chapter 1 - vibrations harmonic motion/ circular motion simple harmonic oscillators –linear,...

Chapter 1 - Vibrations

• Harmonic Motion/ Circular Motion

• Simple Harmonic Oscillators– Linear, Mass-Spring Systems

– Initial Conditions

• Energy of Simple Harmonic Motion

• Damped Oscillations

• Driven/Forced Oscillations

Math Prereqs

dcos

d

dsin

d

cos

sin

2 2 2

0 0 0

cos d sin d sin cos d

0

2 22 2

0 0

1 1cos d sin d

2 2

1

2

Identities (see appendix A for more)

cos cos 2cos sin2 2

2 2sin cos 1

cos cos cos sin sin

2 1 1cos cos 2

2 2

ie cos jsin j j

j e eRe e cos

2

j jj e e

Im e sin2j

Why Study Harmonic Motion

http://www.kettering.edu/~drussell/Demos/waves/wavemotion.html

http://www.falstad.com/mathphysics.html

Relation to circular motion

x A cos A cos t

2

T

j j tx Ae Ae

Or

Math Prereqs

T

0

1f t f t dt

T

"Time Average"

2 2cos t

T

T T2

0 0

1 2 1 1 1 2 1cos t dt cos 2 t dt

T T T 2 2 T 2

Example:

Horizontal mass-spring

• Good model!– Force is linear– Mass is constant– Spring has negligible mass– No losses

f maHooke’s Law: sf sx

2

2

d xsx m

dt

2

2

d x sx 0

dt m

Frictionless(1D constraint)

Solutions to differential equations

• Guess a solution• Plug the guess into the differential equation

– You will have to take a derivative or two• Check to see if your solution works. • Determine if there are any restrictions (required

conditions).• If the guess works, your guess is a solution, but it

might not be the only one.• Look at your constants and evaluate them using

initial conditions or boundary conditions.

Our guess

1 ox A cos t

Check

1 ox A cos t 2

21 o o2

d xA cos t

dt

2

2

d x sx 0

dt m

21 o o 1 o

sA cos t A cos t 0

m

2o o

scos t 0

m

The restriction on the solution

2o

s

m

oo

1 sf

2 2 m

o o

1 2 mT 2

f s

Any Other Solutions?

1 ox A cos t

2 ox A sin t

1 o 2 ox A cos t A sin t

A

A1

A2

o ox A cos cos t Asin sin t

ox A cos t

Or

Definitions

• Amplitude - (A) Maximum value of the displacement (radius of circular motion). Determined by initial displacement and velocity.

• Angular Frequency (Velocity) - Time rate of change

of the phase. Natural Angular Frequency

• Period - (T) Time for a particle/system to complete one cycle.

• Frequency - (fo) The number of cycles or oscillations completed

in a period of time. Natural Frequency

• Phase - t Time varying argument of the trigonometric function.

• Phase Constant - Initial value of the phase. Determined by initial displacement and velocity.

ox A cos t

The constants – Phase Angle ox t 0 x u t 0 0 0

x t 0 0 0u t 0 u 2

o ox x cos t o o ou x sin t

o

2o oa x cos t

ox

ox

o ox

o ox

2o ox

2o ox

o

o

u

o

o

u

Case I:

Case II:

o

o

uA

Note phase relationshipbetween x, u, and a

General Case

ox t 0 x 0u t 0 u

1 o 2 ox A cos t A sin t 1 oA x

1 o o 2 o ou A sin t A cos t o2

o

uA

A

A1

A2

2

2 oo

o

uA x

1 o

o o

utan

x

Energy in the SHO

2 2 2 2K P

1 1 1 1E E E mu sx sA mU

2 2 2 2

2 2su A x

m

Average Energy in the SHO

2 2 2 2P o

1 1 1E s x sA cos t sA

2 2 4

2 2 2 2 2 2 2K o o o

1 1 1 1E m u m A sin t m A sA

2 2 4 4

ox A cos t

o o

dxu A sin t

dt

K PE E

Example

• A mass of 200 grams is connected to a light spring that has a spring constant (s) of 5.0 N/m and is free to oscillate on a horizontal, frictionless surface. If the mass is displaced 5.0 cm from the rest position and released from rest find:

• a) the period of its motion, • b) the maximum speed and • c) the maximum acceleration of the mass.• d) the total energy• e) the average kinetic energy• f) the average potential energy

Complex Exponential Solution

• Check it – it works and is simpler. • Phase relationships are more obvious.• Implied solution is the real part• Are there enough arbitrary constants? What are they?

oj tx Ae

oj to ou j Ae j x

oj t2 2o oa Ae x

o

ot

Re

Im

jA a jb Ae

a

jb

oA cos t

Dashpot

r m

dxf R

dt

2

m2

d x dxm R sx 0

dt dt Equation of Motion:

Solution Guess:

Damped Oscillations

Dissipative forces

22mo2

Rd x dxx 0

dt m dt

tx Ae

Check2 t t 2 tm

o

RAe Ae Ae 0

m

2 2 tmo

RAe 0

m

2 2o

tx Ae

mR

2m

2 2d o dj

d d d d dj t j t j t j t j tt1 2 1 2x Ae A e A e e A e A e

d dj t j tt tdx A e e e Ae cos t

Damped frequency oscillation

2m

d 2

Rs

m 4m

2mR 4ms

B - Critical damping (=)C - Over damped (>)

mR

2m

tAe

Relaxation Time

• Decay modulus, decay time, time constant, characteristic time

• Time required for the oscillation to decrease to 1/e of its initial value

m

1 2m

R

Forced Vibrations

j tf t Fcos t or Fe

2

m2

d x dxm R sx f t

dt dt

f t

• Transient Solution – decays away with time constant, • Steady State Solution

Resonance0

s

m Natural frequencyj tx Ae

m

1 FA

sj R j m

2 j t j tmA m jA R As e Fe

sm 0

0

s

m

make small!!

2 2d o

Mechanical Input Impedance

• Think Ohm’s LawV

ZI

m

fZ

u

j t

j t

m

1 Fex Ae

sj R j m

j tf Fe

j t

m

Feu

sR j m

jm m m m m

sZ R j m R jX Z e

22

m m

sZ R m

1

m

sm

tanR

Significance of Mechanical Impedance• It is the ratio of the complex driving force to the

resulting complex speed at the point where the force is applied.

• Knowledge of the Mechanical Impedance is equivalent to solving the differential equation. In this case, a particular solution.

m

fu

Z

m

u fx

j j Z

V

Electrical Analogs2

2

d q dq qL R V(t)

dt dt C

2

m2

d x dxm R sx f t

dt dt

m

s

Rm

Elec Zelec Mech

V f

I u

L jL m jm

R R Rm Rm

1/C 1/jC s s/jf

How would you electrically model this?

m

s

Rm

f

u um

f 1/s

Rm

m

u um

Transient Responsej t

j t

m

1 Fex Ae

sj R j m

m

Fx sin t

Z

22

m m

sZ R m

1

m

sm

tanR

td

m

Fx Ae cos t sin t

Z

See front cover and figure 1.8.1 (pg 14)

Which is transient, which is steady state?

Instantaneous Power

• Think EE P VIi fu

j t

m

Feu

sR j m

m

Fu cos t

Z

2

im

Fcos t cos t

Z

f Fcos t j tf Fe

Average Power

T

i iT 0

1dt

T

2

T

0m

1 Fcos t cos t dt

T Z

22m

2m m

F RFcos

2Z 2Z

2T 2

0m

1 Fcos t cos cos t sin t sin dt

T Z

Quality (Q) value

• Q describes the sharpness of the resonance peak

• Low damping give a large Q• High damping gives a small Q• Q is inversely related to the

fraction width of the resonance peak at the half max amplitude point.

0 0 0

m

mQ

R 2 2

0 0

u l

Q

Tacoma Narrows Bridge

Tacoma Narrows Bridge (short clip)

tx Ae cos t

t tdxv Ae sin t A e cos t

dt

2

t 2 t t 2 t2

d xa Ae cos t A e sin t A e sin t A e cos t

dt

t 2 2Ae 2 sin t cos t

tAe sin t cos t

2

2

d x b dx kx 0

dt m dt m

t t2 2 tb kAe Ae Ae 0

m mcos t cos t cos2 sin tt sin t

t 2 2

b

2mb k

cAe 0b

2 s oin tm

s tm m

22k b

0m 2m

2k b

m 2m

b

2m