8.3 Solutions
373
Expanding tan (kL) by series
k L
k L +
(k L) 3
3
+ 2
(k L) 5
15
+ · · ·
=
M
m
If k L < 0.2, we may retain only the first term within the brackets:
k
2 L
2
=
M
m
ω 2 L 2
v 2 =
M
m
f =
ω
2π
=
1
2π
v
L
M
m
=
1
2π L
Y
ρ
M
m
But Y =
k L
A
and M = ALρ
∴ f =
1
2π
k
m
8.3.3 Waves in Liquids
8.35 (a) v
2
=
g
k
tanh(kh) =
gλ
2π
tanh
2π
λ
λ
4
=
gλ
2π
tan h
π
2
=
9.8
2π
× 0.917 λ
v = 1.2
√ λ m/s
(b) v =
g
k
=
gλ
2π
= 1.25
√
λ m/s
(c) v =
gh =
gλ
4
=
9.8λ
4
= 1.56
√
λ m/s
8.36 The fractional error introduced by the use of the formula v =
√
gh is
√
gh −
g
k
tanh(kh)
g
k
tanh(kh)
= 0.01
373
Expanding tan (kL) by series
k L
k L +
(k L) 3
3
+ 2
(k L) 5
15
+ · · ·
=
M
m
If k L < 0.2, we may retain only the first term within the brackets:
k
2 L
2
=
M
m
ω 2 L 2
v 2 =
M
m
f =
ω
2π
=
1
2π
v
L
M
m
=
1
2π L
Y
ρ
M
m
But Y =
k L
A
and M = ALρ
∴ f =
1
2π
k
m
8.3.3 Waves in Liquids
8.35 (a) v
2
=
g
k
tanh(kh) =
gλ
2π
tanh
2π
λ
λ
4
=
gλ
2π
tan h
π
2
=
9.8
2π
× 0.917 λ
v = 1.2
√ λ m/s
(b) v =
g
k
=
gλ
2π
= 1.25
√
λ m/s
(c) v =
gh =
gλ
4
=
9.8λ
4
= 1.56
√
λ m/s
8.36 The fractional error introduced by the use of the formula v =
√
gh is
√
gh −
g
k
tanh(kh)
g
k
tanh(kh)
= 0.01