Question

A smooth piston of mass $m$ and area of cross-section $A$ is in equilibrium with the gas in the jar, when the pressure of the gas is $P_0$. Find the angular frequency of oscillation of the piston, assuming adiabatic change of state of the gas.

• A. $\omega =\sqrt{\frac{\gamma P_0{A}^2}{2m{V}_0}}$
• B. $\omega =\sqrt{\frac{2\gamma P_0{A}^2}{m{V}_0}}$
• C. $\omega =\sqrt{\frac{P_0{A}^2}{m{V}_0}}$
• D. $\omega =\sqrt{\frac{\gamma P_0{A}^2}{m{V}_0}}$

Answer 1

The correct option is D $\omega =\sqrt{\frac{\gamma P_0{A}^2}{m{V}_0}}$


Equation of state of an adiabatic process is given by
$P{V}^{\gamma }=k$
Taking logarithm on both sides, we have
$\log P+\gamma \log V=\log k$
Differentiating on both sides, and using the data given in the diagram, we get
$\frac{\Delta P}{P_0}+\frac{\gamma \Delta V}{V_0}=0$
When we disturb the piston by displacing it by a small distance $\Delta x$, the excess pressure $\Delta P$ can be written as
$\Delta P=\frac{-\gamma P_0}{V_0}\Delta V......\left(1\right)$
where $\Delta V$ = change in volume of the gas $=A\Delta x$.
We know that, $\Delta P=\frac{\Delta F}{A}$
Using this in $\left(1\right)$, we get
$\frac{\Delta F}{A}=-\frac{\gamma P_0}{V_0}A\Delta x$
$\Rightarrow \frac{\Delta F}{\Delta x}=-\frac{\gamma P_0{A}^2}{V_0}$
We know , $k_{eff}=\left|\frac{\Delta F}{\Delta x}\right|=\frac{\gamma P_0{A}^2}{V_0}$
Then, by using $\omega =\sqrt{\frac{k_{eff}}{m}}$,
we can say that, $\omega =\sqrt{\frac{\gamma P_0{A}^2}{m{V}_0}}$
Thus, option (d) is the correct answer.