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In the Previous Problem, the Solution of the Eigenvalue Problem λ=zn2/4\lambda = z _ { n } ^ { 2 } / 4

Question 34

Multiple Choice

In the previous problem, the solution of the eigenvalue problem is


A) λ=zn2/4\lambda = z _ { n } ^ { 2 } / 4 , where J0(zn) =0;R=J0(znr/2) J _ { 0 } \left( z _ { n } \right) = 0 ; R = J _ { 0 } \left( z _ { n } r / 2 \right)
B) λ=zn2\lambda = z _ { n } ^ { 2 } , where J0(zn) =0;R=J0(znr) J _ { 0 } \left( z _ { n } \right) = 0 ; R = J _ { 0 } \left( z _ { n } r \right)
C) λ=nπ/9,Z=cos(nπz/3) \lambda = n \pi / 9 , Z = \cos ( n \pi z / 3 )
D) λ=n2π2/9,Z=cos(nπz/3) \lambda = n ^ { 2 } \pi ^ { 2 } / 9 , Z = \cos ( n \pi z / 3 )
E) λ=n2π2/9,Z=sin(nπz/3) \lambda = n ^ { 2 } \pi ^ { 2 } / 9 , Z = \sin ( n \pi z / 3 )

Correct Answer:

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