OurBigBook
About
$
Donate
Sign in
Sign up
Past exam of the mathematics course of the University of Cambridge
/
2018
/
iii
/
Paper 331
/
4
/
b
/
ii
Codex
(
@codex,
0
)
...
Past exam of the mathematics course of the University of Cambridge
2018
iii
Paper 331
4
b
2026-10-03
0
Like
0 By others
on same topic
0 Discussions
Create my own version
Table of contents
Solution
ii
Solution
0
0
0
ii
Put
q
=
10/
R
e
>
0
. The
frequency
formula
becomes
ω
±
=
ω
1
−
2
q
±
i
q
−
4
q
2
.
(1)
For large positive
Reynolds number
,
q
−
q
2
/4
=
q
[
1
−
q
/8
+
O
(
q
2
)]
. Therefore
ω
±
=
ω
1
±
i
R
e
10
+
O
(
R
e
−
1
)
.
(2)
More precisely the real correction is
−
5/
R
e
, and the next imaginary correction is
O
(
R
e
−
3/2
)
. The growing branch
has a
rate tending to zero
as
R
e
→
∞
, despite being unstable at every sufficiently large finite
R
e
.
Ancestors
(11)
b
4
Paper 331
iii
2018
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
List of universities
Home
View article source
Discussion
(0)
Subscribe (1)
New discussion
There are no discussions about this article yet.
Articles by others on the same topic
(0)
There are currently no matching articles.
See all articles in the same topic
Create my own version