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Applied PhysicsIntroduction to Vibrations and Waves
(with a focus on elastic waves)Course Outline
Simple Harmonic Motion
02 =+ xx &&mk /=
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 161167, Courseware pg 5861See 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 Newtons 2nd Law to a segment of string, pinned at both ends and vibrating.
String: pinned at x = 0 and x = L, uniform linear density (mass/length), T constant along length, y small (small angle approximation).
+
x x+xx
y
T
Tsegment of string
+x
+y
( ) += sinsin TTFyNet forces on this segment
( ) += coscos TTFx

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next assume y is small so and + 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 embodies how y changes with x for a given constant t
need to use partial derivatives
1cos =sin 0xF
( ) ===+= yyy axmaTTTTF yxaT =
2
2
tya y
=
1
and
Yep,we know
First, do we have a partial derivative describing ay?

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What about a partial derivative describing ?
Well, we dont have one off of the top of our head but maybe we can create one.
xy
=tanstart with
2
22sec
xy
x =
then differentiate wrt x
1sec x
xy
= 2
2(love those
small angles )
1
2
2
2
2
tyxx
xyT
=
Plug ay and back into
This is the wave velocity we will derive it later
or
Tv =
But
2
2
22
2 1ty
vxy
=
2
2
2
2
ty
Txy
=
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 resonancessee 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 (because we are at resonance)!
All points on the string are moving with a time dependence in the form cost (where 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
( ) ( ) txftxy cos, =
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 cos, =
( ) txfty cos22
2
= ( )
2
2
2
2cos
dxxfdt
xy =
( ) ( )( )txfvdx
xfdt cos1cos 2222
=
( ) ( )xfvdx
xfd2
2
2
2 =
( ) xv
Bxv
Axf
+
= cossin
put NM solution back into wave equation
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( )vxAxf sin=
not a partial derivative because only function of xNow into
wave equation
Now we have to evaluate THIS differential equation but at least it doesnt 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
2=
tBtAx 00 cossin +=
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(x/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 sin=Put into from previous page
vLA sin0 =
nvL =
must = 0gives
where n is an integer
We have now defined what must be for the normal modesof a string fixed at both ends!!
21
=
TL
nn
21
=Tv
Lvn
n = or since
Subscriptmeans normal mode
( ) tvxAtxy nnn
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 cos, =
This is the wave functionfor standing waves on a stretched string.
1
21
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 =
From before
Lvn
n =
nn
n
nn
v
22 ==plugging in L and rearranging gives
n
n
v 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 cos2sin, = Wave function
(version 2!)
121
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 cos, =
1
21
nTL
nn =
=
( )n
xAxf2sin=
( ) txAtxy nn
n cos2sin, =
(called stationary waves, standing waves, or resonances  where 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 1 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 Estring 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?
21
=
TL
nn
called the fundamental
21
21
221)(
=
=
T
LnT
LnHzfn
String fixed at both ends so we can use our previously derived equation
Working in Hz
n=1 (fundamental) and = 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 n = n 1) and n = 1/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)