Chapter 13 Simple Harmonic Motion

In chapter 13 we will study a special type of motion known as: simple harmonic motion (SHM). It is defined as the motion in which the position coordinate x of an object that moves along the x-axis varies with time t as:

x(t) = Asin(t) or x(t) = Acos(t)

The condition for having simple harmonic motion is that the force F acting on the moving object is:

F = -kx (13-1)

Periodic motion is defined as a motion that repeats itself in time.

The period T the time required to complete one repetition Units: s

A special case of periodic motion is known as “harmonic” motion. In this type of motion the position x of an object moving on the x-axis is given by:

x(t) = Asin(t) or x(t) = Acos(t)

The most general expression for x(t) is: The parameters A, , and are constants

(13-2)

x(t) = Asin(t + )

Plots of the sine and cosine functions. Make sure that you can draw these two functions

/2 = 90º

= 180º

3/2 = 270º

2= 360º

Note: The period of the sine and cosine function is 2

(13-3)

Most general form of simple harmonic motion x(t) = Asin(t + )

A amplitude

angular frequency

phase.x(t)

mOx-axis

t

sin(a + b) = (sina)(cosb) +(cosa)(sinb)

x = (Acos )sint + (Asin)cost

The coordinate x is a superposition of sint and cost functions

Special case 1: = 0 x = A sint Special case 1: = /2 x = A cost (13-4)

The effect of the phase on is to shift the Asin(t + ) function with respect to the Asin(t) plot This is shown in the picture below Red curve: Asin(t) Blue curve: Asin(t + )

(13-5)

.x(t)

mOx-axis

t Determination of A and These two parameters are determined from the initial conditions x(t = 0) and v(t = 0)

sin( ) Velocity cos( )

(0) sin (0) cos

Divide e

eqs.

qs.1

1 eqs.2

(0)t by eqs.2

Once we have

an (0)

we

dxx A t v A t

dtx A v

x

v

A

2 2

can get A from eqs.1

The acceleration

(0)

s

- sin( )

in

- dv

a A t xd

x

t

A

(13-6)

Plots of x(t), v(t) and a(t) versus t for = 0

x = Asin(t) v = Acos(t)

a = -2Asin(t)

(13-7)

Consider a point P moving around O on a circular path of radius R win constant angular velocity . The Cartesian coordinates of P are: x and y. The angle = t. From triangle OAP we have: OA = x = Rcost and AP = Rsint

The projection of point P on either the x- or the y-axis executes harmonic motion as point P is moving with uniform speed on a circular orbit

(13-8)

P

A

m

mFs

x axisO

x

x axis

k A mass m attached to a spring (spring constant k) moves on the floor along the x-axis without friction

(13-9)

2

2

2

2

2

,

k and m are constants

k is also a constant

m

net s

k d xF F kx ma a x a

m dt

d x kx

mdtk

m

22

2

d xx

dt

m

mFs

x axisO

x

x axis

k

(13-10)

22

20 is the equation of motion. It is called a

"differential equation" because it involves derivatives

A function x(t) is called a "solution" of a differential equation if

when it is subs

d xx

dt

22

2

tituted into the differential equation it "satisfies" it.

In this case the expression: must be equal to zerod x

xdt

22

2

22

2

2

0

k=

m

d xx

dt

d xx

dt

m

mFs

x axisO

x

x axis

k 22

2

2

0

k=

m

d xx

dt

(13-11)

22

2

22 2 2

2

22

2

Solution candidate: sin( )

sin( )

= cos( ) - sin( )

- sin( ) sin( ) 0

Indeed the proposed function

makes equal t

dx d xA t A t

dt dt

d xx A t A t

dt

d xx

x A t

t

x

d

A t

o zero and thus it is a solution.

m

mFs

x axisO

x

x axis

k Summary

The spring-mass system obeys the following equation:

The most general solution of the equation of motion is:

Note 1: The angular frequency ( and thus the frequency f , and the period T) depend on m and k only

Note 2: The amplitude A and the phase on the other hand are determined from the initial conditions:

x(t = 0) and v(t=0) (13-12)

22

20 , 2

d x kx f

mdt

x = Asin(t + )

m

mFs

x axisO

x

x axis

k Energy of the simple harmonic oscillator

2

sin( )

cos( )

x A t

v A t

k

m

(13-13)

2 22

2 2 2 22 2

22 2

2 22 2

Total energy , sin ( ) 2 2

cos ( ) co

sin co

s ( ) 2 2 2

[sin ( ) cos ( )]2

[sis 1 n ( ) cos ( )

kx kAE U K U t

mv mA kAK t t

kAE U K t t

t t

2

] 1

Energ 2

y is constantkA

E

(13-14)

X

E

Why all this fuss about the simple harmonic oscillator?

Consider an object of energy E (purple horizontal line) moving in a potential U plotted in the figure using the blue solid line

Motion in the x-axis is allowed in the regions where

These regions are color coded green on the x-axis. Motion is forbidden elsewhere (color coded red)

If we have only small departures from the equilibrium positions x01 and x02 we can approximate U by the potential of a simple harmonic oscillator (the blue dashed line in the figure)

(13-15)

U E

We expand the function U(x) around one of the minima (x01 in this example) using a Taylor series

2 21 1

1 1 12

221

1 1 12

( ) ( )( ) ( ) ( ) ( ) ..

1! 2!

( )( ) 0 , ( ) Set ( ) 0 ( )

.

2

o oo o o

oo o o

x x x xdU d UU x U x x x

dx dx

x xdU d Ux x k U x U x k

dx dx

If we choose the origin to be at xo1 U(x) = kx2/2

This approximation for U is indeed the potential of a simple harmonic oscillator (dotted blue line) (13-16)

E

22

2( ) ( ) ( ) ( ) ( ) ( ) ...o o o o o

df d ff x f x x x x x x x

dx dx

Simple pendulum: A mass m suspended from a string of length that moves under the influence of gravity

(13-17)

2

t 2

2

2

2

2

dv dTangential acceleration a = = From Newton's second law

dtwe have: where Ft is the net force along the arc.

- ssin - sin n i

t

t

dtF ma

dF

d g

dtmg m mg

dt

Arc length

Velocity

s

ds dv

dt dt

s

Simple Pendulum

This differential equation is too difficult to solve. For this reason we consider a simpler version making the “small angle approximation”

2

2sin

d g

dt

(13-18)

3 5

2

2

sin - ..3!

Assume: <<1 ( is i

5!

n radians) Taylor expansion

of the function sin around 0 : .

If << 1 then sin The differential equation becomes:

d g

dt

This simpler version we can handle.

2

2

2

Simple pendulum, small angle

approximation: 1

If we call the equation

above takes the form:

d g

dtg

(13-19)

222

22

2 Compare this equation with:

whose solution is: The two equations have

identical form and therefore they have the same solution

x

sin( )

s:

sin( ) wher

e

o

d

d

d x

dtx t

tA

t

The period 2Tg

g

Summary

The simple pendulum (Small angle approximation:) << 1

(13-20)

Angular frequency Period

Note 1: the period T does not depend on the mass m

Note 2: The pendulum can be used as a clock

For example if we set T = 1sec w

2

e g

gT

g

2

2et 0.248

4 A pendulum of that length is a practical proposition

gTm

sin ) ( o t

In the small angle approximation we assumed that << 1 and used the approximation: sin We are now going to decide what is a “small” angle i.e. up to what angle is the approximation reasonably accurate?

(degrees) (radians) sin

5 0.087 0.087

10 0.174 0.174

15 0.262 0.259 (1% off)

20 0.349 0.342 (2% off)

Conclusion: If we keep < 10 ° we make less that 1 % error(13-21)

Energy of a simple pendulum

s =

O

A

B

C

U = 0

sin( )

Arc length

Potential energy

o t

s

U mgh

h OB OA

(13-22)

2 2 2

, From triangle OAC: cos

(1 cos ) (1 cos )

In the small angle approximatio

cos 1

n 1

1 12

2 2

OB OA

h U mg

mgU mg

Energy of a simple pendulum(13-23)

sin( )

Arc length

Velocity cos( )

o

o

t

s

dsv t

dt

222

2 2 222

22 2

2 22 2

Potential energy sin ( )2 2

Kinetic energy cos ( )2 2

cos ( )2

cos ( ) sin ( )2 2

The energy of the pedulum i constants

o

o

o

o o

mglmglU t

mmvK t

mglgK t

mgl mglE U K t t

The Physical Pendulum is an object of finite size suspended from a point other than the center of mass (CM) and is allowed to oscillate under the influence of gravity. The torque about point O due to the gravitational force is:

r

mg

d

O

CA

(13-24)

2

2

2

2

From triangle OAC sin

sin The torque is also given by the equation:

We elimi

sin

nate between

(eqs.1

eqs.1

)

a

nd e

(eqs. qs.2 2)

d m

mgd d r

mgr

dId

gr

Idt

t

O

mg

r

2

2

5

2

3

2

sin

In the small angle approx

si

imation: 1

.

si

n

n

..3! 5!

d mgr

Idt

d mgr

Idt

2

2

2

2

We define: The equation of motion becomes:

This equation is an old friend and we can handle itd

dt

mgr

I

(13-25)

O

mg

r2

2 22

where =

I is the moment of inertia about an axis

that passes through the point of suspension O.

d mgr

Idt

(13-26)

2 22

22

2

22

2

Use the parallel axis theorem:

Compare the equation of the physical pendulum

with that of the simple harmonic oscillator

the two equations have the

cmcm

mgrI I mr

I mr

d

d

d x

t

xdt

same form and therefore have

the same solution sin ) ( o t

mFs

x

x axis

Overall Summary

r

k

(13-27)

22

2

22

2

22

2

,

sin( )

,

sin( )

,

s

Spring-mass system

Simple pendulum

Physical pendulu

i )

m

n(

o

o

d x k

dt mx A t

d g

dtt

d mgr

dt It

Damped Harmonic Oscillator

Consider the spring-mass system in which in addition to the spring force Fs = -kx we have a damping force Fd = -bv. The constant b is known as the “damping coefficient” In this example Fd is provided by the liquid in the lower part of the picture

x

-kx

-bv v

2

2

2

2

x 2

2

net

If we eliminate F form eqs.1 and eqs.2 w

(eqs.1)

e get:

Divide

(eqs.2

both part

)

s by m

xnet xnet

dx d xF kx b F m

dt dt

d x dxm kx bd

d x k b dxx

dt m m t

t t

d

d

(13-28)

A plot of the solution is given in the figure. The damping force . Results in an exponential decrease of the amplitude with time. The mechanical energy in this case is not conserved. Mechanical energy is converted into heat and escapes.

(13-29) 2

2

22 2

2

Solution:

, 4

Damping factor 2

o o

d x k b dxx

dt m m dt

b k

m mb

m

sin( )tx Ae t

mFs

x

x axis

Fosint

Driven harmonic oscillator

k

Consider the spring-mass system that oscillates with angular frequency o = (k/m)1/2 If we drive with a force F(t) = Fosint (where in general o) the spring-mass system will oscillate at the driving angular frequency with an amplitude A() which depends strongly on the driving frequency as shown in the figure to the right. The amplitude has a pronounced maximum when = o This phenomenon is called resonance (13-30)

The effects of resonance can be quite dramatic as these pictures of the Tacoma Narrows bridge taken in 1940 show. Strong winds drove the bridge in oscillatory motion whose amplitude was so large that it destroyed the structure. The bridge was rebuilt with adequate damping that prevents such a catastrophic failure

(13-31)