Question

A smooth piston of mass $m$ and cross - section area $A$ is in equilibrium on a diatomic gas filled in a cylindrical container and there is vaccum above the piston. The piston and container are non - conducting and the height of gas column in the container is $l_0$ at equilibrium. If the piston is displaced slightly down the equilibrium position, then the time period of its oscillation will be:

• A. $2\pi \sqrt{\frac{10l_0}{7g}}$
• B. $2\pi \sqrt{\frac{7l_0}{10g}}$
• C. $2\pi \sqrt{\frac{7l_0}{5g}}$
• D. $2\pi \sqrt{\frac{5l_0}{7g}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{5l_0}{7g}}$

At equilibrium $P_1=\frac{mg}{A}$
Now, lets displace the positon by $x$ distance downward
$P_i{V}_i^{\gamma }=P_t{V}_f^{\gamma }$
$\left(\frac{mg}{A}\right)(l_0{)}^{\gamma }=(P_f\left)\right(l_0-x{)}^{\gamma }$
$P_f=\left(\frac{mg}{A}\right){(1-\frac{x}{l_0})}^{-\gamma }=\frac{mg}{A}(1+\frac{\gamma x}{l_0})$
$P_f=\frac{mg}{A}+\frac{mg}{A}\frac{\gamma }{l_0}x$
$F_{net}=P_{f}A-mg=\frac{\gamma mg}{l_0}x$
$F_{net}=-\left(\frac{\gamma mg}{l_0}\right)x$
As $F_{net}\propto -x$, piston undergoes SHM.
Time period of oscillation is
$T=2\pi \sqrt{\frac{m}{\gamma \left(\frac{mg}{l_0}\right)}}\Rightarrow T=2\pi \sqrt{\frac{l_0}{\gamma g}}\text{where}\gamma =1+\frac{2}{5}$
$T=2\pi \sqrt{\frac{5l_0}{7g}}$

Why this question ? Key Concept: Here time period of piston is independent of area of cross section of piston i.e if the piston were spherical, then also the answer would be the same. Tip: Here gas obeys adiabatic process as cylinder and piston are non-conducting.