Answer
The second least water height is $~~0.375~m$
Work Step by Step
We can find the lengths of a half-open pipe which will produce a resonant frequency:
$f = \frac{nv}{4L},~~$ where $~~n = 1,3,5,...$
$L = \frac{nv}{4f},~~$ where $~~n = 1,3,5,...$
$L = \frac{(n)(343~m/s)}{(4)(686~Hz)},~~$ where $~~n = 1,3,5,...$
$L = \frac{n}{8},~~$ where $~~n = 1,3,5,...$
$L = \frac{1}{8}~m,\frac{3}{8}~m,\frac{5}{8}~m,\frac{7}{8}~m,...$
The second least water height in the pipe is when the length of the air-filled space in the pipe is $\frac{5}{8}~m$
We can find the least water height:
$h = 1.00~m-\frac{5}{8}~m$
$h = \frac{3}{8}~m$
$h = 0.375~m$
The second least water height is $~~0.375~m$
