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Epicyclic motion
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Past exam of the mathematics course of the University of Cambridge
/
2024
/
iii
/
Paper 321
/
3
/
c
/
iv
/
Solution
2026-09-25
View more
When
k
y
=
0
,
k
x
is constant and the system becomes
u
˙
x
=
2
(
Ω
+
Ω
c
s
2
k
x
2
)
u
y
,
u
˙
y
=
−
2
Ω
u
x
.
(1)
Thus
u
¨
x
+
(
Ω
2
+
c
s
2
k
x
2
)
u
x
=
0.
(2)
For
time
dependence
e
−
iω
t
, the
dispersion relation
is
ω
2
=
Ω
2
+
c
s
2
k
x
2
.
(3)
Explicitly,
u
x
=
C
1
cos
(
ω
t
)
+
C
2
sin
(
ω
t
)
,
u
y
=
2
(
Ω
+
c
s
2
k
x
2
/Ω
)
u
˙
x
,
(4)
and
δ
=
2
i
k
x
u
y
/Ω
. This is an axisymmetric
inertial-acoustic wave
:
pressure
supplies the
c
s
2
k
x
2
restoring term and
epicyclic motion
supplies the
Ω
2
term.
Total
articles
:
1