Lecture Note Chapter 15 Harmonic Oscillation 1 Simple harmonics 1.1 Equation of motion

We consider an equation of motion given by

xdt

xd

kxFdt

xdm

22

2

2

2

where

m

k

x

tBtAdt

xd

tBtAdt

dx

tBtAx

2

21

212

2

11

11

)sin()cos(

)cos()sin(

)sin()cos(

We note that x can be rewritten as

)cos()sin()cos( 11 tAtBtAx

where A, w and are constants, independent of time. The quantity A is called the amplitude of the motion and is the maximum displacement of the mass. The time-varying quantity (t + ) is called the phase of the motion and is called the phase constant. The phase constant is determined by the initial conditions.

The angular frequency is a characteristic of the system, and does not depend on the initial conditions. The unit of angular frequency is rad/s.

The period T of the motion is defined as the time required to complete-one oscillation. Therefore, the displacement x(t) must return to its initial value after one period

)()( Ttxtx

This is equivalent to

)cos()cos( TtAtA

Using the relation

)2cos()cos(

it is immediately clear that

2T

The number of oscillations carried out per second is called the frequency of the oscillation. The symbol for frequency is f and its unit is the Hertz (Hz):

1 Hz = 1 oscillation per second = 1 s-1

The period T and the frequency f are related as follows

fT

1

1.2 Energy conservation law

dt

dxkx

dt

xd

dt

dxm

2

2

or

)2

1(]

2

1[ 2

2

kxdt

d

dt

dxm

dt

d

or

0]2

1

2

1[ 2

2

kx

dt

dxm

dt

d

or

222

2

1

2

1

2

1kAkxmvUKE (energy-conservation law)

since v = 0 at x = A. or

2max

22

2

1

2

1

2

1mvkxmvUKE (energy conservation law)

since x = xmax at v = 0. ((Note)) The amplitude A can be estimated from the initial condition (x = x0, v = v0 at t = t0).

2

202

02

02

0 v

xvk

mxA

________________________________________________________________________ Now we consider

)(2

1 22 xAkK

((Note)) Derivation of the equation of motion from the energy conservation law

22

2

1

2

1kxxmE

Taking the derivative of this with respect to t, we have

0

0)(0

kxxm

kxxmxxkxxxmdt

dE

2 Simple harmonics 2.1 Configuration-1

12

222

111

)(

)(

ffxm

LxLkf

Lxkf

or

2221121

1122

)(

)()(

LkLkLkxkk

LxkLxLkxm

or

m

LkLkLkx

m

kkx 2221121 )(

The resultant spring constant k is

21 kkk 2.2 Configuration-2 Simple harmonics Problem 15-24*** (SP-15) (10-th edition)

12

21221

1111

)(

)(

fxm

Lxxkf

Lxkf

or

21

2211

21

221

221122121 )(

kk

LkLk

kk

xkx

LkLkxkxkk

Then we have an equation of motion for the mass m,

)(

)(

)(

)(

21221

21

1121

22111

21

221

1121

2211

21

221

1112

LLxkk

kk

Lkkk

LkLkk

kk

xkk

Lkkk

LkLk

kk

xkk

Lxkxm

The resultant spring constant k is

21

21

kk

kkk

The solution of this differential equation is given by

)cos(212 tALLx where

mkk

kk 1

21

21

2.3 Configuration-3

We now consider the case shown above.

)(

)(

22

11

21

Lxkf

Lxkf

ffxm

or

)()()( 221 LxLx

m

kkx

where

m

kk 21

The period T is given by

21

22

kk

mT

2.4 Mass hanging from the ceiling (advanced topics)

Equation of motion

)(

)()2(

3

21

321

axkf

xakaxakff

mgfffxm

or

mgaxkmgaxkxakxm )(3)()(2 At equilibrium,

k

mgax

mgaxk

3

0)(3

0

0

)(3

0xxm

kx

This indicates the simple harmonic oscillation with an angular frequency

m

k3

The solution for x is

)cos(0 tAxx

3. Two-body oscillations 3.1

We imagine the molecules to be represented by two particles of masses m1 and m2 connected by a spring force constant k, as shown below. Here we examine the motion of this system.

We apply the Newton’s second law to the system with two masses m1 and m2.

fxm 11 (1)

fxm 22 (2)

)( 12 Lxxkf From the calculation of m2 x Eq(1) –m1 x Eq(2), we have

))(()( 12212121 Lxxmmkxxmm Here we introduce a new variable x defined by

Lxxx 21

and the reduced mass defined as

21

21

mm

mm

Then we get

xxk

x 2

where

k

This indicates that two particles connected (with a spring constant k) can be replaced by a single particle (with a spring constant k) with a mass equal to the reduced mass of the system. 3.2 Longitudinal oscillations of two coupled masses We find the modes and their frequencies for the coupled springs and masses sliding on a frictionless surface. At equilibrium the springs are relaxed.

We set up equations of motion for mass m1 (= m) and m2 (= m).

)3(

)(

)(

213

1222

111

2322

1211

xakf

axxkf

axkf

ffxm

ffxm

or

akkxkxkkaxxkxakxm

akkxkkxkaxkaxxkxm

)3()()()3(

)()()()(

21122211222122

21121221112211

First we find the solutions of 021 xx

0)3()(

0)()(

210

120

221

210

1210

22

akkxkxkk

akkxkkxk

From these equations, we have

21

2102

21

2101

2

)53(

2

)3(

kk

kkax

kk

kkax

For convenience, we introduce new variables y1 and y2 defined by

0222

0111

xxy

xxy

Then

1222122

1212211

)(

)(

ykykkym

ykkykym

We assume that

22

2

12

1

yy

yy

which leads to

1222122

2

1212212

1

)(

)(

ykykkym

ykkykym

or

0)]([

0)]1([

2212

212

22122

1

ykkmyk

ykykkm

For the nontrivial solutions of y1 and y2, we have the condition

0)(

)(det

212

22

2212

1

kkmk

kkkm

When m1 = m2 = m, we have two mode frequencies

m

kkm

k

2122

121

2

((Physical meaning)) We start from the differential equations

122212

121221

)(

)(

ykykkym

ykkykym

These equations can be rewritten as

)()()2(

)(

)()()(

212

22121

212

2

212

1211

212

2

yyyym

kkyy

dt

d

yyyym

kyy

dt

d

or

)cos(

)cos(

22221

11121

tAyy

tAyy

Finally we get

)cos(2

)cos(2

)cos(2

)cos(2

222

111

2

222

111

1

tA

tA

y

tA

tA

y

In conclusion, there are two modes, since there are two degrees of freedom. 4 Simple pendulum

Ls

mgdt

sdm

sin2

2

or

sinsin 2

02

2

L

g

dt

d

In the limit of →0,

202

2

dt

d

or

)cos( 00 tA In the large , we have a nonlinear differential equation. We can solve the problem numerically using Mathematica (see the Section 12). ((Note)) A different derivation of the equation of motion for the simple pendulum is presented in Section 12. 5 Physical Pendulum 5.1

In the real world pendulums are far from simple. In general, the mass of the pendulum is not concentrated in one point, but will be distributed. Figure shows a physical pendulum. The physical pendulum is suspended through point O. The effect of the force of gravity can be replaced by the effect of a single force, whose magnitude is m g, acting on the center of gravity of the pendulum (which is equal to the center of mass if the gravitational acceleration is constant). The resulting torque (with respect to O) is given by

sinmgh

where h is the distance between the rotation axis and the center of gravity. In the limit of small angles ( ≈ 0), this torque can be rewritten as

mgh

The angular acceleration of the pendulum is related to the torque and the rotational inertia I

I

We therefore conclude that

I

mgh

dt

d

2

2

This is again the equation for harmonic motion with an angular frequency given by

I

mgh2

and a period equal to

mgh

IT

22

Note that the simple pendulum is a special case of the physical pendulum: h = L and I = m L2. The period of the oscillation is then given by

g

L

mgh

IT 22

5.2 Ring (hula hoop)

sinRMgI

where R is the distance between the pivot point and the center of mass. I is the moment of inertia around the pivot point P,

2222 2MRMRMRMRII cm (parallel-axis theorem)

In the limit of small angle , the motion undergoes a simple harmonic oscillation,

2 Here we have

I

RMg .

g

R

MgR

MR

MgR

IT

22

222

2

.

5.3 Bifilar pendulum

For simplicity, the extended mass is described by a simple rod in this figure. We have

2

Lh , and MgT

where = 0 or = 0. The equation of motion for the bifilar pendulum is described by

hMgL

h

LMgL

LMg

LTI

2

2sin

sin2

2

2)

2cos(2

2

or

2I (simple harmonic oscillation) where

Ih

MgL

T 2

2 2

.

The moment of inertia I can be derived as

h

TMgLI

2

22

8 .

6 The torsion pendulum

The operation of a torsion pendulum is associated with twisting a suspension wire. The motion described by the torsion pendulum is called angular simple harmonic motion. The restoring torque is given by

where is a torque constant that depends on the properties of the suspension wire (its length, diameter and material). This equation is essentially a torsional equivalent to Hooke's law. For a given torque we can calculate the angular acceleration

I or

IIdt

d

2

2

where I is the moment of inertia of the disk (about a perpendicular axis through its centre). Comparing this equation with the relation between the linear acceleration and the linear displacement of an object, we conclude that

I

2

The period of the torsion pendulum is given by

I

T 22

7 Damped oscillation

So far we have discussed systems in which the force is proportional to the displacement, but pointed in an opposite direction. In these cases, the motion of the system can be described by simple harmonic motion. However, if we include the friction force, the motion will not be simple harmonic anymore. The system will still oscillate, but its amplitude will slowly decrease over time.

We now consider the simple harmonics with a damping constant b,

0)()()( tkxtxbtxm with the initial conditions

0)0( vx and 0)0( xx

The differential equation can be written as

0)()()( txm

ktx

m

btx

We introduce

m

k0 ,

m

b

2 .

Then we get

0)()(2)( 20 txtxtx

The solution of this differential equation is classed into three types,

(1) underdamping: 020

2

(2) critical damping 020

2

(3) overdamping 020

2

The solution for the overdamping is given by

)]sin()cos([)( 1211 tCtCetx t with

2201

From the initial condition, we have

01 xC

1

002

xvC

The final form is given by

)cos(

)]sin()cos([)(

1max

11

0010

txe

txv

txetx

t

t

where is the phase factor and xmax is the amplitude. The period T1 is

2201

1

22

T

((Mathematica))

second-order differential equation for a simple harmonics with damping

Clear"Global`";eq1 x''t 2 x't 02 xt 0, x'0 v0, x0 x0;eq2 DSolveeq1, xt, t Simplify

xt 1

2 2 02t 202

1 2 t 202 v0

x0 1 2 t 202 1 2 t 202 2 02

xt_ Simplifyxt . eq21 . 2 02 12, 1 0;x1t_ Exp t xt ExpToTrig Simplify

x0 1 Cost 1 v0 x0 Sint 11

x11t_ Exp t x1tt x0 1 Cost 1 v0 x0 Sint 1

1

Limitx11t . 1 02 2 , 0 Simplifyt 0 x0 t v0 x0 0

v11t_ Dx11t, t Simplifyt v0 1 Cost 1 v0 x0 2 12 Sint 1

1

Limitv11t . 1 02 2 , 0 Simplify

t 0 v0 t v0 0 t x0 02

x2t_ x11t . 1 02 2 Simplify;

v2t_ v11t . 1 02 2 Simplify;

rule1 0 1, x0 1, v0 1;x3t_, _ x2t . rule1 Simplify;v3t_, _ Dx3t, , t . rule1 Simplify;PlotEvaluateTablex3t, , , 0.001, 1, 0.05,t, 0, 8 , PlotStyle TableHue0.051 i, Thick, i, 0, 20,AxesLabel "time", "amplitude",PlotRange 0, 8 , 1.5, 1.5, Background LightGray

5 10 15 20 25time

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5amplitude

PlotEvaluateTablex3t, , , 0.001, 1, 0.05,t, 0, 3 , PlotStyle TableHue0.051 i, Thick, i, 0, 20,AxesLabel "time", "amplitude",PlotRange 0, 3 , 1.5, 1.5, Background LightGray

2 4 6 8time

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5amplitude

PlotEvaluateTablex3t, , , 1.1, 3, 0.1, t, 0, 3 ,PlotStyle TableHue0.051 i, Thick, i, 0, 20,AxesLabel "time", "amplitude",PlotRange 0, 3 , 0, 1.5, Background LightGray

0 2 4 6 8time0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

amplitude

PlotEvaluateTablev3t, , , 0.001, 1, 0.1, t, 0, 4 ,PlotStyle TableHue0.051 i, Thick, i, 0, 10,AxesLabel "time", "velocity", Prolog AbsoluteThickness2,Background LightGray

2 4 6 8 10 12time

-1.0

-0.5

0.5

1.0

velocity

ParametricPlotEvaluateTablex3t, , v3t, , , 0.001, 1, 0.1,t, 0, 4 , PlotStyle TableHue0.051 i, Thick, i, 0, 10,AxesLabel "amplitude", "velocity", Background Gray,

AspectRatio 1

-1.0 -0.5 0.5 1.0amplitude

-1.0

-0.5

0.5

1.0

velocity

ParametricPlotEvaluateTablex2t, v2t . 1.1, 0 1, x0 Cos, v0 Sin,, 0, 2 , 10, t, 0, 4 ,

PlotStyle TableHue0.051 i, Thick, i, 0, 10,AxesLabel "t", "v", Background Gray, PlotRange All

-1.0 -0.5 0.5 1.0t

-1.0

-0.5

0.5

1.0

v

ParametricPlotEvaluateTablex2t, v2t . 1.1, 0 1, x0 Cos, v0 Sin,, 0, 2 , 10, t, 0, 4 ,

PlotStyle TableHue0.051 i, Thickness0.01, i, 0, 10,AxesLabel "x", "v", Prolog AbsoluteThickness2,Background Gray, PlotRange All

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

v

ParametricPlotEvaluateTablex2t, v2t . 0.5, 0 1, x0 Cos, v0 Sin,, 0, 2 , 10, t, 0, 4 ,

PlotStyle TableHue0.051 i, Thick, i, 0, 10,AxesLabel "x", "v", Background Gray, PlotRange All

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

v

ParametricPlotEvaluateTablex2t, v2t . 2, 0 1, x0 Cos, v0 Sin,, 0, 2 , 10, t, 0, 4 ,

PlotStyle TableHue0.051 i, Thick, i, 0, 10,AxesLabel "x", "v", Background Gray, PlotRange All

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

v

Total energy which cahnges with time

E1 1

2m v3t, 2

1

2k x3t, 2 . k 1, m 1;

PlotEvaluateTableE1 . rule1, , 0.01, 1.5, 0.1,t, 0, 4 , PlotStyle TableHue0.08 i, Thick, i, 0, 10,AxesLabel "t", "E", Background LightGray, PlotRange All

2 4 6 8 10 12t

0.2

0.4

0.6

0.8

1.0

E

Overdamping, phase space

ratiot_ :v2tx2t . 2 02 p TrigToExp FullSimplify

ratiotp v0 Coshp t v0 x0 02 Sinhp tp x0 Coshp t v0 x0 Sinhp t

A1 Limitratiot, t , Assumptions p 0 . p 2 02

v0 x0 02 v0 2 02

v0 x0 2 02

Solvea A1 . v0 a x0, aa 2 02 , a 2 02

8 Forced oscillation (steady state solution)

)cos()()('2)(" 02

0 ttxtxtx

We assume that x(t) can be given by

]Re[)( tiAetx Re denotes a real part. A is in general a complex number. i (= 1 ) is a pure imaginary ((Note)) Euler’s equation

sincos iei R.P. Feynman: This is the most remarkable formula. This is our jewel (22-10 volume-1, Feynman’s lecture on physics.) Then we have

]Re[])2Re[()()(2)( 02

022

0titi eAeitxtxtx

or

iei

A

22220

2

02

02

0

4)(2)(

Then x is obtained as

)cos(4)(

]4)(

Re[]Re[)(

22220

2

0

)(

22220

2

0

t

eAetx titi

or

20

2

20

22

20

22

0

22220

20

4)1(

1

4)(

1

A

or

2222

20

2

20

22

20

2

20

0 4)1(

1

4)1(

1

xx

AY

Now we calculate the value of Y as a function of x when a parameter is changed.

0

0

x

((Mathematica))

Y 1

x2 12 4 2 x2

1

1 x22 4 x2 2

PlotEvaluateTableY, , 0, 1, 0.02, x, 0.5, 1.5,PlotRange 0.5, 1.5, 0, 20,PlotStyle TableHue0.1 i, Thick, i, 0, 10,Background LightGray, AxesLabel "

0", "Y"

0.6 0.8 1.0 1.2 1.4

w

w0

5

10

15

20Y

Y vs 0

where (= 0

) is changed as a parameter. = 0 – 1.0.

((Note)) Simple explanation for the resonance

We consider the special case (b = 0, or 0 ).

ttxtx cos)()( 02

0

We assume that the solution of x(t) is given by

tAtx cos)( 0

Then we have

02

002

0 AA

or

220

00

A

We note that

220

00

A

becomes divergent as approaches 0. 9. Energy consideration in the forced oscillation

We start from

)cos()()(2)( 02

0 ttxtxtx .

Multiplying )(tx on both sides, we have

)()cos()()()]([2)()(' 02

02 txttxtxtxtxtx ,

or

)()cos()]([2})]([2

)(2

1{ 0

222

02 txttxtxtxdt

d

.

This equation can be rewritten as

tt

dttxtdttxtxtxtxtx0

1110

0

12

12

2022

202 )()cos()]([2)]0([

2)0(

2

1)]([

2)(

2

1{

Here we introduce the instantaneous energy (t) which is defined by

22

02 )]([2

)(2

1)( txtxt

.

We take an average of the above equation over a one period T,

0)()cos(1

)]([21

)}0()({1

0

1110

0

12

1 TT

dttxtT

dttxT

tTtT

.

We now calculate the second and third terms using our steady-state solution

])(Re[)(

]Re[)(ti

ti

eiAtx

Aetx

,

with

"'4)(

]2)[(

2)( 22220

2

20

20

20

20

ii

iA

and

22220

2

20

20

20

20

4)(

]2)([

2)(

i

i

iiA

where ' and " are the real part and imaginary part of A. The calculation of the second term:

"4

4)(2

2

)]()(2[4

2

4

])()(][)()([2)]('[2

1

20

22220

2

32

0

32

22

*

0

1

**

0

12

1

1111

A

AT

iAiAT

dteiAeiAeiAeiA

Tdttx

T

T titititiT

The calculation of the third term:

"4

4)(2

4)(

2

2

)](Re[2

)]()([4

4

])()()[()(')cos(

1

20

22220

2

32

0

22220

200

0

*0

0

1

*0

0

1110

1111

T

iAT

iAiAT

dteiAeiAee

Tdttxt

T

T titititiT

Then it is found that the second term is equal to the third term. These terms are proportional to " (imaginary part of A). The energy absorbed by the system from the external force is dissipated through the resistive damping. Then we have

)0()( tTt .

The sum of the kinetic energy and the potential energy is a periodic function of t with a period of T. 10 Example 10.1 Example-1: Problem 15-26***(SP-15) (10-th edition)

In Fig., two blocks (m = 1.8 kg and M = 10 kg) and a spring (k = 200 N/m) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.40. What amplitude of simple harmonic motion of the spring–blocks system puts the smaller block on the verge of slipping over the larger block?

FIGURE 15-34 Problem 24.

m = 1.8 kg, M = 10 kg, k = 200 N/m

For the small mass,

mgNf

mgN

fxm

ss

11

1

1

(1)

For the large mass,

gmMMgNN

kxfxM

)(12

1

(2)

From Eqs.(1) and (2),

xx

kxxmM2

)(

with

mM

k

2

The solution of the differential equation is

)cos( tAx with srad /11.4

mgmM

mkxxmf

mgN

s

1

1

or

mmMk

gx s 23.0)(

10.2 Example-2 Problem 15-25*** (10-th edition) In Fig., a block weighing 14.0 N, which can slide without friction on an incline at angle = 40°, is connected to the top of the incline by a massless spring of un-stretched length 0.450 m and spring constant 120 N/m. (a) How far from the top of the incline is the

block’s equilibrium point? (b) If the block is pulled slightly down the incline and released, what is the period of the resulting oscillations?

mg = 14.0 N, = 40°, x0 = 0.450 m, k = 120 N/m

(a)

)(

sin

01 xxkf

fmg

or

mk

mgxx 525.0sin01

(b)

)(

sin

0xxkf

fmgxm

or

)()sin( 10 xxm

k

k

mgxx

m

kx

The solution of this equation is

)cos(1 tAxx

with m

k

The period T is sT 686.02

10.3 Example-3 Simple harmonics in fluid mechanics

In equilibrium

MgghAFb 0

or

A

Mh 0

Simple harmonics

MgAgyhMgGyM b )( 0

where Gb is the buoyant force.

yyM

Agy

AgyMgAgyhyM

2

0 )(

(simple harmonics)

where

Ag

MT

M

Ag

2

10.4 Simple harmonics in fluid mechanics

V: volume of liquid with the density m = V: mass of liquid L = V/A: total length of the liquid column

LAVm

AgyFym

)2(

or

yyL

g

LA

Agyy 22)2(

where

L

g2 ,

g

LT

22

2

((Example)) Oscillation of liquid in a U-tube

The mass m (= 9 kg) of mercury is poured into a glass U-tube as shown in Fig. The tube’s inner diameter is 1.2 cm and the mercury oscillates freely up and down about its position of equilibrium (x = 0). Compute (a) the effective spring constant k for the oscillation, and (b) the period of oscillation. The density of mercury is = 13.6 x 103 kg/m3. Ignore frictional and surface tension effects.

((Solution)) r = 0.6 cm. = 13.6 x 103 kg/m3. m = 9.0 kg. When the mercury is displace x from its equilibrium position, the restoring force is the weight of the unbalanced column of mercury with height 2x. The restoring force F is described by

gxAF )2( , where A is the area of U-tube (A = r2). Then the mercury undergoes a simple harmonic oscillation,

kxgxAFxm )2( , where

m

gr

m

Agk

222 = 30 N/m.

This equation can be rewritten as

xx 2

where w is the angular frequency,

m

k .

The period of the oscillation, T, is

k

mT 2 = 3.4 s.

10.5 Example-4 (Serway 12-75)

Imagine that a hole is drilled through the center of the Earth to the other side. An object of mass m at a distance r from the center of the Earth is pulled toward the center of the Earth only by the mass within the sphere of radius r (the reddish region in Fig.). (a) Write Newton’s second law of gravitation for an object at the distance r from the

center of the Earth, and show that the force on it is of Hooke’s law form krF , where the effective force constant is Gmk )3/4( . Here r is the density of the Earth assumed uniform, and G is the gravitational constant.

(b) Show that a sack of mail dropped into the hole will execute simple harmonic motion if it moves without friction.

(c) When will it arrive at the other side of the Earth.

3

3

3

3

3

43

4

R

r

M

M

rM

RM

E

r

r

E

The force is directed toward the center.

rFrrR

GmMr

R

r

r

GmMr

r

GmMr

E

E

E

Er ˆˆˆˆ33

3

22F

The equation of motion for the particle on the tunnel along the x-axis.

xR

GmMr

R

GmMFxm

E

E

E

Er 33 coscos

or

xx 2 (Simple harmonics) where

sg

RT

R

g

R

GM

E

EE

E

43.506122

3

or

min2.427.25302

sT

((Note)) Period of satellite

sec21241506122

/17640/9.4/910.7

initueshoursg

R

v

rT

hmilessmilesskmR

GMv

E

E

E

10.6 Example-5 Problem 15-41 (SP-15) (10-th edition)

In Fig., the pendulum consists of a uniform disk with radius r = 10.0 cm and mass 500 g attached to a uniform rod with length L = 500 mm and mass 270 g. (a) Calculate the rotational inertia of the pendulum about the pivot point. (b) What is the distance between the pivot and the center of mass of the pendulum? (c) Calculate the period of oscillation.

2222

222

2

205.0)(2

1

3

1

)(2

1

212

1

kgmrLMMrmL

rLMMrL

mmLI

where M is the mass of disk and m is the mass of the rod The equation of motion

gL

mrLM

gL

mrLMI

]2

)([

sin]2

)([

10.7 Example-6 Problem 15-51** (SP-15) (10-th edition)

In Fig. a stick of length L = 1.85 m oscillates as a physical pendulum. (a) What value of distance x between the stick’s center of mass and its pivot point O gives the least period? (b) What is the least period?

220

0

12

1

sin

MxMLI

MgxI

or

2I

mgx (simple harmonics)

where is the angle between the vertical axis and the thin rod, and is defined by

2222

12

1

12

1xL

gx

mxmL

mgx

.

10.8 Example-7 Problem 15-53** (HW-15) (10-th edition)

In the overhead view of Fig., a long uniform rod of mass 0.600 kg is free to rotate in a horizontal plane about a vertical axis through its center. A spring with force constant k = 1850 N/m is connected horizontally between one end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. What is the period of the small oscillations that result when the rod is rotated slightly and released?

4)2sin(

8

)2sin(2

1

4cos

2sin

222

2

kLkLI

LkL

Lk

II

10.9 Example-8

Equation of motion

)( 0xxkf

mgfxm

or

)( 1xxkxm where

k

mgxx 01

The solution of the differential equation is given by

)cos(

)(

1

12

tAxx

xxx

with

m

k

10.10 Problem 15-52** (SP-15) (10-th edition)

The 3.00 kg cube in Fig. has edge lengths d = 6.00 cm and is mounted on an axel through its center. A spring (k = 1200 N/m) connects the cube’s upper corner to a rigid wall. Initially the spring is at its rest length. If the cube is rotated 3° and released, what is the period of the resulting simple harmonic oscillation?

((Solution)) k = 1200 N/m M = 3 kg The moment of inertia is

222

6

1)(

12

1MdddMI

around the axis through its center for the cube with the side d. The equation of motion:

2

2

1)

2

2()

2

2()

2

2( dkddkdkxI

or

22

2

1

6

1dkMd

or

23

M

k (simple harmonics)

where

M

k3 = 34.641 rad/s

The period T is given by

sk

MT 181.0

32

The initial condition: (0) = 3° and 0)0( Using this condition, we get the time dependence of as

0)0(

)0(

)cos()sin()(

)sin()cos()(

B

A

tBtAt

tBtAt

or

)cos()0()( tt 12. Advanced problems 12.1 Serway 15-73 ((numerical calculation))

Consider a bob on a light stiff rod, forming a simple pendulum of length L = 1.20 m. It is displaced from the vertical by an angle max and then released. Predict the subsequent angular positions if max is small or if it is large. Proceed as follows: Set up and carry out a numerical method to integrate the equation of motion for the simple pendulum;

)(sin)(

2

2

tL

g

dt

td

.

Take the initial conditions to be = max and d/dt = 0 at t = 0. On one trial, choose max = 5.00 , and on another trial take max = 100 . In each case, find the position (t) as a function of time t. Using the same values of max, compare your results for with those obtained from (t) = max cos(t). How does the period for large values of max compares with that for the small value of max? ((Solution)) We use the Mathematica for the solution.

2 4 6 8 10t

-1.0

-0.5

0.5

1.0

qtqmax

qmax = 100°

Fig.1 Red: (t)/max for max = 100°: Green: cos(0 t), where L

g0 2.8577 rad/s.

2 4 6 8 10t

-1.0

-0.5

0.5

1.0

qtqmax

qmax = 100°

qmax = 5°

Fig.2 Red: (t)/max for max = 100°. Blue: (t)/max for max = 5.00°:, where

L

g0 2.8577 rad/s.

12.2 Serway 15-56

A solid sphere (radius = R) rolls without slipping in a cylindrical trough (radius = 5 R) as shown in Fig. Show that, for a small displacements from equilibrium perpendicular to the length of the trough, the sphere executes simple harmonics with a period.

g

RT

5

282

((Solution))

Fig. OO1 = 4R, O1H = R. s = R = 4R . Kinetic energy K;

22

2

1

2

1 ImvK c .

I is the moment of inertia for the sphere with radius R, 2

52 mRI . is the angular

velocity, = d/dt. From the rollinc of the ball without slipping, we have

RRs 4 , or

RRdt

dsvc 4 .

Then K can be rewritten as

222222222

5

56)4(

10

7

10

7

5

2

2

1)(

2

1 mRmRmRmRRmK .

The potential energy U is given by

)cos45( RRmgU , where the reference point is at the bottom of the cylindrical trough. Since the total energy E is independent of t, we have

)cos45(5

56 22 RRmgmRUKE =const

or

0)]sin4(5

112[ 2 RmgmR

This is a differential equation for the simple harmonics, when sin .

02 with

R

g

28

5

The period T is

g

RT

5

282

2

13. Exact measurement of the period for the simple pendulum with small

maximum angle "The history of the physics of the pendulum stretches back to the early moments of modern science itself. We might begin with the story, perhaps apocryphal, of Galileo’s observation of the swinging chandeliers in the cathedral at Pisa. By using his own heart

rate as a clock, Galileo presumably made the quantitative observation that, for a given pendulum, the time or period of a swing was independent of the amplitude of the pendulum’s displacement. Like many other seminal observations in science, this one was only an approximation of reality. Yet it had the main ingredients of the scientific enterprise; observation, analysis, and conclusion. Galileo was one of the first of the modern scientists, and the pendulum was among the first objects of scientific enquiry." (G.L. Baker and J.A. Blackburn, Oxford University Press, 2005).

q

qmax

A

O

Fig. Simple pendulum with a point mass m. max is the maximum of the angle .

We consider the motion of the simple pendulum. The kinetic energy is given by

2)(2

1 lmK ,

The potential energy:

)cos1( mglU

The energy conservation

)cos1(2

1 22 mglmlUKE

or

constl

g )cos1(

2

1 2

with

We now start with

)cos1()cos1(2

1max

2

where

l

g ,

We note that (i) = 0 for = max (ii) = 0 for = So we have

)cos1(2

1)cos1(

2

1max

20

2 .

From this, we get

)cos1()cos1(2

1max

2

or

)cos(cos max22

Using a formula,

2sin21cos 2

we have

2sin

2sin2 2max2

dt

d

Then the period T is given by

max

0 2max2

2sin

2sin

2

1

4

dT

.

Here we put

2sin

1

2sin

2sin

max

kz

with

2sin max

k .

Note that

dzk

kd

kd

kdz 222 1

2

1

2sin1

2

1

2cos

2

1

Then we have

)(4

11

4

11

222 21

0222

1

0222

kKzkz

dz

zkzk

kdzT

or

4

T EllipticK[k2]

where

l

g ,

2sin max22

k

Note that K() is the complete elliptic integral of the first kind and is defined by

1

022 11

)(zz

dzK

In the Mathematica, this function corresponds to EllipticK[]. When 0k , we have

g

lT

22

The series expansion of T around k = 0 is given by

....65536

3969.

16384

1225

256

25

64

9

4

11(

2 108642 kkkkkT

We make a plot of the deviation (T - T0)/T0 (denoted by %) as a function of max, where

g

lT

22

0

It is clear that the deviation starts to occur when max is larger than max 2°.

TT0-1 %

qmaxdegrees5 10 15 20

0.2

0.4

0.6

0.8

Fig. The deviation T/T0 x 100 (%) as a function of the maximum angle max . The

deviation is defined by that 000 /)(/ TTTTT .

We note that the period is independent of max only for a few degrees. The period

becomes dependent of max as max increases. The deviation is 0.19 % for max = 10° and

0.77% for max = 20°. So if one want to measure the exact period (T0), one needs to use

the small value of max below a few degrees.

________________________________________________________________________ Table The value of the deviation 0/TT x 100 % for typical values of max where

0TTT

max (T/T0 - 1) x 100 %

_______________________________________________________________________ 14. Simple pendulum in an accelerated reference frame

We consider a simple pendulum in an accelerated frame using the two examples. These examples come from (P.T. Tipler and G. Mosca, Physics for Scientists and Engineers, 6-th edition W.H. Freeman and Company, 2008) Chapter 14. ((Example-1)) A simple pendulum suspended in the moving cart (with

constant acceleration) A simple pendulum of length L suspended from the ceiling of a cart (C) that has

acceleration CGa . Find the period of oscillation for small oscillations of this pendulum.

We start with an equation of motion given by

BGmm agT

where T is the tension and m is the mass of bob of the simple pendulum. The acceleration of the bob relative to the ramp is equal to the acceleration of the bob(B) relative to the cart plus the acceleration of the cart (C) relative to the ground (G),

CGBCBG aaa .

Then we get

)( CGBCmm aagT ,

or

BCCG mm aagT )( .

We define effg as

CGeff agg .

Then we have

BCeff mm agT

geff g

aCR

O

PQ

Fig. CGeff agg . 22CGeff agg . QOP .

The magnitude of geff is obtained as

22CGeff agg

where

g

aCGtan

T

mgeff

An equation for the motion of the bob draw the free-body diagram is given by

sineffmgml

where l is the length of the string. This can be written as

2sin l

g

l

g effeff

where

l

geff2

Then the period T is obtained as

effg

lT 2

with

22CGeff agg

((Example-2))

Simple pendulum on the moving cart (with constant acceleration) on the ramp

A simple pendulum of length L is attached to a massive cart that slides without friction down a plane inclined at angle with the horizontal, as shown in Figure. Find the period of oscillation for small oscillations of this pendulum (P.T. Tipler and G. Mosca, Physics for Scientists and Engineers, 6-th edition W.H. Freeman and Company, 2008) Problem 14-65 p.491

The cart accelerates down the ramp with a constant acceleration of sing . This

happens because the cart is much more massive than the bob, so the motion of the cart is unaffected by the motion of the bob oscillating back and forth. The path of the bob is quite complex in the reference frame of the ramp, but in the reference frame moving with the cart the path of the bob is much simpler—in this frame the bob moves back and forth along a circular arc. To solve this problem we first apply Newton’s second law (to the

bob) in the inertial reference frame of the ramp. Then we transform to the reference frame moving with the cart in order to exploit the simplicity of the motion in that frame.

T

mgeff

Apply Newton’s 2nd law to the bob, labeling the acceleration of the bob (B) relative to

the ramp (R), BRa

BRmm agT

The acceleration of the bob relative to the ramp is equal to the acceleration of the bob relative to the cart plus the acceleration of the cart relative to the ramp,

CRBCBR aaa

Then we get

)( CRBCmm aagT

or

BCCR mm aagT )(

We define effg as

cReffg ag

Then we have

BCeff mm agT .

The magnitude of geff is obtained as

)cos(ggeff ,

from the geometry of the figure below.

geffg

aCR

O

P

Q

Fig. CReff agg . singPQ CR a . cosggOQ eff . 90OQP .

Suppose that 0BCa

effmgT

An equation for the motion of the bob draw the free-body diagram is given by

sineffmgml ,

where l is the length of the string. This can be written as

2cossin

l

g

l

geff ,

where

l

g cos2 .

Then the period T is obtained as

cos22

g

l

g

lT

eff

.

When = 0, T is equal to the conventional value of T,

g

lT 2 .

15 Link Tacoma Narrows Bridge collapse (forced oscillation) http://www.youtube.com/watch?v=j-zczJXSxnw Simple harmonics motion Wikipedia http://en.wikipedia.org/wiki/Simple_harmonic_motion Simple pendulum: Physics of simple pendulum, a case study of nonlinear dynamics http://physics.binghamton.edu/Sei_Suzuki/suzuki.html Lecture Note (University of Rochester) http://teacher.pas.rochester.edu/phy121/LectureNotes/Contents.html Appendix Nonlinear oscillation (Challenging topics) A.1 Formulation

We consider the motion of mass m hanging from a ceiling with a string. The mass of the string is neglected. We set up an equation of motion,

sinmglI (1) where I is the moment of inertia and is given by

2mlI Equation (1) can be rewritten as

0sin where

20

l

g. or

l

g0

For small ,

0...)1206

(33

20

The solution of this equation is given by the nonlinear terms

...)5cos()3cos()cos( 531 ttt

where 321 and is a phase factor;

19224

1

9

1

)16

1()16

1(22

)8

11(

313

12

020

23

2

0

2

0

21

20

2

TT

with

g

lT

22

00 .

In the limit of →0,

020

corresponding to a simple harmonics,

)cos( 01 t

A.2 Energy conservation

The multiplication of on both sides and integration with respect to t lead to

0sin or

0)]cos1(2

1[ 2

dt

d

or

)cos1(2

1

)cos1(2

1

2

2

v

E

where the first term is the kinetic energy and the second term is a potential energy defined by

)cos1( U

-3p -2p -p 0 p 2p 3pq

1

2

3Ue

For small , U can be approximated by

...)40320720242

(8642

U

Here we use an initial condition

0

0

)0(

)0(

vt

t

Then the total energy E is expressed by

)cos1(2

10

20 vE

In other words, E increases as v0 increases and it also increases as 0. increases. A.3 Phase space of vs v = d/dt

From v sinv

we have

sinv

v

d

dv

This is a differential equation (separation variable type). So we can solve easily as

Cv

dvdv

cos2

1

sin

2

From the energy conservation law,

CvE )cos1(2

1 2

or

EC (a) ContourPlot of

Ev )cos1(2

1 2

in the vs v plane, where E = 2 and the parameter is varied as a parameter. The character of the orbit inside and outside of the curve are rather different. Such a family of curve is called sepratrices.

(b) ContourPlot of

Ev )cos1(2

1 2

in the vs v plane, where = 1 and the parameter E is changed as a parameter.

Fig, Phase space (v vs ) for = 1. E = 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, and 3.5. E = 2 = 2 is

the highest energy of the potential. B. Numerical calculation using Mathematica for simple pendulum B.1 Formulation

For convenience, the differential equation is separated into two parts

v sinv 1. First we solve this differential equation using numerical method (Mathematica,

NDSolve) 2. We analyze the Fourier component by using fast Fourier transform (FFT)

(Mathematica , Fourier).

a. We choose the N (= 2n) data, typically the data between 0 and T (= tmax). b. The minimum time division is NT / , which corresponds to the maximum

wmax =

NT

2max

c. The Fourier spectrum is plotted as a function of the channel number scaled by (2p)/T.

For simplicity, we consider the case of 0 = 0. Then we have

202

1vE

When v0<2 (orE<2, the value of is limited between – and . When

v0>2 (orE>2, the value of is unlimited. B.2 Time dependence ( vs t)

We show the time dependence of as a function of t with v0 being changed as a parameter. Here we choose. = 1 (a) v0 = 0.1, 0.2, 0.3, 0.4, 0.5

10 20 30 40 50t

-p6

p6

q

(b) v0 = 1.4, 1.5, 1.6

10 20 30 40 50 60 70 80 90 100t

-p4

p4

q

(c) v0 = 1.91, 1.92, 1.93

10 20 30 40 50 60 70 80 90 100t

-3 p4

-p2

-p4

p4

p2

3 p4

q

(d) v0 = 1.982, 1.984, 1.986, 1.988, 1.990

10 20 30 40 50 60 70 80 90 100t

-3 p4

-p2

-p4

p4

p2

3 p4

q

(e) v0 = 1.9992, 1.9993, 1.9994, 1.9995, 1.9996, 1.9997

10 20 30 40 50 60 70 80 90 100t

-p

-3 p4

-p2

-p4

p4

p2

3 p4

p

q

(f) v0 = 1.999992, 1.999993, 1.999994, 1.999995, 1.999996, 1.999997, 1.999998,

1.999999

As v0 approaches 2.0, the shape of vs t curve becomes square-like. This means that the effect of the nonlinearity is enhanced.

10 20 30 40 50 60 70 80 90 100t

-p

-p2

p2

p

q

(g) v0 = 2

In this case, the total energy E is equal to the peak height of the potential energy.

100 200 300 400 500t

-5p

-4p

-3p

-2p

-p

p

2p

3p

q

(h) v0 = 2.00001, 2.00002, 2.00003, 2.00004, 2.0005, 2.0006, 2.0007, 2.00008,

2.00009, 2.0001 For v0>2, the value of increases with increasing t. In other words, the pendulum

rotates around the rotational axis.

10 20 30 40 50 60 70 80 90 100t

5p

10p

15p

q

(i) v0 = 2.02, 2.03, 2.04, 2.05, 2.07. 2.08,

10 20 30 40 50 60 70 80 90 100t

10p

20p

30p

q

B.3 Phase space (v vs )

We make a plot of the phase space for = 1, when v0 is changed as a parameter. (a) v0 = 1.0, 1.1, 1.2, 1.3, 1.4, 1.5

-p4

p4

q

-1

1

v

(b) v0 = 1.6, 1.7, 1.8, 1.9

-3 p4

- p2

- p4

p4

p2

3 p4

q

-2

-1

1

2v

(c) v0 = 1.91, 1.92, 1.93, 1.94, 1.95, 1.96, 1.97

- 3 p4

- p2

- p4

p4

p2

3 p4

q

-2

-1

1

2

v

(d) v0 = 1.980, 1,981, 1.982, 1.983, 1.984, 1.985, 1.986, 1.987, 1.988, 1.989, 1.990.

- 3 p4

- p2

- p4

p4

p2

3 p4

q

-2

-1

1

2

v

(e) v0 = 2.0

-p p 2p 3pq

-1

1

v

(f) v0 = 2.01

p 2 p 3p 4p 5 pq

0

1

2

v

B.4 Fourier analysis using Fast Fourier transform (FFT)

The time dependence of q can be analyzed using the FFT analysis. The Fourier spectrum is the square of th absolute amplitude as a function of angular frequency . In the small limit of vo = 0,

0 .

In the present case, 0 = 1 since = 1.

As vo approaches 2 , the time dependence of changes from sinusoidal wave to a square wave, indicating the enhancement of nonlinearity in the system. There appear 3, 5, 7, .. components. The frequency of the fundamental mode rapidly decreases. = 1. (a) v0 = 0.5 (≈ 1) component is mainly observed.

0 1 2 3 4 510-5

0.001

0.1

10

1000

Fig. Fourier spectrum obtained from the FFT analysis of the vs t. The x axis is the

angular frequency w and the y axis is the squares of the absolute value of complex amplitude. The fundamental harmonic is observed at = 1.

(b) v0 =1.6 and 3 component are mainly observed.

0 1 2 3 4 50.001

0.01

0.1

1

10

100

1000

(c) v0 = 1.9 , 3, and 5 component are mainly observed.

0 1 2 3 4 50.001

0.01

0.1

1

10

100

1000

(d) v0 = 1.981 , 3, 5, and 7 component are mainly observed.

0 1 2 3 4 50.001

0.01

0.1

1

10

100

1000

(e) v0 = 1.9962 , 3, 5, 7, and 9 component are mainly observed.

0 1 2 3 4 50.001

0.01

0.1

1

10

100

1000

(f) v0 = 1.9991 , 3, 5, 7, 9, 11, and 13 component are mainly observed.

0 1 2 3 4 50.001

0.01

0.1

1

10

100

1000

(g) v0 = 1.999995 , 3, 5, 7, 9, 11, 13, 15, and 17 component are mainly observed.

0 1 2 3 4 50.001

0.01

0.1

1

10

100

1000

(h) v0 = 1.9999985

0 1 2 3 4 510-4

0.01

1

100

(i) v0 = 1.9999999

0 1 2 3 4 510-5

0.001

0.1

10

1000

The Fourier spectrum drastically changes at v0 = 2. The transition of the behavior occurs from oscillatory to non-oscillatory. (j) v0 = 2.001

0 1 2 3 4 5

1μ104

2μ104

100

200

500

1000

2000

5000