The dispersion equation of a waveguide, which relates the wavenumber k to the frequency \omega, is k(\omega)=(1 {/} c) \sqrt{\omega^{2}-\omega_{0}^{2}} where the speed of light c=3 \times 10^{8} {m} {/} {s}, and \omega_{0} is a constant. If the group velocity is 2 \times 10^{8} {m} {/} {s}, then the phase velocity is

GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 50 of 65

Marks: +1, -0

The dispersion equation of a waveguide, which relates the wavenumber

k

to the frequency

ω

, is

k(ω)=(1 {/} c) √{ω^{2}-ω_{0}^{2}}

where the speed of light

c=3 × 10^{8} {m} {/} {s}

, and

ω_{0}

is a constant. If the group velocity is

2 × 10^{8} {m} {/} {s}

, then the phase velocity is

Solution:

f(-1)=0

So only option (B) and (C) are possible
Let's try option (B)

f(x)=2|x+1|

∴ f(x)=\{\table 2(x+1) \;\text " for "\; x+1>0 ; -2(x+1) \;\text " for "\; x+1<0.

∴ f(x)=\{\table 2(x+1) , \;\text " for "\; x>-1 ; -2(x+1) , \;\text " for "\; x<-1.

∴ f^{'}(x)=\{\table 2 , \;\text " for "\; x>-1 ; -2 , \;\text " for "\; x<-1.

∴ |f^{'}(x)| ≤ 2

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