Answer
$T = 0.0653~s$
Work Step by Step
We can find an expression for the magnitude of the linear acceleration of the endpoint for any displacement $x$:
$\tau = I~\alpha$
$\frac{L}{2}~F = \frac{1}{12}mL^2~\frac{a}{L/2}$
$F = \frac{1}{3}ma$
$kx = \frac{1}{3}ma$
$a = \frac{3k}{m}~x$
Then $\omega = \sqrt{\frac{3k}{m}}$
We can find the period:
$T = 2\pi~\sqrt{\frac{m}{3k}}$
$T = 2\pi~\sqrt{\frac{0.600~kg}{(3)(1850~N/m)}}$
$T = 0.0653~s$
