then
For non-trivial solutions,
Symmetric mode: and
Antisymmetric mode: and
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Department of PhysicsHong Kong Baptist University
L
g2
1
m
k
L
g 22
2
Consider two pendula of length L and mass m coupled by a spring with force
constant k. Find the normal frequencies, the normal modes and the general
solution.
Symmetric mode:
Antisymmetric mode:
, A = B
Examples
, A = B
m
LL
km
x1 x2
12
m
LL
km
x1 x2
12 LL
km
x1 x2
12
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