Piston SHM

a) Use Newton's second law to get the pressure inside as a function of the atmospheric pressure. Then, by an adiabatic process and its equation, find pressure for the displaced piston. Use Newton's second law again to get an equation similar to a harmonic motion.

b) The period can be found directly from the equation that relates it with the angular frequency.

a) The force exerted due to pressure is:

\begin{equation*}
F=PA.
\end{equation*}

By Newton's second law we have:

\begin{equation}
P_1 A - P_0 A - mg = 0.
\end{equation}

Solving for \(P_1\) we obtain:

\begin{equation*}
P_1=P_0+\frac{mg}{A}.
\end{equation*}

As an adiabatic process, after the piston is displaced, the equation to relate the variables is:

\begin{equation*}
P_1V_0^{\gamma}=P_2(V_0+Ay)^{\gamma},
\end{equation*}

and since \(V_0 = Ah \) we get:

\begin{equation*}
P_2=P_1\frac{1}{\left(1+\frac{y}{h}\right)^{\gamma}}.
\end{equation*}

We can make the following approximation:

\begin{equation*}
P_2\approx P_1\left(1-\gamma\frac{y}{h}\right).
\end{equation*}

By Newton's second law when the piston is displaced we get:

\begin{equation}
P_2 A - P_0 A - mg = -ma.
\end{equation}

By replacing the pressure \(P_2\) and \(P_1\) in terms of the atmospheric pressure, which is can be written as:

\begin{equation*}
\left(P_0+\frac{mg}{A}\right)\left(1-\gamma\frac{y}{h}\right)A-P_0A-mg=-ma.
\end{equation*}

After some algebra, we obtain:

\begin{equation*}
\frac{(P_0A+mg)\gamma}{mh}y=a.
\end{equation*}

The coefficient of the \(y\) variable is the angular frequency. Then:

\begin{equation*}
\omega=\sqrt{\frac{(P_0A+mg)\gamma}{mh}}.
\end{equation*}

b) The period is \(T = \frac{2 \pi}{\omega} \). Then:

\begin{equation*}
T=\frac{2\pi}{\sqrt{\frac{(P_0\pi R^2+mg)\gamma}{mh}}}.
\end{equation*}

Using the numerical given values:

\begin{equation*}
T\approx 0.047\,\text{s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) They ask us to show that the piston motion will resemble a simple harmonic motion. First, let us find the initial pressure inside the cylinder. In order to do this, we must consider the forces exerted on the piston as shown in the force diagram in the next figure.

The forces \(F\) exerted due to pressure \(P\) are written using the relation

\begin{equation}
F=PA,
\end{equation}

where \(A\) is the transverse area, in this case of the piston.

\begin{figure}[H] \centering
\includegraphics[scale=0.8]{Nico y Jose/Img/Ondas/cilindro.png}
\caption{Force diagram for the piston }
\label{fig:cilindro}
\end{figure}

We denote with \(P_0\) the atmospheric pressure, with \(P_1\) the pressure inside the cylinder, \(m\) is the mass of the piston and \(g\) is the gravitational acceleration on earth. The product \(mg\) is the weight of the piston.

We can now use Newton's second law

\begin{equation}
\label{newton}
\sum\vec{F}=m\vec{a},
\end{equation}

where \(\vec{a}\) is the acceleration, \(m\) is the mass of the piston and \(\sum\vec{F}\) is the sum of all forces exerted on the piston. From the direction of the forces as seen in the previous figure we can write equation \eqref{newton} as

\begin{equation}
\label{newton2}
P_1A\,\hat{\textbf{j}}-P_0A\,\hat{\textbf{j}}-mg\,\hat{\textbf{j}}=m\vec{a}.
\end{equation}

Noticing that the piston is in equilibrium, then the acceleration is zero, \(\vec{a}=\vec{0}\). Thus we can rewrite equation \eqref{newton2} as

\begin{equation}
\label{newton3}
P_1A\,\hat{\textbf{j}}-P_0A\,\hat{\textbf{j}}-mg\,\hat{\textbf{j}}=\vec{0},
\end{equation}

which we can write without the vector notation (since all the forces are along the same axis) as

\begin{equation}
P_1A-P_0A-mg=0.
\end{equation}

Solving for \(P_1\) in the equation above we obtain

\begin{equation}
P_1A=P_0A+mg,
\end{equation}

\begin{equation}
\label{p1}
P_1=P_0+\frac{mg}{A}.
\end{equation}

When the piston is displaced a distance \(y\) upwards, the gas inside the cylinder is expanded adiabatically and reaches a pressure \(P_2\) and a volume \(V=V_0+Ay\), where \(V_0\) is the initial volume. Because the process is adiabatic, we can write
the following expression for the pressure and volume before and after the adiabatic compression

\begin{equation}
\label{adiabat}
P_1V_0^{\gamma}=P_2(V_0+Ay)^{\gamma},
\end{equation}

where \(\gamma\) is the ratio of the heat capacity at constant pressure and the heat capacity at constant volume. For a diatomic gas \(\gamma=7/5\). Solving for \(P_2\) in equation \eqref{adiabat} we get

\begin{equation}
\label{adiabat2}
P_2=P_1\frac{V_0^{\gamma}}{(V_0+Ay)^{\gamma}}.
\end{equation}

The initial volume \(V_0\) is equal to product of the transverse area and the initial height \(h\), thus

\begin{equation}
\label{volumen}
V_0=Ah.
\end{equation}

Using the result for the initial volume given in equation \eqref{volumen} into equation \eqref{adiabat2} we obtain

\begin{equation}
P_2=P_1\frac{(Ah)^{\gamma}}{(Ah+Ay)^{\gamma}},
\end{equation}

where the transverse area \(A\) can be cancelled out to get

\begin{equation}
P_2=P_1\frac{h^\gamma}{(h+y)^{\gamma}}.
\end{equation}

We will write the above equation in the convenient form

\begin{equation}
P_2=P_1\frac{1}{\left(1+\frac{y}{h}\right)^{\gamma}}.
\end{equation}

If the displacement \(y\) is much smaller than the height \(h\) we can make the following approximation

\begin{equation}
\label{p2}
P_2\approx P_1\left(1-\gamma\frac{y}{h}\right).
\end{equation}

Using the result for \(P_1\) given in equation \eqref{p1} into the expression above we get

\begin{equation}
\label{p2best}
P_2\approx \left(P_0+\frac{mg}{A}\right)\left(1-\gamma\frac{y}{h}\right)
\end{equation}

We make the force diagram once again when the piston is slightly displaced by \(y\) as seen in the following figure.

\begin{figure}[H] \centering
\includegraphics{Nico y Jose/Img/Ondas/cilindro.png}
\caption{Es el mismo diagrama pero cambiar P1 por P2}
\label{fig:cilindro2}
\end{figure}

Using again Newton's second law we can write in this case

\begin{equation}
\label{newton4}
P_2A\,\hat{\textbf{j}}-P_0A\,\hat{\textbf{j}}-mg\,\hat{\textbf{j}}=m\vec{a}.
\end{equation}

Because the pressure \(P_2\) is less than the equilibrium pressure \(P_1\) from relation \eqref{p2}, then the acceleration is directed towards the negative Y axis, that is

\begin{equation}
\vec{a}=-a\,\hat{\textbf{j}}.
\end{equation}

Using the above result into equation \eqref{newton4} we obtain

\begin{equation}
\label{newton5}
P_2A\,\hat{\textbf{j}}-P_0A\,\hat{\textbf{j}}-mg\,\hat{\textbf{j}}=-ma\,\hat{\textbf{j}},
\end{equation}

or omitting the vector notation for simplicity

\begin{equation}
\label{newton6}
P_2A-P_0A-mg=-ma.
\end{equation}

Using the results for \(P_2\) given by equation \eqref{p2best} into equation \eqref{newton6} we get

\begin{equation}
\label{newton7}
\left(P_0+\frac{mg}{A}\right)\left(1-\gamma\frac{y}{h}\right)A-P_0A-mg=-ma.
\end{equation}

Expanding the terms in parenthesis in equation \eqref{newton7} we can write

\begin{equation}
P_0A+mg-(P_0A+mg)\gamma\frac{y}{h}-P_0A-mg=-ma,
\end{equation}

which after cancelling out some terms is

\begin{equation}
-(P_0A+mg)\gamma\frac{y}{h}=-ma,
\end{equation}

or equivalently

\begin{equation}
\label{final}
\frac{(P_0A+mg)\gamma}{mh}y=a.
\end{equation}

The relation between acceleration and displacement for a simple harmonic oscillator in the Y axis is

\begin{equation}
\label{ho}
\omega^2 y=a,
\end{equation}

where \(\omega\) is the angular frequency of the oscillatory motion. Contrasting equations \eqref{final} and \eqref{ho} we can conclude the effectively , when \(y\ll h\) the motion of the cylinder is that of a simple harmonic oscillator with angular frequency given by

\begin{equation}
\omega^2=\frac{(P_0A+mg)\gamma}{mh},
\end{equation}

or equivalently

\begin{equation}
\label{omega}
\omega=\sqrt{\frac{(P_0A+mg)\gamma}{mh}}.
\end{equation}

b) Finally, we need to find the period of this motion. In terms of the angular frequency, the period is given by

\begin{equation}
T=\frac{2\pi}{\omega}.
\end{equation}

Using the result from equation \eqref{omega} int the equation above we get

\begin{equation}
T=\frac{2\pi}{\sqrt{\frac{(P_0A+mg)\gamma}{mh}}},
\end{equation}

where we can include the explicit expression for the transverse area \(A=\pi R^2\), where \(R\) is the radius of the cylnder to get

\begin{equation}
T=\frac{2\pi}{\sqrt{\frac{(P_0\pi R^2+mg)\gamma}{mh}}}.
\end{equation}

Using the numerical given values in SI units we get

\begin{equation}
T=\frac{2\pi}{\sqrt{\frac{((101325\,\text{Pa})\pi(0.2\,\text{m})^2+(2\,\text{kg})(9.81\,\text{m/s}^2))(7/5)}{(2\,\text{kg})(0.5\,\text{m})}}},
\end{equation}

\begin{equation}
T\approx 0.047\,\text{s}.
\end{equation}