Answer
The smallest value at which a harmonic frequency of B matches one of the harmonic frequencies of A is $~~n_B = 2$
Work Step by Step
We can write an expression for the harmonic frequencies of pipe A:
$f = \frac{nv}{4L_A},~~$ where $~~n = 1,3,5,...$
We can write an expression for the harmonic frequencies of pipe B:
$f = \frac{nv}{2L_B},~~$ where $~~n = 1,2,3,...$
$f = \frac{nv}{(2)(4~L_A)},~~$ where $~~n = 1,2,3,...$
$f = \frac{nv}{8~L_A},~~$ where $~~n = 1,2,3,...$
When $n_B = 2$, then $f = \frac{2v}{8~L_A} =\frac{v}{4~L_A}$
The smallest value at which a harmonic frequency of B matches one of the harmonic frequencies of A is $~~n_B = 2$
