Answer
f=39.6Hz
Work Step by Step
We know that for a single spring
$F=-Kx$
for two identical springs, it will be doubled that is
$F=-2Kx$
but$F=ma$
so $ma=-2Kx$
also $a=-\omega^2x$
so $m(-\omega^2)x=-2Kx$
or $\omega^2=\frac{2K}{m}$
$\omega=\sqrt{\frac{2K}{m}}$
put the values
$\omega=\sqrt{\frac{2\times7580}{0.245}}$
$\omega=248.75$
or $\omega=249 \frac{rad}{s}$
Now to find the frequency, we use the following formula
$\omega=2\pi$$f$
or $f=\frac{\omega}{2\pi}$
putting the values
$f=\frac{249}{6.28}$
$f=39.6Hz$
