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.

$\left(a\right)$At what point in the cycle does the penny first lose contact with the piston?

$\left(b\right)$What is the maximum frequency for which the penny just barely remains in place for the full cycle?

Short answer

Part $\left(a\right)$

$\left(a\right)$As a result,the oscillation at its maximum height, the oscillator begins to speed downhill, and the penny lost contact with the oscillator.

Part $\left(b\right)$

$\left(b\right)$The maximum frequency at penny remains piston at$f=2.5Hz.$

Step-by-step solution

Step; 1 Penny conactact with oscillator: (part a)

In the case of the oscillation of piston in upward direction, the coin remains on the surface due to normal force $n$.

The coin in red oscillates with the piston with acceleration $a,m$is its mass, and $g$is gravity.

$ma=n-mg$

$n=m(a+g)$

The piston accelerating upwards.and the normal force zero.

$0=m(a+g)$

$a=-g$

As a result, as the piston begins to accelerate downwards, the penny will loses contact area of the oscillating piston.During the oscillation at its maximum height, the oscillator begins to speed downhill, and the penny lost contact with the oscillator.

Step: 2 Equating angular frequency: (part b)

The acceleration at highest point is,

$\begin{array}{l}a=-g\\ -{\omega }^{2}A=-g\\ \omega =\sqrt{\frac{g}{A}}\end{array}$

The angular frequency at

$\begin{array}{l}\omega =2\pi f\\ 2\pi f=\sqrt{\frac{g}{A}}\\ f=\frac{1}{2\pi }\sqrt{\frac{g}{A}}\end{array}$

It remains contact for frequency penny.

Step: 3 Finding frequency: (part b)

Substituting the values of $g=9.8m/s^2$and $A=4.0cm$in above equation as,

$\begin{array}{l}f=\frac{1}{2\pi }\sqrt{\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\left(4.0\mathrm{cm}\right)}}\\ f=\frac{1}{2\pi }\sqrt{\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\left(4.0\mathrm{cm}\right)}\left(\frac{100\mathrm{cm}}{1\mathrm{m}}\right)}\\ f=2.5\mathrm{Hz}.\end{array}$