SOLUTION:
(a) Equation 2-3,
C = — tanh
2k
Equation 2-1,
Therefore, equating 2-1 and 2-3,
T
2k
L
ST
k
— = — tanh
and multiplying both sides by (2k)2/LT
Hence,
(2k)2 L
(2tt)2 gT
, /27rd
------ — = ------ — tanh I----LT T
LT 2ît
\ L ?
/2k \2
2Kg
/2ird\
~
= ---- tanh I-----I
\T /
L
\ L /
(b) Equation 2-13 may be written
gTH cosh [2k(z+d)/L]
/2kx
2îrt\
u = ----- ----------------------- cos I---- — ---- I
2L
cosh (2nd/L)
U
T ] ’
1 gH cosh [27r(z + d)/L]
/2kx
2nt\
= — — ------- ---------------- cos ---- — ---- I
C 2
cosh (27rd/L)
y L
T /
since
T
1
L " C
Since
gT
, /27rd\
C = — tanh | —— j,
2k
\ L /
jrH
1
cosh [2k(z+d)/L]
/2kx
2kî
u = — ---------------- --------------- :------ cos I----- — ——
T tanh(2ird/L)
cosh (27rd/L)
\ L
T
2 -1 9