10/22/2012PHY 113 A Fall 2012 -- Lecture 211 PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan...

PHY 113 A Fall 2012 -- Lecture 21 110/22/2012

PHY 113 A General Physics I9-9:50 AM MWF Olin 101

Plan for Lecture 21:

Chapter 15 – Simple harmonic motion

1. Object attached to a spring

Displacement as a function of time

Kinetic and potential energy

2. Pendulum motion


iclicker exercise:Why did your textbook develop a whole chapter on oscillatory motion?

A. Because the authors like to torture students.B. Because it is different from Newton’s laws and

needs many pages to explain.C. Because it is an example of Newton’s laws and

needs many pages to explain.

iclicker exercise:Simple harmonic motion has a characteristic time period T. What determines T?

A. The characteristics of the physical system.B. The initial displacements or velocities.C. It is not possible to know T.


Hooke’s lawFs = -k(x-x0)

x0 =0

Behavior of materials:

Young’s modulus

LL

EAF

FFLL

AFE

material

appliedmaterial

applied

/

/


Microscopic picture of material with springs representing bonds between atoms

Measurement of elastic response:

LL

EAF

FFLL

AFE

material

appliedmaterial

applied

/

/

k


F s (N

)

x

Us (

J)

x

Fs = -kxUs = ½ k x2

General potential energy curve:U

(J)

x

Us = ½ k (x-1)2

U(x)

)1(2

2

xdx

Udk

Potential energy associated with Hooke’s law:

Form of Hooke’s law for ideal system:


In addition to the spring-mass system, Hooke’s law approximates many physical systems near equilibrium.


Motion associated with Hooke’s law forces

Newton’s second law:

F = -k x = m a

xmk

dt

xd

dt

xdmkxF

2

2

2

2

“second-order” linear differential equation


How to solve a second order linear differential equation:

Earlier example – constant force F0 acceleration a0

00

2

2

am

F

dt

xd

x(t) = x0 +v0t + ½ a0 t2

2 constants (initial values)


Hooke’s law motion:

xmk

dt

xd

dt

xdmkxF

2

2

2

2

Forms of solution:

where:)ωsin()ωcos()(

)φωcos()(

tBtAtx

tAtx

mkω

2 constants (initial values)

PHY 113 A Fall 2012 -- Lecture 21 11

Verification:

Differential relations:

Therefore:

10/22/2012

)φω(sin ω)φωcos(

)φωcos(ω)φωsin(

tdt

td

tdt

td

)φω(cos ω)φωcos( 2

2

2

tA

dt

tAd

mk

thatprovidedxmk

dt

xd

satisfiestAtx

22

2

ω

)φω(cos)(


) s(determine ω

)φω(cos)φω(cos ω)φωcos(

:equation thesatisfies guessthat Condition

2

22

2

m

k

tAm

ktA

dt

tAd

constantsunknown are and and where)cos()(

:form thehas )(for solution that Guess

:system spring-massfor law sNewton' :Recap

2

2

AtAtx

tx

xm

k

dt

xd


iclicker exercise:Which of the following other possible guesses would provide a solution to Newton’s law for the mass-spring system:

above. theof one than More E.

)exp( D.

C.

)sin()cos( B.

sin A.

221

0

φωt

gttv

tBtA

)t(A

)sin()sin()cos()cos(os

)cos()sin()sin()cos(sin : thatNote

tAtA)t(cA

tAtA)t(A


tm

kxtxxA

)(Atdt

dx) (Axtx

tdt

dxxtx

A

cos)( and 0

sin0)0( cos)0(

0)0( and )0(

that know weSuppose

:conditions initial from and Finding

00

0

0

constantsunknown are and wherecos)(

:form thehas )(for solution that theknow now We

:system spring-massfor law sNewton'for solution Complete

2

2

Atm

kAtx

tx

xm

k

dt

xd


tm

kvtx

vA

)(m

kAvt

dt

dx) (Atx

vtdt

dxtx

A

mk

mk

sin)( and 2

sin)0( cos0)0(

)0( and 0)0(

that know weSuppose

:conditions initial from and Finding

00

0

0

constantsunknown are and wherecos)(

:form thehas )(for solution that theknow now We

:system spring-massfor law sNewton'for solution Complete

2

2

Atm

kAtx

tx

xm

k

dt

xd


“Simple harmonic motion” in practice

A block with a mass of 0.2 kg is connected to a light spring for which the force constant is 5 N/m and is free to oscillate on a horizontal, frictionless surface. The block is displaced 0.05 m from equilibrium and released from rest. Find its subsequent motion.

x(t)= A cos ( t+f) x(0) = A cos (f) = 0.05 m

v(t)=-A sin ( t+f) v(0)=-A sin (f) = 0 m/s

f 0 and A = 0.05 m

rad/s 5rad/s 2.0/5/ mk


iclicker exercise:

A certain mass m on a spring oscillates with a characteristic frequency of 2 cycles per second. Which of the following changes to the mass would increase the frequency to 4 cycles per second?

(a) 2m (b) 4m (c) m/2 (d) m/4


Simple harmonic motion:

Note that:

xmk

dt

xd

dt

xdmkxF

2

2

2

2

mk

tAtx ω);φω(cos)(

φ)ωcos(ω)(

)φωsin(ω)(

2

tAdtdv

ta

tAdtdx

tv

Conveniently evaluated in radians

Constants

Summary --


Energy associated with simple harmonic motion

22222

22

)cos(2

1)sin(

2

1

)sin()(

2

1

2

1

:Energy

constants are and and where)cos()(

:ntdisplaceme of Form

tkAtAmE

tAdt

dxtv

kxmvE

Am

kωtAtx


Energy associated with simple harmonic motion

2222

2

22222

22

2

1)cos()sin(

2

1

But

)cos(2

1)sin(

2

12

1

2

1

:Energy

kAttkAE

m

k

tkAtAmE

kxmvE


Energy diagram:

x

U(x)=1/2 k x2

E=1/2 k A2

A

K


Effects of gravity on spring motion

x=0

k

mgd

mgkd

0 :mequilibriuAt

tm

kAdtx

dxm

k

k

mgx

m

k

dt

dxd

gxm

k

dt

xd

dt

xdmmgkxF

cos)(

:Rewriting

2

2

2

2

2

2


Simple harmonic motion for a pendulum:

Q

L ) (since sinsin

αsinτ

22

2

2

2

mLILg

ImgL

dt

d

dt

d-IImgL

QQQ

QQ

Approximation for small Q:

Lg

ω );φωcos()

:Solution

sin

2

2

Q

QQ

QQ

tA(t

Lg

dt

d


Pendulum example:

Q

L

Lg

ω );φωcos() Q tA(t

Suppose L=2m, what is the period of the pendulum?

s 84.2ω2π

T

T2π

rad/s 2135.22m

9.8m/sLg

ω2

10/22/2012PHY 113 A Fall 2012 -- Lecture 211 PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan...

Documents

Transcript of 10/22/2012PHY 113 A Fall 2012 -- Lecture 211 PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan...