Answer
$\approx 26.28$Hz
Work Step by Step
General equation for inverse variation is given by $y=\dfrac{k}{x}$
Here, we have $f=k \dfrac{\sqrt T}{Ld}$ ...(1)
Plugging in the data, we have $262=k \dfrac{\sqrt {670}}{(62)(0.1025)}$
and $K \approx 64.3249482$
Thus, equation (1) becomes:
$f=64.3249482 \dfrac{\sqrt {1629}}{(201.6)(0.49)} \approx 26.28$Hz
