Answer
For the $4^{th}$ harmonics of B, the frequency match the frequency of A’s first harmonic. The harmonic belongs to the the first eight harmonics of string B.
Work Step by Step
For a stretched string of length $L$ with fixed ends, the resonant frequencies are
$f=\frac{v}{\lambda}$
or, $f=\frac{n}{2L}\sqrt {\frac{T}{\mu}}\;\;\;\text{for}\;n=1, 2, 3, . . . .$
For string A
$f_A=\frac{n}{2L}\sqrt {\frac{T}{\mu}}$
For string B
$f_B=\frac{n}{2\times(4L)}\sqrt {\frac{T}{\mu}}$
Let, For the $n^{th}$ harmonics of B, the frequency match the frequency of A’s first harmonic. Therefore,
$\frac{n}{2\times(4L)}\sqrt {\frac{T}{\mu}}=\frac{1}{2L}\sqrt {\frac{T}{\mu}}$
or, $n=4$
Therefore, for the $4^{th}$ harmonics of B, the frequency match the frequency of A’s first harmonic. The harmonic belongs to the the first eight harmonics of string B.
