$\pi {r^2}\rho gx = n{\omega ^2}x$
$\therefore \omega \sqrt {\frac{{\pi {r^2}\rho g}}{{\rho v}}} $
$ = r\sqrt {\frac{{\pi g}}{v}} = 2.5 \times {10^{ - 2}}\sqrt {\frac{{3.14 \times 10}}{{310 \times {{10}^{ - 6}}}}} $
$ = 2.5 \times {10^{ - 2}} \times {10^2}\sqrt {10} $
$\omega = 2.5 \times \sqrt {10} $
$\therefore f = \frac{{2.5 \times \sqrt {10} }}{{2\pi }} = 1.25$
$(A)$ The force is zero $t=\frac{3 T}{4}$
$(B)$ The acceleration is maximum at $t=T$
$(C)$ The speed is maximum at $t =\frac{ T }{4}$
$(D)$ The $P.E.$ is equal to $K.E.$ of the oscillation at $t=\frac{T}{2}$
