simple harmonic motion - hong kong physics …hkpho.phys.ust.hk/protected/past_notes/4...

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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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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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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

4



Since the motion returns to its initial value after one period T,

)])(cos[)cos( Ttxtx mm

)(2 Ttt

2T

fT

22

5



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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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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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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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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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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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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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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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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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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.

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,


Since

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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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,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


Since

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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


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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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

23


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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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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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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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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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,


The period should be

Therefore

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,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.

30


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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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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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


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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)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,


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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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

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simple harmonic motion - hong kong physics …hkpho.phys.ust.hk/protected/past_notes/4...

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