Download Solid State III, Lecture 23

Lecture 4 - Stretched string
Transverse waves on a stretched string
 Aims:
 Derivation of the wave equation:
obtain the wave velocity, etc.
 Wave impedance:
characteristic impedance of a stretched
string.
 Polarization.
Linear;
circular;
Polarization and coherence.
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Waves 4
Derivation of the wave equation
 General approach:
 Consider a small segment;
 Find the difference in forces on the two ends;
 Find the element’s response to this imbalance.
 We need the following useful result:

dg
g  x  x   g  x   x
dx
easy to see graphically:
 Waves on a string:
 Neglect extension of the string i.e. tension is
constant everywhere and unaltered by the wave.
 Displacements are small so
Y<<l.
Yx and q are << 1.
 Neglect gravity
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Stretched string
 Forces on an element of the string:
 Net force in the y-direction is
T sin q  q   T sin q  T tan q  q   T tan q
Y
tan q 
But
x
Y   Y 
tan

q


q


 
so
x
x x  x 
 Net transverse force = T
 2Y
x
2
x
 To apply Newton’s 2nd law we need mass and
acceleration. These are:
mass of element =
x
acceleration (transverse) =
3
 2 Y t 2
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Derivation of wave equation
 Newton’s second law:
force
T
 2Y
x
2

  2Y
T t
2
 2Y
x
2
acceleration
mass
x  x
 2Y
t 2
wave velocity  T 
 The wave velocity is not the transverse velocity
of the string, which is Y t.
 Two Polarization's for a transverse wave:
Horizontal (z) ;
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Vertical (y).
Waves 4
Wave impedance
 The general concept:
 Applying a force to a wave medium results in a
response and we can therefore define an
impedance = applied force/velocity response.
 Expect the impedance to be real (force and
velocity are in-phase) - since energy fed into
the medium propagates (without loss) away
from the source of the excitation.
Imaginary impedance - no energy can be
transported (examples later: waveguide
below cut-off).
Complex impedance - lossy medium.
 Characteristic impedance:
 Transverse driving force:
Y
F  T sin q  T tan q  T
x
 Transverse velocity:
Y
t
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Impedance of a stretched string
 Impedance
driving force
TY x

 Z
transverse velocity
Y t
 For a wave in +ve x-direction, recall
Y f
Y
f

;
 v
x u
t
t
Y
1 Y
so,

x
v t
 Impedance (for wave in +ve x-direction)
Z   T v  T   v
 Wave in -ve x-direction, Z has the opposite sign:
Z    T v   T    v
 Note that the impedance is real. i.e. the medium
is lossless (in this idealised picture).
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Energy in a travelling wave
 Energy is both kinetic and potential.
 Kinetic energy associated with the velocity of
elements on the string.
 Potential energy associated with the elastic
energy of stretching during the motion.
 Potential energy density (P.E./unit length)
 Calculate the work done increasing the length
of segment x against a constant tension T.
 Increase in length of segment is
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 2  Y  2 


x  
x 
 x


 x
 

2
 1  Y  2 
1

Y

 x

 x1  


x



 2  x  
2  x 


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Energy density
 Increase in P.E (= force x extension)
extension
force
1  Y  2
T. 
 x
2  x 
 Potential energy density
1  Y  2
T. 

2  x 

 Kinetic energy density
 K.E. of length x
1/2
 Density
2
1

Y

x

2

t


m
v2
1  Y  2


2  t 
 Note: instantaneous KE and PE are equal since
2
Y x   1 v  Y t and   T v
 Total energy density
 Sum of KE and PE
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2

Y

 


t


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Energy: harmonic wave
 Y Y  
1

 Average energy density  
2  t t 
(section 1.1.3)

2


 Y  A ei t  kx  ; Y
Y


A
o
o
 
 average energy density is
1  Ao 2
2
 consider each element, length x, as an
oscillator with energy
1 xAo 2
2
 In time t we excite a length x=vt. So the
energy input is
1 vt Ao 2
2
 Since Z=v,
 Mean power
1 Z 2 Ao2
2
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