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10/22/2012 PHY 113 A Fall 2012 -- Lecture 21 1 PHY 113 A General Physics I 9-9:50 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

# 10/22/2012PHY 113 A Fall 2012 -- Lecture 211 PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 21: Chapter 15 – Simple harmonic motion

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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

PHY 113 A Fall 2012 -- Lecture 21 210/22/2012

PHY 113 A Fall 2012 -- Lecture 21 310/22/2012

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.

PHY 113 A Fall 2012 -- Lecture 21 410/22/2012

Hooke’s lawFs = -k(x-x0)

x0 =0

Behavior of materials:

Young’s modulus

LL

EAF

FFLL

AFE

material

appliedmaterial

applied

/

/

PHY 113 A Fall 2012 -- Lecture 21 510/22/2012

Microscopic picture of material with springs representing bonds between atoms

Measurement of elastic response:

LL

EAF

FFLL

AFE

material

appliedmaterial

applied

/

/

k

PHY 113 A Fall 2012 -- Lecture 21 610/22/2012

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:

PHY 113 A Fall 2012 -- Lecture 21 710/22/2012

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

PHY 113 A Fall 2012 -- Lecture 21 810/22/2012

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

PHY 113 A Fall 2012 -- Lecture 21 910/22/2012

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)

PHY 113 A Fall 2012 -- Lecture 21 1010/22/2012

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

mk

thatprovidedxmk

dt

xd

satisfiestAtx

22

2

ω

)φω(cos)(

PHY 113 A Fall 2012 -- Lecture 21 1210/22/2012

) s(determine ω

)φω(cos)φω(cos ω)φωcos(

:equation thesatisfies guessthat Condition

2

22

2

m

k

tAm

ktA

dt

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

PHY 113 A Fall 2012 -- Lecture 21 1310/22/2012

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

PHY 113 A Fall 2012 -- Lecture 21 1410/22/2012

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

PHY 113 A Fall 2012 -- Lecture 21 1510/22/2012

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

PHY 113 A Fall 2012 -- Lecture 21 1610/22/2012

“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

PHY 113 A Fall 2012 -- Lecture 21 1710/22/2012

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

PHY 113 A Fall 2012 -- Lecture 21 1810/22/2012

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

ta

tv

Constants

Summary --

PHY 113 A Fall 2012 -- Lecture 21 1910/22/2012

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

dxtv

kxmvE

Am

kωtAtx

PHY 113 A Fall 2012 -- Lecture 21 2010/22/2012

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

PHY 113 A Fall 2012 -- Lecture 21 2110/22/2012

Energy diagram:

x

U(x)=1/2 k x2

E=1/2 k A2

A

K

PHY 113 A Fall 2012 -- Lecture 21 2210/22/2012

Effects of gravity on spring motion

x=0

k

mgd

mgkd

0 :mequilibriuAt

tm

dxm

k

k

mgx

m

k

dt

dxd

gxm

k

dt

xd

dt

xdmmgkxF

cos)(

:Rewriting

2

2

2

2

2

2

PHY 113 A Fall 2012 -- Lecture 21 2310/22/2012

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

PHY 113 A Fall 2012 -- Lecture 21 2410/22/2012

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