#### Answer

(a) The penny would first lose contact with the piston when the piston reaches its highest point.
(b) The maximum frequency such that the penny remains in contact with the piston is 0.40 Hz.

#### Work Step by Step

(a) The penny would first lose contact with the piston when the piston reaches its highest point. At this point, the piston has its maximum acceleration directed downward.
(b) The magnitude of the maximum acceleration is $\omega^2A$, which occurs when the piston reaches its highest and lowest points. If this maximum acceleration is greater than $g$ at the highest point, then the penny will lose contact with the piston. We can find the maximum value of $\omega$ such that the penny remains in contact with the piston.
$\omega^2A = g$
$\omega = \sqrt{\frac{g}{A}}$
$\omega = \sqrt{\frac{9.80~m/s^2}{0.040~m}}$
$\omega = 15.65~rad/s$
We can find the maximum frequency.
$f = \frac{2\pi}{\omega}$
$f = \frac{2\pi}{15.65~rad/s}$
$f = 0.40~Hz$
The maximum frequency such that the penny remains in contact with the piston is 0.40 Hz.