Question

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface.
1) At what point in the cycle does the penny first lose contact with the piston? (answer choices:)

A. midpoint moving up

B. midpoint losing down

C. highest point

D. lowest point


2) What is the maximum frequency for which the penny just barely remains in place for the full cycle?

Answer 1

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

$a = -\omega^2 A$

Where,

a = Acceleration

A = Amplitude

$\omega$= Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

$a = -g$

$-\omega^2 A = -g$

$\omega = \sqrt{\frac{g}{A}}$

From the definition of frequency and angular velocity we have to

$\omega = 2\pi f$

$f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}$

$f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}$

$f = 2.5Hz$

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz