Answer
The sound produced by the string will cause resonance in pipe d, and the resonance will be the fundamental frequency $n=1$
Work Step by Step
We can find an expression for the fundamental frequency of the string:
$f = \frac{v}{\lambda} = \frac{v}{2L} = 0.5~\frac{v}{L}$
We can find an expression for the resonant frequencies of pipe a:
$f = \frac{nv}{4L},$ where $n = 1,3,5,...$
$f = 0.25~\frac{v}{L}, 0.75~\frac{v}{L}, 1.25~\frac{v}{L},...$
We can find an expression for the resonant frequencies of pipe b:
$f = \frac{nv}{4(2L)},$ where $n = 1,3,5,...$
$f = 0.125~\frac{v}{L}, 0.375~\frac{v}{L}, 0.625~\frac{v}{L},...$
We can find an expression for the resonant frequencies of pipe c:
$f = \frac{nv}{2(L/2)},$ where $n = 1,2,3,...$
$f = \frac{v}{L}, 2~\frac{v}{L}, 3~\frac{v}{L},...$
We can find an expression for the resonant frequencies of pipe d:
$f = \frac{nv}{4(L/2)},$ where $n = 1,3,5,...$
$f = 0.5~\frac{v}{L}, 1.5~\frac{v}{L}, 2.5~\frac{v}{L},...$
The sound produced by the string will cause resonance in pipe d, and the resonance will be the fundamental frequency $n=1$
