Answer
$7.68cm$
Work Step by Step
We can find the require distance as follows:
$T=\frac{56.7s}{102 \space oscillaitons}=0.556s$
We know that
$d=\frac{mg}{K}=(\frac{m}{K})g$.....eq(1)
We also know that
$T=2\pi \sqrt{\frac{m}{K}}$
This simplifies to:
$\frac{m}{K}=(\frac{T}{2\pi})^2$
Now from eq(1) $d=(\frac{T}{2\pi})^2g$
We plug in the known values to obtain:
$d=(\frac{0.556}{2\pi})^2(9.81)$
$d=0.0768m=7.68cm$
