IPHO 2003 Theoretical Question 2 ( Solution )
(2003/07/17)
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Solution- Theoretical Question 2
A Piezoelectric Crystal Resonator under an Alternating Voltage
Part A
(a) Refer to Figure A1. The left face of the rod moves a distance vt while the
pressure wave travels a distance ut with u  Y /  . The strain at the left face is

S


 vt  v

ut
u
(A1a)*1
From Hooke’s law, the pressure at the left face is
p  YS  Y
v
  uv
u
ut
(A1b)*
p
Figure A1
t=0
p
p
p
t/2
p
t
vt
(b) The velocity v is related to the displacement  as in a simple harmonic motion
(or a uniform circular motion, as shown in Figure A2) of angular frequency
  ku . Therefore, if  ( x, t )   0 sin k ( x  u t ) , then
v( x, t )  ku 0 cos k ( x  u t ) .
(A2)*
The strain and pressure are related to velocity as in Problem (a). Hence,
S ( x, t )  v( x, t ) / u  k 0 cos k ( x  u t )
(A3)*
p( x, t )   uv( x, t )  k u 2 0 cos k ( x  u t )
(A4)*
 YS ( x, t )  kY 0 cos k ( x  u t )
------------------------------------------------------------------------------------------------Alternatively, the answers may be obtained by differentiations:
x
v ( x, t ) 

 ku 0 cos k ( x  u t ) ,
t
S ( x, t ) 

 k 0 cos k ( x  u t ) ,
x
p( x, t )  Y
0 
Figure A2

 kY 0 cos k ( x  u t ) .
x
------------------------------------------------------------------------1
An equations marked with an asterisk contains answer to the problem.

v
kxt
IPHO 2003 Theoretical Question 2 ( Solution )
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Part B
(c) Since the angular frequency  and speed of propagation u are given, the
wavelength is given by = 2 / k with k =  / u. The spatial variation of the
displacement  is therefore described by
b
b
g ( x)  B1 sin k ( x  )  B2 cos k ( x  )
(B1)
2
2
Since the centers of the electrodes are assumed to be stationary, g(b/2) = 0. This
leads to B2 = 0. Given that the maximum of g(x) is 1, we have A = ±1 and

b
(B2)*
g ( x)   sin ( x  )
u
2
Thus, the displacement is

b
(B3)
 ( x, t )  2 0 sin ( x  ) cos  t
u
2
(d) Since the pressure p (or stress T ) must vanish at the end faces of the quartz slab
(i.e., x = 0 and x = b), the answer to this problem can be obtained, by analogy,
from the resonant frequencies of sound waves in an open pipe of length b.
However, given that the centers of the electrodes are stationary, all even
harmonics of the fundamental tone must be excluded because they have antinodes,
rather than nodes, of displacement at the bisection plane of the slab.
Since the fundamental tone has a wavelength = 2b, the fundamental
frequency is given by f1  u /( 2b) . The speed of propagation u is given by
7.87  1010
 5.45  10 3 m/s
(B4)
3

2.65  10
and, given that b =1.0010-2 m, the two lowest standing wave frequencies are
u
3u
(B5)*
f1 
 273 (kHz ) , f 3  3 f1 
 818 (kHz )
2b
2b
------------------------------------------------------------------------------------------------[Alternative solution to Problems (c) and (d)]:
u
Y

A longitudinal standing wave in the quartz slab has a displacement node at x
= b/2. It may be regarded as consisting of two waves traveling in opposite
directions. Thus, its displacement and velocity must have the following form
b
 ( x, t )  2 m sin k ( x  ) cos  t
2
(B6)
b
b
  m [sin k ( x   ut )  sin k ( x   ut )]
2
2
b
b
v( x, t )  ku m [cos k ( x   ut )  cos k ( x   ut )]
2
2
(B7)
b
 2 m sin k ( x  ) sin  t
2
where  = k u and the first and second factors in the square brackets represent
waves traveling along the +x and –x directions, respectively. Note that Eq. (B6) is
identical to Eq. (B3) if we set  m = ± 0.
IPHO 2003 Theoretical Question 2 ( Solution )
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For a wave traveling along the –x direction, the velocity v must be replaced
by –v in Eqs. (A1a) and (A1b) so that we have
v
and p   uv
(waves traveling along +x)
(B8)
S
u
v
and p    uv
(waves traveling along –x)
(B9)
S
u
As in Problem (b), the strain and pressure are therefore given by
b
b
S ( x, t )  k m [ cos k ( x   ut )  cos k ( x   ut )]
2
2
(B10)
b
 2k m cos k ( x  ) cos  t
2
b
b
p ( x, t )    u m [cos k ( x   ut )  cos k ( x   ut )]
2
2
(B11)
b
 2  u m cos k ( x  ) cos  t
2
Note that v, S, and p may also be obtained by differentiating  as in Problem (b).
The stress T or pressure p must be zero at both ends (x = 0 and x = b) of the
slab at all times because they are free. From Eq. (B11), this is possible only
if cos( kb / 2)  0 or
2f

(B12)
kb  b 
b  n ,
n  1, 3, 5, 
u
f
In terms of wavelength , Eq. (B12) may be written as
2b
(B13)
 ,
n  1, 3, 5,  .
n
The frequency is given by
u nu n Y
(B14)
f  

,
n  1, 3, 5,  .
 2b 2b 
This is identical with the results given in Eqs. (B4) and (B5).
--------------------------------------------------------------------------------------------------(e) From Eqs. (5a) and (5b) in the Question, the piezoelectric effect leads to the
equations
T  Y (S  d p E )
(B15)
  Yd p S   T (1  Y
d p2
T
)E
(B16)
Because x = b/2 must be a node of displacement for any longitudinal standing
wave in the slab, the displacement  and strain S must have the form given in Eqs.
(B6) and (B10), i.e., with   ku ,
b
(B17)
 ( x, t )   m sin k ( x  ) cos( t   )
2
b
(B18)
S ( x, t )  k m cos k ( x  ) cos( t   )
2
where a phase constant  is now included in the time-dependent factors.
By assumption, the electric field E between the electrodes is uniform and
IPHO 2003 Theoretical Question 2 ( Solution )
(2003/07/17)
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depends only on time:
V (t ) Vm cos  t

.
(B19)
h
h
Substituting Eqs. (B18) and (B19) into Eq. (B15), we have
d
b
(B20)
T  Y [km cos k ( x  ) cos( t   )  p Vm cos  t ]
2
h
The stress T must be zero at both ends (x = 0 and x = b) of the slab at all times
E ( x, t ) 
because they are free. This is possible only if  = 0 and
kb
V
k m cos  d p m
(B21)
2
h
Since  = 0, Eqs. (B16), (B18), and (B19) imply that the surface charge density
must have the same dependence on time t and may be expressed as
 ( x, t )   ( x) cos  t
(B22)
with the dependence on x given by
b
2
 ( x)  Yd p k m cos k ( x  )   T (1  Y
d p2 Vm
)
T h
(B23)*
d p2
d p2 Vm
b
 [Y
cos k ( x  )   T (1  Y )]
kb
2
T h
cos
2
(f) At time t, the total surface charge Q(t) on the lower electrode is obtained by
integrating  ( x, t ) in Eq. (B22) over the surface of the electrode. The result is
Q (t )
1 b
1 b

 ( x, t ) wdx    ( x) w dx

V (t ) V (t ) 0
Vm 0
d p2
d p2
w b
b
  [Y
cos k ( x  )   T (1  Y )] dx
kb
h 0
2
T
cos
2
2
d 2
d2
bw
kb
 ( T
)[Y p ( tan )  (1  Y p )]
h
 T kb
2
T
 C0 [ 2 (
(B24)
2
kb
tan )  (1   2 )]
kb
2
where
bw
,
C0   T
h
 Y
2
d p2
T

(2.25) 2  102
 9.82  10 3
1.27  4.06
(B25)*
(The constant  is called the electromechanical coupling coefficient.)
Note: The result C 0 =  T bw / h can readily be seen by considering the static
limit k = 0 of Eq. (5) in the Question. Since tan x  x when x << 1, we have
lim Q(t ) / V (t )  C0 [ 2  (1   2 )]  C0
k 0
(B26)
Evidently, the constant C 0 is the capacitance of the parallel-plate capacitor formed
by the electrodes (of area bw) with the quartz slab (of thickness h and permittivity
IPHO 2003 Theoretical Question 2 ( Solution )
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 T) serving as the dielectric medium. It is therefore given by  T bw / h.
(B47)*