lecture21 - university of rochester

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester

Physics 121, April 8, 2008.Harmonic Motion.


Physics 121.April 8, 2008.

• Course Information

• Topics to be discussed today:

• Simple Harmonic Motion (Review).

• Simple Harmonic Motion: Example Systems.

• Damped Harmonic Motion

• Driven Harmonic Motion


Physics 121.April 8, 2008.

• Homework set # 8 is due on Saturday morning, April 12, at8.30 am.

• Homework set # 9 will be available on Saturday morning at8.30 am, and will be due on Saturday morning, April 19, at8.30 am.

• Requests for regarding part of Exam # 1 and # 2 need to begiven to me by April 17. You need to write down what Ishould look at and give me your written request and yourblue exam booklet(s).

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Harmonic motion (a quick review).Motion that repeats itself at regular intervals.


Simple Harmonic Motion (a quick review).

x(t) = xm

cos(! t + ")

Amplitude

Phase Constant

Angular Frequency


Simple Harmonic Motion (a quick review).

• Other variables frequently used todescribe simple harmonicmotion:

• The period T: the time required tocomplete one oscillation. Theperiod T is equal to 2π/ω.

• The frequency of the oscillation isthe number of oscillations carriedout per second:

ν = 1/T

The unit of frequency is the Hertz(Hz). Per definition, 1 Hz = 1 s-1.

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Simple Harmonic Motion (a quick review).What forces are required?

• Using Newton’s second law we can determine the forceresponsible for the harmonic motion:

F = ma = -mω2x

• The total mechanical energy of a system carrying out simpleharmonic motion is constant.

• A good example of a force that produces simple harmonicmotion is the spring force: F = -kx. The angular frequencydepends on both the spring constant k and the mass m:

ω = √ ( k/m)


Simple Harmonic Motion (SHM).The torsion pendulum.

• What is the angular frequency ofthe SHM of a torsion pendulum:• When the base is rotated, it twists

the wire and a the wire generateda torque which is proportional tothe the angular twist:

τ = -Kθ

The torque generates an angularacceleration α:

α = d2θ/dt2 = τ/I = -(K/I) θ

The resulting motion is harmonicmotion with an angular frequencyω = √ (K/I).


Simple Harmonic Motion (SHM). The simple pendulum.

• Calculate the angular frequencyof the SHM of a simplependulum.• A simple pendulum is a pendulum

for which all the mass is located ata single point at the end of amassless string.

• There are two forces acting on themass: the tension T and thegravitational force mg.

• The tension T cancels the radialcomponent of the gravitationalforce.

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Simple Harmonic Motion (SHM). The simple pendulum.

• The net force acting on he mass isdirected perpendicular to thestring and is equal to

F = - mg sinθ

The minus sign indicates that theforce is directed opposite to theangular displacement.

• When the angle θ is small, we canapproximate sinθ by θ:

F = - mgθ = - mg x/L

• Note: the force is againproportional to the displacement.


Simple Harmonic Motion (SHM). The Simple Pendulum.

• The equation of motion for thependulum is thus

F = m d2x/dt2 = -(mg/L) x

or

d2x/dt2 = - (g/L) x

• The equation of motion is thesame as the equation of motionfor a SHM, and the pendulum willthus carry out SHM with anangular frequency ω = √ (g/L).

• The period of the pendulum isthus 2π/ω = 2π √ ( L/g). Note: theperiod is independent of the massof the pendulum.


Simple Harmonic Motion (SHM).The physical pendulum.

• In a realistic pendulum, not allmass is located at a single point.

• The motion carried out by thisrealistic pendulum around itsrotation point O can bedetermined by determining thetotal torque with respect to thispoint:

• If the angle θ is small, we canapproximate the torque by

! = "mgh sin#

! = "mgh#

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• The angular acceleration α isrelated to the torque:

• The equation of motion for theangular acceleration α is givenby

• This again is an equation forSHM with an angular frequencyω where

! = I"

! =d2"

dt2=#

I= $

mgh

I"

!2=mgh

I



• The period of the physicalpendulum is equal to

• We can double check our answerby requiring that the simplependulum is a special case of thephysical pendulum (h = L,I = mL2):

T =2!

"= 2!

I

mgh

T = 2!I

mgh= 2!

mL2

mgL= 2!

L

g


Physics 121.Quiz lecture 21.

• The quiz today will have 3 questions!

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Simple Harmonic Motion (SHM).The equation of motion.

• All examples of SHM were derived from he followingequation of motion:

• The most general solution to the equation is

d2x

dt2= !"

2x

x t( ) = A cos !t +"( ) + B sin !t + #( )


Simple Harmonic Motion (SHM).The equation of motion.

• If A = B

which is SHM.

x t( ) = A cos !t +"( ) + B sin !t + #( ) =

= A sin1

2$ %!t %"&

'()*++ sin !t + #( )

&'(

)*+=

= 2A sin1

4$ +

#2%"2

&'(

)*+cos

1

4$ %!t %

#2%"2

&'(

)*+


Damped Harmonic Motion.

• Consider what happens when in addition to the restoringforce a damping force (such as the drag force) is acting onthe system:

• The equation of motion is now given by:

F = !kx ! bdx

dt

d2x

dt2+

b

m

dx

dt+

k

mx = 0

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• The general solution of this equation of motion is

• If we substitute this solution in the equation of motion wefind

• In order to satisfy the equation of motion, the angularfrequency must satisfy the following condition:

x t( ) = Ae

i!t

!"2Ae

i"t+ i"

b

mAe

i"t+

k

mAe

i"t= 0

!2" i!

b

m"

k

m= 0



• We can solve this equation and determine the two possiblevalues of the angular velocity:

• The solution to the equation of motion is thus given by

SHM TermDamping Term

! =1

2i

b

m± 4

k

m" b

2

m2

#

$%%

&

'((!

1

2i

b

m±

k

m

x t( ) ! x

me!

bt

2meit

k

m



x t( ) ! xm

e!

bt

2meit

k

m

E t( ) !1

2kx

m

2e!

bt

m

The general solution contains a SHM term, with an amplitude that decreases as function of time

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Damped Harmonic Motion has many practicalapplications.

Damping is not always a curse.


Driven Harmonic Motion.

• Consider what happens when we apply a time-dependentforce F(t) to a system that normally would carry out SHMwith an angular frequency ω0.

• Assume the external force F(t) = mF0sin(ωt). The equationof motion can now be written as

• The steady state motion of this system will be harmonicmotion with an angular frequency equal to the angularfrequency of the driving force.

d2x

dt2= !"

0

2x + F

0sin "t( )



• Consider the general solution

• The parameters in this solution must be chosen such that theequation of motion is satisfied. This requires that

• This equation can be rewritten as

x t( ) = Acos !t +"( )

!" 2

Acos "t +#( ) +"0

2Acos "t +#( )! F

0sin "t( ) = 0

!0

2 "! 2( ) Acos !t( )cos #( ) "!

0

2 "! 2( ) Asin !t( )sin #( )" F0

sin !t( ) = 0

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• Our general solution must thus satisfy the followingcondition:

• Since this equation must be satisfied at all time, we mustrequire that the coefficients of cos(ωt) and sin(ωt) are 0.This requires that

and

!0

2 "! 2( ) Acos !t( )cos #( ) " !0

2 "! 2( ) Asin #( )" F0{ }sin !t( ) = 0

!

0

2 "! 2( ) Acos #( ) = 0

!

0

2 "! 2( ) Asin #( )" F0= 0



• The interesting solutions are solutions where A ≠ 0 andω ≠ ω0. In this case, our general solution can only satisfythe equation of motion if

and

• The amplitude of the motion is thus equal to

cos !( ) = 0

!

0

2 "! 2( ) Asin #( )" F0= !

0

2 "! 2( ) A" F0= 0

A=F

0

!0

2"!

2( )



• If the driving force has afrequency close to the naturalfrequency of the system, theresulting amplitudes can be verylarge even for small drivingamplitudes. The system is said tobe in resonance.

• In realistic systems, there willalso be a damping force.Whether or not resonancebehavior will be observed willdepend on the strength of thedamping term.

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Done for today!Thursday: Temperature and Heat!

Unusually Strong Cyclone Off the Brazilian Coast: A lot of Rotational Motion!Credit: Jacques Descloitres, MODIS Land Rapid Response Team, GSFC, NASA

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