# Modeling

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```					                   Wave Equations and Functions
Transverse Waves          Summarized      Longitudinal Waves
(eg. stretched string)                    (eg. vibrating rod)

2 y 1 2 y                                            2 1  2
                                                     
x 2 v 2 t 2                                         x 2 v 2 t 2
T                                                     Y
v                     wave speed                  v
                                                     
normal mode                                           normal mode
solutions                                             solutions
(standing                                             (standing
waves)                                                waves)

y  x, t   f  x cos t          general       x, t   f x  cost
solution

Boundary           both ends fixed                   Boundary                one end fixed
Conditions:                                          Conditions:             one end free

n x                                               n x
y  x, t   An sin          cos  n t               x, t   An sin             cos nt
v                                                     v
where
1                                                 1

n      T     2                              n  1   Y 
    2n  11
2
n             n1                     n           2
 
L                                               L        
Also (in Hz)
1                                                1

fn 


n T 
 
2
fn   
n  1   Y 
2
 
2     1

2 2 L   
                                               2L   
 
Week 8 Lecture 1: Problems 47, 48, F2000Q3a
Travelling wave solutions to the wave
equation
Recall the Plan…..
Earlier we derived the wave equation for both longitudinal
and transverse waves. There are lots of variations on this
but here is an example:

 y  y
2            2

x 2
T t 2
Remember that this describes the general
behaviour for a wave on a string. However, if
we specifically want to know where (in y) some
point x is along a string at a given time t, we
have to get a solution to the wave equation

We said there were two types of solutions

Normal modes                 Travelling waves
aka                           aka
(stationary waves)             Progressive waves
(standing waves)
We have been working                Now need to look
on these for the last few             at this type of
2
lectures                        solution
Traveling Wave (Progressive Wave) Solutions to the
Wave Equation
(French pg 202-209)
 So far we have considered only the stationary wave
(normal mode) solution to the wave equation.
 Now we will look at the traveling wave solution.
 In fact – a stationary wave is just a traveling wave
which gets reflected from one end of a medium.
The incident and reflected waves superimpose, and
if conditions are right (i.e. appropriate frequencies
and wavelengths then a standing wave ( stationary
wave or normal mode) develops.

Traveling Wave Solution to the Wave Equation

y  y0 sinkx  t          For a
transverse wave

Note:
1. There is no phase constant in this equation as
written (but you could add one).
2. k is not the spring constant but the wave number
3. Since a stationary vibration is actually created from
a travelling wave, you should be able to show that
these 2 solutions are the same and you can (see
French pg 203-204).                                3
What exactly is the Wave Number?

“k” in a y vs x plot is the same as “” in a y vs t plot!

Explanation:
 If you plot the y displacement of a single point x on a
string as a function of time you get:

y
T = time for 1 complete
oscillation on a y vs t plot
t
and
T
 = angular
frequency

 Now, if we look at a single snapshot in time you are
essentially plotting y as a function of x:
y                           = wavelength = distance for
1 complete oscillation on a y
vs x plot
x


k = wave number = 2/ (rad/m)
4
Note here: on pg 214 French the wave number is
defined as k = 1/, but then French corrects for the
missing 2 by multiplying it through later. We will
use k = 2/ but be aware of the little catch (note:
most other books use 2/)

2                           2
so                      and         k
T                            

traveling wave solution is    y  y0 sinkx  t 

subbing in  and k gives
      x t 
y  y0 sin 2    
     
      T 

what fraction of a             what fraction of
wavelength is this position?   a period has
passed?

5
Wave Speed relationships
 The wave speed is the speed at which the wave
travels along the medium (in m/s).

 put the traveling wave solution back into the wave
equation

into
2 y     2 y
y  y0 sin kx  t                       v2 2
t 2     x
gives

 y 0 2 sinkx  t    v 2 k 2 y 0 sinkx  t 


 v k
2      2 2       or          v
k

So there IS a relationship between 
and k – they are related by the wave
velocity                           6
How everything fits in…

          22
v , k   , 
k           T

 2    
   f
f is
 v    
 T  2  T                frequency
in Hz

Keep handy -                     
will use this     v                  f
over and over            k       T

7
One more thing to clean up…
Recall that earlier we stated that               We stated this when
T        we wrote down the
Wave speed =            v                  wave equation but
        we didn’t derive it

Mass/length

Can we derive this? (of course, or I wouldn’t be asking..)
Consider the following travelling wave pulse on a rope:
If the pulse is small enough then we
can assume the tension T in the rope
doesn’t change, and we can consider                        Speed v
the wave a circular arc.        s

R
Sum forces (tension):

Horizontal:     SFh=0
s
Vertical (also the net radial force):
½
T                   T
SFv=2Tsin½  2T ½ T                                         R

Mass of segment         m = s=R

v2
T    R
R
T
v                                       8
Voilla!!

Remember this slide??

 General Wave Equation (1D):
2 y     2 y
 v2 2
t 2     x
where v is the wave speed

Can write this for all sorts of situations!
 stretched string      transverse displacement

(transverse             2 y T 2 y               T      tension
                v
oscillations):         t 2  x 2                     mass/
length

longitudinal displacement
 longitudinal waves       Y
2         2
Y
Young’s
                v          modulus
in a solid:             t 2  x 2                     density

 transverse (shear)      2 y Y 2 y               n      shear
               v          modulus
t 2
 x 2             
waves in a solid:

 2 B  2               B      bulk
 longitudinal waves                          v          modulus
t 2
 x 2             
in a gas or liquid:
The velocity terms are
here. We derived the
top one and could9
derive the others if we
wanted…. Naw…
Particle Velocity and Pressure
 So far the only parameter we have looked at is
particle displacement
y  y0 sinkx  t       (transverse wave)
  0 sin kx  t      (longitudinal wave)

 However, other parameters such as particle velocity
and pressure (stress) are also important.

 Considering longitudinal waves in a thin rod (so Y is
the appropriate modulus) (you could also do for transverse)

 particle displacement:             0 sin kx  t 

 particle velocity:                

          0 coskx  t 
(NOT same as wave speed)
t

 pressure:             strain      0 k coskx  t 
x

stress  Y      Y0 k coskx  t 
x
pressure  P  compressiv stress
e
 Y0 k coskx  t 

10
OR if you have a gas modulus B:
P   B0 k coskx  t 
Note this is
analogous to
Specific Acoustic Impedance                    refractive index in
(Courseware pg 72-73)                E/M waves
     Very important property of materials that are transmitting
sound waves (but “impedance” generally is applicable to
any wave in any medium).
pressure       P
Defined as                            z
particle velocity 


(Effectively a resistance term – high zs mean that it takes a large pressure to
induce a given particle velocity  high “stiffness”).
(since Y is a
 Y0 k coskx  t  Yk
plug in equations
for pressure and                                                      stiffness this
z                         
 0 coskx  t                        fits!)

 Now we really need z in a form that we can associate
with material properties, so recall some previous
relationships…                 then z 
Y
since v 
k               v
v
Y                        v 2
also

or   Y  v  z 
2
 v
v

z  v               units  kg  m 
 3  
kg
 2
 m  s  m s

 So materials with higher zs (stiffer) have higher wave
velocities.
P                         So particle speed 
also note             z  v

                     is inversely proportional
to wave speed v
11

See table of z and v values 2 pages along!
Traveling Waves: Summary Page
Wave Velocities
T                                              Y
transverse           v                   longitudinal oscillation v 
oscillation (string)                     (solid rod)                       

transverse                  n             longitudinal oscillation          B
v                                             v
oscillation (solid)                      (gas or liquid)                   

2                  2
wave number  k                             
                   T    All of these
            
v                          f
equations appear
on the formula
k        T                   sheet of my exams
Longitudinal Wave: (solid)
 particle displacement:   0 sinkx  t                               But
Not these
 particle velocity:                          0 coskx  t 


 pressure:                        P  Y           Y0k coskx  t 
x

Specific Acoustic Impedance:
P Yk Y
z  v       
  v

Intensity:                                      2                  Note we haven’t
I  zm 2  m  2 
1        1P     watts
                                           done intensity yet –
2        2 z  m 
                                next lecture
I
decibels dB  10 log10 12 dB                                            12
10
Calculated from
long. velocity

13

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