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tesi R. Valiante.pdf - EleA@UniSA

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30<br />

⎛<br />

⎜<br />

⎜−<br />

k<br />

⎝<br />

2<br />

−<br />

K<br />

F<br />

2<br />

ω ⎞<br />

+ 2 ⎟<br />

⎟e<br />

c0<br />

⎠<br />

( kx−ωt<br />

)<br />

i<br />

= 0<br />

(1.20)<br />

This equation must fulfill Eq.1.21 in order to admit non-trivial solutions<br />

2 ⎛<br />

= c0<br />

⎜k<br />

⎝<br />

or, alternatively, Eq.1.22.<br />

K ⎞<br />

+ ⎟<br />

F ⎠<br />

2<br />

2<br />

ω ω(<br />

k)<br />

2<br />

2 ω<br />

k = − 2<br />

c<br />

0<br />

K<br />

F<br />

Here the phase velocity, p<br />

ω = (1.21)<br />

( ω)<br />

c , is different from the 0<br />

k = k (1.22)<br />

c of the taut string. Indeed,<br />

substituting ω = kc p in Eq.1.21 or Eq.1.22, the result of Eq.1.23 is obtained.<br />

⎛ K<br />

⎜1+<br />

⎝ Fk<br />

2 2<br />

= c0<br />

2<br />

⎞<br />

⎟<br />

⎠<br />

c p c ( k)<br />

c p p<br />

Alternatively, one can obtain the results in Eq.1.24.<br />

k<br />

2<br />

=<br />

K<br />

F<br />

2 2 ( c c ) −1<br />

p<br />

0<br />

= (1.23)<br />

( c )<br />

k = k (1.24)<br />

Another set of relations can be obtained by eliminating k from Eq.1.22 or ω<br />

from Eq.1.24, to give Eq.1.25<br />

and Eq. 1.26<br />

K<br />

Fc<br />

2<br />

p<br />

ω =<br />

ω(<br />

c<br />

2 2<br />

p )<br />

( c c ) −1<br />

p<br />

0<br />

2<br />

p<br />

ω = (1.25)

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