physics 131: lecture 23 today’s agenda - u of t physicsameyerth/phy131f14/lect...physics 131:...

Physics 131: Lecture 23

Today’s Agenda Finish up Ch 12 Simple harmonic motion

D fi iti Definition Period and frequency Position, velocity, and accelerationPosition, velocity, and acceleration Period of a mass on a spring Vertical spring

Energy and simple harmonic motion Energy of a spring force

Physics 201: Lecture 1, Pg 1

Clicker Question 1:A circular merry-go-round of mass 200 kg and 2 m radius

rotates freely with an initial angular velocity of 0.5 rad/s y g yabout a vertical axis passing through its center. A 50 kg student is initially standing on the rim of the merry-go-round. Treat the merry-go-round as a solid disk and the y gstudent as a point particle. Student

i f


initial final

Clicker Question 1:

i f

initial final


Clicker Question 1:

Which of the following best describes what is conserved as theWhich of the following best describes what is conserved as the student walks from the rim of the merry-go-round to the center?

A. Only the angular momentum of the system is conserved.B Only the kinetic energy of the system is conservedB. Only the kinetic energy of the system is conserved.C. Both the angular momentum and the kinetic energy is

conserved.


Periodic Motion

Motion that repeats itself over and over

Characterized by two pieces of information:

Period ( T ): time it takes to complete one cycle Unit: seconds

Frequency ( f ): number of cycles per unit of timeU it H d 1 Unit: Hz = seconds-1

T 1


fT

Clicker Question 2:

An object is undergoing periodic motion and takes 10 sAn object is undergoing periodic motion and takes 10 s to undergo 20 complete oscillations. What is the period and frequency of the object?

(a) T = 10 s, f = 2 Hz(b) T = 2 s f = 0 5 Hz(b) T = 2 s, f = 0.5 Hz(c) T = 0.5 s, f = 2 Hz(d) T = 0.5 s, f = 20 Hz(d) T 0.5 s, f 20 Hz(e) T = 10 s, f = 0.5 Hz


Simple Harmonic Motion

Particular type of periodic motion Particular type of periodic motion Very common type of motion Motion due to a springo o due o a sp g Motion of a pendulum (small angles) Motion of atoms in molecules

SHM requires a restoring force


Restoring Force: Hooke’s Law F = -kx x is distance spring is displaced from its relaxed length k is the spring constant (how stiff the spring is) Restoring force is proportional to displacement Restoring force is opposite in direction to displacement

FSPRING

F = -k(-x) = +kx

F = -k(+x) = -kxFSPRING-x


( )

+x

Simple Harmonic Motion

Simple Harmonic MotionLet’s investigate this SHM a bit:

t

TAy 2cos

A

A

A


A

Position 2

Drawing of:

A lit d f ill ti

t

TAx 2cos

A = amplitude of oscillationT = period

A

tA

2

0

AT

Ax

02cos

2


22cos2cos t

TATt

TAx

t

TA 2cos

SHM


Final Exam Final Exam: Tuesday 12/15 at 2:00pm Final Exam: Tuesday 12/15 at 2:00pm Review Session: Thursday 12/4 12pm 2pm in SS2135 (Harlow and Thursday 12/4 12pm-2pm in SS2135 (Harlow and

Meyertholen) Exam Jam

We will keep office hours until the finalOffice hours: Meyertholen M 2 3 F 12 2 Office hours: Meyertholen M 2-3 F 12-2

Meyertholen extra office hours: Monday 12/12: 10-12pm and 1-4pm12pm and 1 4pm

Office hours: Harlow T 2-3 RF 10-11

Physics 201: Lecture 1, Pg 12 Drop in center: Please check schedule

Vertical Mass and SpringIf we include gravity, there are two forces acting on mass. With mass, new equilibrium position has spring stretched d. F = 0 Fy = 0

kd – mg = 0d = mg/k Let this point be y = 0

F = mak(d y) mg = mak(d - y) – mg = mak(mg/k) - ky – mg = ma

-k y = may

Same as horizontal! SHO


New equilibrium position y = -d

Clicker Question 3: A block on a spring oscillates back & forth with simple harmonic motion of amplitude A A plot of displacement (x)harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the speed of the block biggest?

a) When x = +A or -A (i.e. maximum displacement) b) When x = 0 (i e zero displacement)b) When x = 0 (i.e. zero displacement) c) The speed of the mass is constant

+Ax

A

tA

x


-A

Clicker Question 4: A block on a spring oscillates back & forth with simple harmonic motion of amplitude A A plot of displacement (x)harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the magnitude of the acceleration of the block biggest?biggest?(a) When x = +A or -A (i.e. maximum displacement) (b) When x = 0 (i.e. zero displacement) ( ) ( p )(c) The acceleration of the mass is constant

x+A

tx


-A

Angular Frequency vs. FrequencyFrequency = number of oscillations per second

f 2 f

q y p

f 22

f

d radsradt

s

2


t

TAx 2cos tA cos

Example 1A block is pulled 0.30 m from its equilibrium position and let go.A block is pulled 0.30 m from its equilibrium position and let go. (a) Find the block’s equation of motion. (b) What is the block’s position after 0.5 s? The frequency of oscillation for the block/spring system is 0.77 Hz.The frequency of oscillation for the block/spring system is 0.77 Hz.

+ 0.30 m

srad /83.4Hz77.02 sT 311

tx s

rad83.4cosm30.0(a)Remember to use radians!

sT 3.1Hz77.0


m22.0s5.082.4cosm30.0 sradx(b)

SHM and Circles

A peg on a rotating A peg on a rotating turntable undergoes periodic motion

There is a close

AA

There is a close relationship between circular motion and simple harmonicsimple harmonic motion


SHM and Circles


Position

Angular position: Angular position:

t

Call A the vector AApointing toward the peg

The x-position of A is

The x position of A is

tAAx coscos


This is our equation for SHM!

Clicker Question 5:

What is the time derivative of the function f(t) = A CosWhat is the time derivative of the function f(t) = A Cos (t)?

(a) -A Cos (t)(b) - A Cos (t)(c) A Sin (t)(d) A Sin (t)(e) A Sin (t)(e) - A Sin (t)


Velocity

The x component of v The x-component of v is: vv

vx

sinvvx

AA

AArv

tAv sint

tAv sin

Av


Av max

Position and Velocity

t

x is zero when v = +A or - AcosAx x is zero when v = +A or - A v is zero when x = +A or -A

cosAx

sinAv


Acceleration

The x component of The x-component of a is:

aa ax

aaAA

Ara 22

cosaax

Ara

tAa cos2

t

tAa cos

Aa 2


Aamax tAv sin

Simple Harmonic Motion:Simple Harmonic Motion:

x(t) = [A]cos(t)v(t) = -[A]sin(t)

x(t) = [A]sin(t)v(t) = [A]cos(t)ORv(t) [A]sin(t)

a(t) = -[A2]cos(t)v(t) [A]cos(t)a(t) = -[A2]sin(t)

OR

xmax = Av = A

Period = T (seconds per cycle)Frequency = f = 1/T (cycles per second)vmax = A

amax = A2

Frequency f 1/T (cycles per second)Angular frequency = = 2f = 2/T

A


tAtx cos

Clicker Question 6Object A is attached to spring A and is moving in simple harmonic motion. Object B is attached to spring B and is moving in simple harmonic motion. The period and the amplitude of object B are both two times the corresponding values for object A. How do the maximum speeds of the two objects compare?j p

a) The maximum speed of A is one fourth that of object B.b) The maximum speed of A is one half that of object B.c) The maximum speed of A is the same as that of object B.d) Th i d f A i t ti th t f bj t Bd) The maximum speed of A is two times that of object B.e) The maximum speed of A is four times that of object B.


Simple Harmonic Motion:Simple Harmonic Motion:

x(t) = [A]cos(t) x(t) = [A]sin(t)v(t) = -[A]sin(t)a(t) = -[A2]cos(t)

v(t) = [A]cos(t)a(t) = -[A2]sin(t)

OR

xmax = Avmax = A

Period = T (seconds per cycle)Frequency = f = 1/T (cycles per second)

A l f 2 f 2 /Tamax = A2 Angular frequency = = 2f = 2/T

tAtx cos


physics 131: lecture 23 today’s agenda - u of t physicsameyerth/phy131f14/lect...physics 131:...

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