b
(b) When the piston is moved through a distance of $8.75cm$, the path difference produced is $2 \times 8.75\,cm = 17.5\,cm$. This must be equal to$\frac{\lambda }{2}$ for maximum to change to minimum.
$\therefore$ $\frac{\lambda }{2} = 17.5 cm$
==> $\lambda = 35cm = 0.35m$
So, $v = n\lambda $==> $n = \frac{v}{\lambda } = \frac{{350}}{{0.35}} = 1000Hz$