Answer
The frequency of the oscillations is $~~18.2~Hz$
Work Step by Step
We can find the equivalent spring constant $K$ of the two springs in combination.
When the springs are stretched a total distance of $x$ then each spring is stretched a distance of $\frac{x}{2}$
We can find $K$:
$Kx = k~\frac{x}{2}$
$Kx = (6430~N/m)(\frac{x}{2})$
$k = 3215~N/m$
We can find the frequency of the oscillations:
$f = \frac{1}{2\pi}~\sqrt{\frac{K}{m}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{3215~N/m}{0.245~kg}}$
$f = 18.2~Hz$
The frequency of the oscillations is $~~18.2~Hz$
