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Applied PhysicsIntroduction to Vibrations and Waves

(with a focus on elastic waves)Course Outline

Simple Harmonic Motion

02 =+ xx ω&&

mk /=ω

“k = elastic property of the oscillator

Elastic properties of materialsStretching, bending, twisting

Damped OscillatorsDriven Oscillators

Shock absorbers, resonances in mechanical systems

Natural vibrationalfrequenciesVibrations of a solidReflecting wavesUltrasonic waves/testing

Coupled OscillatorsNormal ModesWaves and the wave equation

Fourier Analysis(introduction)

Breakdown of waves into their components

Optical wavesMultiple source interferencediffraction

Thin film interference techniques

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Week 7 Lecture 1: Problems 44, 45, 46The Wave Equation

(French pg 161-167, Courseware pg 58-61See Math and Physics demos – website link)

→ The wave equation is derived in a similar manner to what we did for coupled equations, except now we consider the string to be continuous and have mass.

Deriving the Wave Equation for Transverse Oscillations on a String

→ Apply Newton’s 2nd Law to a segment of string, pinned at both ends and vibrating.

→ String: pinned at x = 0 and x = L, uniform linear density µ (mass/length), T constant along length, y small (small angle approximation).

θ+∆θ

x x+∆xx

y

T

Tsegment of string

θ

+x

+y

( )∑ −∆+= θθθ sinsin TTFy

Net forces on this segment

( )∑ −∆+= θθθ coscos TTFx

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→ next assume y is small so θ and θ + ∆θ are small

→ Need to rewrite this equation in terms of x, y, and t

Now: ay embodies how y changes with t for a given constant x∆θ embodies how y changes with x for a given constant t

∴ need to use partial derivatives

1cos ≈∴ θ θθ =sin∑ ≈∴ 0xF

( )∑ ∆==∆=−∆+= yyy axmaTTTTF µθθθθ

yxaT ∆=∆ µθ

2

2

tya y ∂

∂=

1

and

Yep,we know

First, do we have a partial derivative describing ay?

4

What about a partial derivative describing ∆θ ?

Well, we don’t have one off of the top of our head but maybe we can create one….

xy

∂∂=θtanstart with

2

22sec

xy

x ∂∂=⎟

⎠⎞

⎜⎝⎛

∆∆θθthen differentiate wrt x

1sec ≈θx

xy ∆

∂∂=∆∴ 2

2θ(love those

small angles )

1

2

2

2

2

tyxx

xyT

∂∂∆=∆

∂∂ µ

Plug ay and ∆θ back into

This is the wave velocity – we will derive it later

or

µTv =

But

2

2

22

2 1ty

vxy

∂∂=

∂∂

2

2

2

2

ty

Txy

∂∂=

∂∂ µ

WAVE EQUATION FOR TRANSVERSE WAVES ON A STRETCHED STRING

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So this is the wave equation(written for waves on a string):

2

2

22

2 1ty

vxy

∂∂=

∂∂

It describes the general behaviour of any wave on a string. However, if we specifically want to know the y position at some point xalong a string at a given time t, we have to get a solution to the wave equation (similar to what we did for SHM)

Two types of solutions

Normal modesAlso known as

stationary wavesstanding waves (these are the

resonance conditions)

Travelling wavesAlso known as

Progressive waves

For the remainder of this lecture and the next one we will look only at the we will look a the NORMAL MODE solutions to the wave equation. Next week (week 8) we will look at travelling waves.

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Normal Mode Solutions to the Wave Equation

(also called standing waves or stationary waves or resonances–see the website links under “Math and Physics Demos” under harmonics -for interesting

demonstrations of this behaviour)

→ All points on the string are moving with the same frequency ω (because we are at resonance)!

→ All points on the string are moving with a time dependence in the form cosωt (where ω is the normal mode frequency)

→ The amplitude at any point is a function of the distance x from the end of the string f(x)

→ Therefore the Normal mode solution looks like:

∴ need to evaluate f(x) and also ω

( ) ( ) txftxy ωcos, =

amplitude at a given point x

common frequency along string

we know…

[This proof is 3 pages long and somewhat painful but not difficult…]

Evaluating f(x):2

2

22

2 1ty

vxy

∂∂=

∂∂

( ) ( ) txftxy ωcos, =

( ) txfty ωω cos22

2

−=∂∂ ( )

2

2

2

2cos

dxxfdt

xy ω=

∂∂

( ) ( )( )txfvdx

xfdt ωωω cos1cos 222

2

−=

( ) ( )xfvdx

xfd2

2

2

2 ω−=

( ) xv

Bxv

Axf ⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛= ωω cossin

put NM solution back into wave equation

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( )vxAxf ωsin=

not a partial derivative because only function of xNow into

wave equation

Now we have to evaluate THIS differential equation but at least it doesn’t have “t” in it! This is actually our familiar form of a differential equation (thank goodness)

Boundary Conditions for a string fixed at both ends

at x=0 , f(x)=amplitude = 0 into solution gives B=0

So the solution is

xdt

xd 202

2ω−=

tBtAx 00 cossin ωω +=

recall:

solution:

diff wrt t diff wrt x

we have evaluated f(x) ... the amplitude at any point x along the string!!

Put B back into f(x) gives...

[note that A is the maximum amplitude of the string since f(x) has the largest value when sin(ωx/v) = 1]

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We have only used one of our Boundary Conditions… what about the other one? f(x)=0 at x=L

( )vxAxf ωsin=Put into from previous page

vLA ωsin0 =

πω nvL =∴

must = 0gives

where n is an integer

We have now defined what ω must be for the normal modesof a string fixed at both ends!!

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⎟⎟⎠

⎞⎜⎜⎝

⎛=

µπω TL

nn

21

⎟⎠

⎞⎜⎝

⎛=µTv

Lvn

nπω = or since

Subscriptmeans normal mode

( ) tvxAtxy n

nn ωω cossin, =

We started off with to find out what they position of a string would be at any position x. We now knowwhat f(x) is, and we also know what the normal mode frequencies have to be.

( ) ( ) txftxy ωcos, =

This is the wave functionfor standing waves on a stretched string.

1

21

ωµ

πω nTL

nn =⎟⎟

⎠

⎞⎜⎜⎝

⎛=

where

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→ It is also useful to describe this relationship in terms of wavelength to do this we note that for a string fixed at both ends the total length must accommodate an integral number of half sin curves in any given mode n.

2nnL λ=∴

From before

Lvn

nπω =

nn

n

nn

v λπ

λπω 22 =×=plugging in L and

rearranging gives

n

n

v λπω 2=

Note this expression is generic and holds for all normal modes (standing waves) no matter the boundary conditions

Subbing this back into the equation on the previous page…

( ) txAtxy nn

n ωλπ cos2sin, = Wave

function(version 2!)

121

ωµ

πω nTL

nn =⎟

⎠

⎞⎜⎝

⎛=

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The Wave Equation: summary → Sum forces on a small piece of a continuous body

(in this case a string)

2

2

22

2 1ty

vxy

∂∂=

∂∂

yx

WAVE EQUATION(for a stretched string)

the solution for normal modes

( ) ( ) txftxy ωcos, =

1

21

ωµ

πω nTL

nn =⎟⎟

⎠

⎞⎜⎜⎝

⎛=

( )n

xAxfλπ2sin=

( ) txAtxy nn

n ωλπ cos2sin, =

(called stationary waves, standing waves, or resonances - where ω is the same for all points along string)

evaluating thisusing BCs for a string fixed both ends

along the way:

mass/ length

frequency of normal mode or standing wave on stringNM amplitude

The y position at any point x along a string and at time t

∴ final solution for standing waves on a string:

WAVE FUNCTION for standing waves (normal modes) of a string!

Amplitude along the string at any point x

Normal mode frequency

This holds for BC of f(x) =0 at each end of the string

For transverse waves

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Examining the Normal Modes→ ω1 is the lowest possible standing wave frequency

(normal mode 1, n = 1).

→ The frequency of the fundamental defines what we recognize as the characteristic pitch of a vibrating string defines the tension required to obtain a note from a string mass (m) and length (L).

Ex. The E-string of a violin is supposed to be tuned to a frequency of 640 Hz (this is the fundamental). L=33cm, m=0.125g. What tension is required?

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⎟⎟⎠

⎞⎜⎜⎝

⎛=µ

πω TL

nn

called the fundamental

21

21

221)( ⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

µµπ

πT

LnT

LnHzfn

String fixed at both ends so we can use our previously derived equation

Working in Hz

n=1 (fundamental) and µ = m/L

21

1 21

⎟⎠⎞

⎜⎝⎛=

mLT

Lf

( ) 21421 )640)(33.0)(1025.1(44 −−== smkgxfmLT

T ≈ 68N (equivalent to hanging a 7kg mass off of the end)

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Harmonics→ n = 1 is the fundamental (1st NM)

→ higher values of n harmonics

→ for 1 cycle of the fundamental, the nth harmonic completes n cycles (since ωn = n ω1) and λn = λ1/n

→ pluck a string – motion consists of fundamental plus a few of the lower harmonics

→ touch a string along length – all modes will stop except those having a node at that point (see next slide).

(See diagram in next slide)

what you see here is basically how the wavelength changes for upper harmonics with t= constant

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Normal Modes are Often Called Harmonics (particularly in musical instruments)

1st Harmonic (normal mode 1): the fundamental

upper harmonics – note that for 1 full cycle of the fundamental, the nth harmonic has completed “n” cycles

touch 1/3 of the way along the string length – all vibrations stop except

(all harmonics having a node [zero point] at that position)

n = 3n = 6

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Tone Quality→ When orchestra instruments all tune to “concert A”

which is 440 Hz, they still all sound different. Why? Because although each instrument is vibrating at the same fundamental frequency, each is also producing harmonics whose relative amplitudesdepend on the instrument and how it is played. If each instrument produced only the fundamental frequency, the sound would be the same for each!

waveforms

Harmonic analysis (which harmonic and amplitude)