Answer
The required speed of the detector is $~~\frac{4}{5}~v$
Work Step by Step
We can find an expression for frequency of the third harmonic in tube 1:
$f = \frac{nv}{4L},~~$ where $n = 1,3,5,...$
$f = \frac{5v}{4L}$
Note that the fundamental frequency is $\frac{1}{5}$ of the third harmonic frequency.
We can find the required speed of the detector:
$f' = f~\frac{v-v_D}{v}$
$\frac{f'}{f} = \frac{v-v_D}{v}$
$\frac{1}{5} = \frac{v-v_D}{v}$
$\frac{1}{5}~v = v-v_D$
$v_D = v-\frac{1}{5}~v$
$v_D = \frac{4}{5}~v$
The required speed of the detector is $~~\frac{4}{5}~v$
