Answer
a) $v = \Delta x \sqrt{\frac{k}{m}}$'
b) $\omega = \frac{m\Delta x \sqrt{\frac{k}{m}}b}{I}$
Work Step by Step
a) We know energy is conserved, so:
$\frac{1}{2}k\Delta x^2 = \frac{1}{2}mv^2 \\ v = \Delta x \sqrt{\frac{k}{m}}$
b) We know angular momentum is conserved, so:
$mvb=I\omega$
$m\Delta x \sqrt{\frac{k}{m}}b=I\omega$
$\omega = \frac{m\Delta x \sqrt{\frac{k}{m}}b}{I}$
