91Ó°ÊÓ

Newton's Third law: The net force$F$on the body with mass $m$is relevant to the body's acceleration by

The maximum transverse acceleration of a particle in simple harmonic motion can be found in areas of angular speed and amplitude

The angular frequency of the wave is relevant to frequency by

The amplitude of piston is $A=(4.0cm)(1/100cm)=0.4m$

The frequency of oscillation of piston will be increasing.

In part (a)

The objective is to find out the point at which the penny first loses it contact with piston

In part (b)

The objective is to determine the

maximum frequency for which the penny just

barely remains in place for the full cycle

The free body diagram in figure 1 displays the force acting on the penny:$m\stackrel{→}{\mathrm{g}}$is the gravitational force exerted by the earth on the penny and$\stackrel{→}{n}$ is the normal contact force

Applying Newton's law in Equation $\left(1\right)$in the vertical direction to penny

$∑F_y=n−mg=ma$

Solve

$n=m(g+a)$The penny loses contact with the surface of the oscillating piston when the normal force $n$exerted by the piston is zero. So

$0=m(g+a)$

$a=-g$

As a result, the valve loses contact with the piston as the piston begins to accelerate downward. First, the piston accelerates downward at its highest point and so the coin loses contact at the highest point of the piston.

The maximum acceleration of the penny at highest point can be obtained from Equation $\left(2\right)$

$\begin{array}{c}a=−{\mathrm{Ó\neg }}^{2}A\end{array}$

therefore, $a=-g$

$\begin{array}{c}\\ −g=−{\mathrm{Ó\neg }}^{2}A\\ g={\mathrm{Ó\neg }}^{2}A\end{array}$

Solve for $\begin{array}{c}\mathrm{Ó\neg }\end{array}$

$\begin{array}{c}\mathrm{Ó\neg }=\sqrt{\frac{g}{A}}\end{array}$

$\begin{array}{c}\begin{array}{c}\mathrm{Ó\neg }\end{array}\end{array}$in the equation $\left(3\right)$

$\begin{array}{c}2\mathrm{Ï€}f=\sqrt{\frac{g}{A}}\end{array}$

Solve for $f$

$\begin{array}{c}f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{g}{A}}\end{array}$

Substitute numerical values

$\begin{array}{r}f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{9.80\mathrm{m}/{\mathrm{s}}^2}{0.04\mathrm{m}}}\end{array}$

$=2.5Hz$