8 6

PHY 5347 Homework Set 2 Solutions – Kimel 1. 8.5 a)

R

L For the TM modes, we saw in class the resonance frequencies are TM:

ω mnp =

1 μ

x 2mn p2π2 + R2 L2

p = 0, 1, 2, . . . . m = 0, 1, 2, . . . . n = 1, 2, 3, . . . .

TE:

ω ′mnp =

1 μ

p2π2 x ′2 mn + L2 R2

p = 1, 2, 3, . . . . m = 0, 1, 2, . . . . n = 1, 2, 3, . . . .

Thus ω mnp 1 μ

ω ′mnp 1 μ

=

x 2mn p2π2 + 2 R L2

=

x ′2 p2π2 mn + R2 L2

The lowest four frequencies are (in these units) ω 010 = 2. 405 ω 110 = 3. 832 ω ′111 =

1. 841 2 + π 2

R L

2

ω ′211 =

3. 054 2 + π 2

R L

2

2. 405, 3. 832, 1. 841 2 + π 2 x 2 , 3. 054 2 + π 2 x 2

10 8 6 4 2

0

0.2

0.4

0.6

0.8

1 x

1.2

1.4

1.6

1.8

2

where x = R/L. The answer is ”No.” ω ′111 and ω 010 cross when 1. 841 2 + π 2 x 2 = 2. 405 or x = 0. 492 58. For frequencies smaller than this cross over frequency, ω ′111 is lowest, whereas for larger frequencies, ω 010 is lowest.