Ch 13 notes - FCPS · 2011. 12. 8. · 13.3 Summary Energy in simple harmonic motion: Energy is conserved in SHM. The total energy can be expressed in terms of the force constant

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures forUniversity Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 13

Periodic Motion


Goals for Chapter 13

• To outline periodic motion

• To quantify simple harmonic motion

• To explore the energy in simple harmonic motion

• To consider angular simple harmonic motion

• To study the simple pendulum

• To examine the physical pendulum

• To explore damped oscillations

• To consider driven oscillations and resonance


Homework

Read 476 to 486

�3, 5, 9, 15

Read 486 to 494

�25, 27

Read 495 to 505 (Skip Molecular Vibration)

�31, 35, 45, 49, 51, 61, 69


Introduction

• If you look to the right, you’ll see a time-lapse photograph of a simple pendulum. It’s far from simple, but it is a great example of the regular oscillatory motion we’re about to study.


Describing oscillations


In Oscillations . . . Displacement: the distance the body is from its equilibrium position.Restoring Force: the force that tends to restore the body to its equilibrium position.Amplitude : the maximum displacement from equilibriumPeriod: (T) the time for one cycle.Frequency: ( ) the number of cycles per unit of time.Angular Frequency: (ω) 2π times the frequency.

(13.1)

(13.2)

f

Tfππω 22 ==

Tf1= fT 1=


13.1 SummaryPERIODIC MOTION: PERIODIC MOTION IS MOTION THAT REPEATS

ITSELF IN A DEFINITE CYCLE. IT OCCURS WHENEVER A BODY HAS A

STABLE EQUILIBRIUM POSITION AND A RESTORING FORCE THAT ACTS

WHEN IT IS DISPLACED FROM EQUILIBRIUM. PERIOD T IS THE TIME FOR

ONE CYCLE. FREQUENCY IS THE NUMBER OF CYCLES PER UNIT TIME.

ANGULAR FREQUENCY IS TIMES THE FREQUENCY.


Period, Frequency, and Angular FrequencyAn ultrasonic transducer (loudspeaker) used for medical diagnosis oscillates at a frequency of 6.7 MHz = 6.7 x 106 Hz. How much time does each oscillation take, and what is the angular frequency?


Simple harmonic motion

• An ideal spring responds to stretch and compression linearly, obeying Hooke’s Law.

• For a real spring,Hookes’ Law is a goodapproximation.


Definition of Simple Harmonic Motion

The easiest way to view SHM is with a mass on a spring. It obeys Hooke’s Law: Fx = -kx (13.3)

When the restoring force is directly proportional to the displacement from equilibrium, as given by Eq. (13.3), the oscillation is called simple harmonic motion abbreviated SHM.

The acceleration ax = d2x/dt2 = Fx/m of a body in SHM is given by:

(13.4)xm

k

dt

xdax −== 2

2


Simple harmonic motion viewed as a projectionIf you illuminate uniform circular motion (say by shining a flashlight on a ball placed on a rotating lazy-Susan spice rack), the shadow projection that will be cast will be undergoing simple harmonic motion, like in figures 13.4 below.The SHM and circular motion analogy should hold true if the amplitude, A, equals the disk radius and the angular frequency equals the disk’s angular velocity.


Comparing acceleration of shadow at P to SHMThe circle the ball moves in so that its shadow matches SHM is called the circle of reference as we call Q the reference point.At a certain time, t, the vector OQ makes an angle,θ, with the horizontal. As the point Q moves around the reference circle with a constant angular speed ω, the vector rotates with the same speed, called a phasor.

The x-component of the phasor at time t is just the x-coordinate of the point Q:x = A cos θ

This position also refers to the position of the shadow at P, which is the projection of Q onto the x-axis. The acceleration of P is just the x-component of the acceleration of Q.


Comparing acceleration of shadow at P to SHMFrom centripetal acceleration, the acceleration of Q is aQ = ω2A, directed toward O, as shown below.The x-component is just ax = -aQ cos θCombining these two equations gives ax = - ω2A cos θFrom before x = A cos θ, so ax = - ω2x

This gives exactly what is to be expected from equation 13.4, if the angular speed of the reference point Q is related to the force constant k and mass m of the oscillating body by

or

m

k=2ωm

k=ω


Equations on 481

Both omegas, one for angular velocity and one for angular frequency are the same thing.

k

m

fT

m

kf

mk

πωπ

ππω

ω

221

2

1

2

===

==

=


Angular frequency, frequency, and period in SHMA spring is mounted horizontally, with its left end held stationary. By attaching a spring balance to the free end and pulling toward the right, we determine that the stretching forceis proportional to the displacement and that a force of 6.0 N causes a displacement of 0.030 m. We remove the spring balance and attach a 0.50 kg body to the end, pull it a distance of 0.020 m, release it, and watch it oscillate. Find the:

Force constant of the spring.

Angular frequency.

Frequency.

Period of the oscillation.


X versus t for SHO

If at time t = 0 the phasor OQ makes an angle Φ with the x-axis then at any later time t the angle is (ωt + Φ). The position of the oscillating object can be found by the equation:

(13.13))cos( φω += tAx


Variations on a theme—write equations on board

Changing either k or m changes the period of oscillation.


SHM phase and motion• SHM can occur with

various phase angles, the angle at which t = 0.

• For a given phase we can examine position, velocity, and acceleration.


SHM equations

)cos(

)sin(

)cos(

2 φωωφωω

φω

+−=

+−=+=

tAa

tAv

tAx

x

x

Do these equations fit with equation 13.4:

xm

k

dt

xdax −== 2

2

When t = 0:

2

22

2 cos

sin

cos

ω

φω

φωφ

x

x

x

oo

o

o

o

vxA

Aa

Av

Ax

+=

−=

−==


Watch variables change for a glider example Keep problems solving strategy on 485 in mind when solving the following problem.Let’s return to the system of mass and horizontal spring we considered previously, with k = 200 N/m and m = 0.50 kg. This time we give the body an initial displacement of +0.015 m and an initial velocity of +0.40 m/s.

a) Find the period, amplitude, and phase angle of the motions.b) Write equations for the displacement, velocity, and

accelerations as functions of time.


13.2 SummarySimple harmonic motion: If the restoring force in periodic motion is directly proportional to the displacement x, the motion is called simple harmonic motion (SHM). In many cases this condition is satisfied if the displacement from equilibrium is small.

The angular frequency, frequency, and period in SHM do not depend on the amplitude, but only on the mass m and force constant k.

The displacement, velocity, and acceleration in SHM are sinusoidal functions of time; the amplitude A and phase angle of the oscillation are determined by the initial position and velocity of the body. (See Examples 13.2, 13.3, 13.6, and 13.7.)

Homework: Read pages 486 to 492

On page 507: 3, 5, 9, 15


Energy in SHM

• Energy is conserved during SHM and the forms (potential and kinetic) interconvert as the position of the object in motion changes.

(13.21)

• Page 487 goes through a proof of this relationship.

• We can rewrite equation 13.21 for a velocity at any displacement:

(13.22)

• Using this equation we can find the maximum velocity (when displacement is zero):

(13.23)

2 2 21 1 12 2 2E= xmv kx kA+ =

2 2xv

kA x

m= ± −

2maxv

k kA A A

m mω= = =


Energy in SHM

• You can use equation 13.22 to calculate the velocity at each point in the figure below.


Energy in SHM II• Figure 13.13 shows the interconversion of kinetic and potential energy

with an energy versus position graphic.


Energy in SHM II• Refer to Problem-Solving Strategy on

page 488 while solving the following problem.

• In the oscillation described in the previous example, k = 200 N/m, m = 0.50 kg, and the oscillating mass is released from rest at x = 0.020 m.

a) Find the maximum and minimum velocities attained by the oscillating body.

b) Compute the maximum acceleration.c) Determine the velocity and acceleration

when the body has moved halfway to the center from its original position.

d) Find the total energy, potential energy, and kinetic energy at this position.


Energy and momentum are related in SHMA block with mass M attached to a horizontal spring with force constant k is moving with simple harmonic motion having amplitude A1. At the instant when the block passes through its equilibrium position, a lump of putty with mass m is dropped vertically onto the block from a very small height and sticks to it.Find the new amplitude and period.Find the amplitude and period of the putty is dropped onto the block at its maximum displacement A1.


13.3 SummaryEnergy in simple harmonic motion: Energy is conserved in SHM. The total energy can be expressed in terms of the force constant k and amplitude A. (See Examples 13.4 and 13.5.)

Homework: Read pages 495 to 505 (Skip molecular vibration)

On page 508: 25 and 27

2 2 21 1 12 2 2E= xmv kx kA+ =


Applications of SHMWhen we have a mass hanging from a spring, or a mass compressing a spring, the new equilibrium position is when the spring’s restoring force balances out the force of gravity on the mass: k∆l = mg


Vertical SHM

We can set x = 0 at this new equilibrium position and the equation for the new net force is:

Fnet = k(∆l – x) + (–mg) = –kx


Vertical SHM in an old carThe shock absorbers in an old car with mass 1000 kg are completely worn out. When a 980 N person climbs slowly into the car to its center of gravity, the car sinks 2.8 cm. When the car, with the person aboard, hits a bump, the car starts oscillating up and down in SHM. Model the car and person as a single body on a single spring, and find the period and frequency of the oscillation.


Angular SHMWatches keep time based on regular oscillations of a balance wheel initially set in motion by a spring.

Figure 13.18 illustrates the delicate inner mechanism of a clock or watch.

The wheel has an moment of inertia I about its axis. A coil spring exerts a restoring torque τ that is proportional to the angular displacement θfrom the equilibrium position.

We write τz = -κθ where κ is called the torsion constant.

Using the rotational analog to Newton’s second law:

-κθ = Iα or

This is very similar to our definition of SHM.

2 2z zI Id dtτ α θ= =∑

2

2d

Idtθ κ θ= −


Angular SHM

Angular simple harmonic motion: In angular SHM, the frequency and angular frequency are related to the moment of inertia I and the torsion constant .

Homework on 508: 31 and 37

( )

1

2

cos

I

fI

t

κω

κπ

θ ω φ

=

=

= Θ +


The Simple Pendulum A simple pendulum consists of a point mass m at the end of a massless string of length L.

Its motion is approximately simple harmonic for sufficiently small amplitude; the angular frequency, frequency, and period then depend only on g and L, not on the mass or amplitude. (See Figure 13.20.)


The Simple Pendulum Equations

1

2 2

2 12

k mg L g

m m L

gf

L

LT

f g

ω

ωπ ππ π

ω

= = =

= =

= = =

The equations on the left are only true for small amplitude oscillations, as shown in the graph on the right.


A simple pendulumFind the frequency of a simple pendulum 1.000 m long at a location where g = 9.800 m/s2.


Physical PendulumsLike a simple pendulum, the physical pendulum’s angular frequency depends on the distance from the axis of rotation to the mass.

With the physical pendulum, the mass is spread out. It is not a point mass. The frequency (and hence period) depends on the distance between the axis of rotation and the center of gravity, d, and the moment of inertia of the body, I.

mgd

Iω = 2 IT

mgdπ=


The physical pendulum• A physical pendulum is any

real pendulum that uses an extended body in motion. Figure 13.22 illustrates a physical pendulum and you may wish to refer to it as you consider the next example.

• Suppose the object to the right is a uniform rod with length L pivoted at one end. Find the period of its motion.


Dinosaurs, long tails, and the physical pendulumAll walking animals, including humans, have a natural walking pace, a number of steps per minute that is more comfortable than a faster or slower pace. Suppose this natural pace is equal to the period of the leg, viewed as a uniform rod pivoted at the hip joint.

a) How does the natural walking pace depend on the length Lof the leg, measured from hip to foot?

b) Fossil evidence shows that Tyrannosaurus rex, had a leg length L = 3.1 m and a stride length of 4 m. Estimate the walking speed of Tyrannosaurus rex.


Damped oscillations

Consider an oscillating system with a damping force due to friction: Fx = -bvx where vx = dx/dt and b is constant that describes the strength of the damping force.

The net force would include this damping force and the oscillatory restoring force: -kx – bvx

From Newton’s second law: -kx – bvx = max

or

Solving with differential calculus yields:( ) ( )2

2

2

cos

4

b m tx Ae t

k b

m m

ω φ

ω

− ′= +

′ = −

2

2dx d xdt dt

kx b m− − =


Kinds of Damping

The amplitude with damped oscillations decreases with time.

The angular frequency ω’is zero when

This situation is called critically damped.

( ) ( )22

2

cos

4

b m tx Ae t

k b

m m

ω φ

ω

− ′= +

′ = −2b km=

An oscillation is overdamped when

and underdamped when 2b km<

2b km>


Damped oscillationsClassify either line as overdamped, underdamped, or critically damped.


When a force Fx = -bvx proportional to velocity is added to a simple harmonic oscillator, the motion is called a damped oscillation. If (called underdamping), the system oscillates with a decaying amplitude and an angular frequency ω’ that is lower than it would be without damping. If (called critical damping) or (called overdamping), when the system is displaced it returns to equilibrium without oscillating.

Damped oscillations II

2b km<

2b km=2b km>


Forced (driven) oscillations and resonanceA force applied “in synch”with a motion already in progress will resonate and add energy to the oscillation.

A singer can shatter a glass with a pure tone in tune with the natural “ring” of a thin wine glass.

( )max

22 2 2

d d

FA

k m bω ω=

− +


Forced (driven) oscillations and resonance II

• The Tacoma Narrows Bridge suffered spectacular structuralfailure after absorbing too much resonant energy (refer to Figure13.28).


Homework

On Page 509:

45, 49, 51, 61, 69


Homework

Read 476 to 486

�3, 5, 9, 15

Read 486 to 494

�25, 27

Read 495 to 505 (Skip Molecular Vibration)

�31, 35, 45, 49, 51, 61, 69

Documents

Ch 13 notes - FCPS · 2011. 12. 8. · 13.3 Summary Energy in simple harmonic motion: Energy is conserved in SHM. The total energy can be expressed in terms of the force constant