Question

An ideal gas enclosed in a verticle cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is Vo and its pressure is Po. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surroundings, the piston executes a simple harmonic motion with frequency

A)$\sf\dfrac{1}{2\pi}\sqrt{\dfrac{MV_o}{A\gamma P_o}}$
B)$\sf\dfrac{1}{2\pi}\dfrac{A\gamma P_o}{V_oM}$
C)$\dfrac{1}{2\pi}\dfrac{V_oMP_o}{A^2\gamma}$
D)$\dfrac{1}{2\pi}\sqrt{\dfrac{A^2\gamma P_o}{MV_o}}$

▶ Don't spam !!​

Answer 1

Pls Refer the attachment

I hope this helps ( ╹▽╹ )

Answer 2

Solution 1:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(5,0){2}{\line(0,1){20}}}\put(0,7.5){\framebox(5,5){\sf{M}}}\put(1.3,-4){ \sf{A} }\put(7,15){\vector(0,-1){10}}\put(8,9.5){ \sf{\Delta x} }\put(0,0){\line(1,0){5}}\end{picture}$

The fig. shows the cylindrical container with a freely moving piston of mass M. Assume there's no friction between the walls of the container and the piston.

The free body diagram of the piston at equilibrium is shown below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(5,5){\sf{M}}}\put(2.5,0){\vector(0,-1){10}}\put(0,-14){ \sf{Mg} }\put(2.5,5){\vector(0,1){10}}\put(0,18){ \sf{P_0A} }\end{picture}$

Here Mg is the weight of the piston and $\sf{P_0A}$ is the force exerted by the gas inside the container to the piston.

From this we get,

$\longrightarrow\sf{Mg=P_0A\quad\quad\dots(1)}$

Let the piston be displaced by a distance $\sf{\Delta x}$ downwards. Let the pressure exerted by the gas at this time be P.

The free body diagram of the piston at this moment is,

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(5,5){\sf{M}}}\put(2.5,0){\vector(0,-1){10}}\put(0,-14){ \sf{Mg} }\put(2.5,5){\vector(0,1){10}}\put(1,18){ \sf{PA} }\end{picture}$

The net force acting on the piston at this time, provides restoring force to the piston since it executes SHM.

Hence the restoring force $\sf{k\,\Delta x}$ is given by (k is the force constant),

$\longrightarrow\sf{k\,\Delta x=PA-Mg}$

Restoring force acts upward since $\sf{\Delta x}$ is acting downwards.

From (1),

$\longrightarrow\sf{k\,\Delta x=PA-P_0A}$

$\longrightarrow\sf{k\,\Delta x=(P-P_0)A\quad\quad\dots(2)}$

After M being displaced, the volume of the gas is decreased by $\sf{A\,\Delta x.}$ Hence new volume is,

$\longrightarrow\sf{V=V_0-A\,\Delta x}$

Since the system is completely isolated from surroundings, the system undergoes adiabatic process, thus we have,

$\longrightarrow\sf{PV^{\gamma}=a\ constant}$

$\Longrightarrow\sf{P_0(V_0)^{\gamma}=PV^{\gamma}}$

$\longrightarrow\sf{P_0(V_0)^{\gamma}=P(V_0-A\,\Delta x)^{\gamma}}$

$\longrightarrow\sf{P=P_0\left(\dfrac{V_0}{V_0-A\,\Delta x}\right)^{\gamma}}$

$\longrightarrow\sf{P=P_0\left(1-\dfrac{A\,\Delta x}{V_0}\right)^{-\gamma}}$

We assume $\sf{\Delta x}$ very small, so that $\sf{\dfrac{A\,\Delta x}{V_0}}$ can be neglected,

Since $\sf{(1+x)^n=1+nx}$ for $\sf{|x|\ \textless\textless\ 0,}$

$\longrightarrow\sf{P=P_0\left(1+\dfrac{A\gamma\,\Delta x}{V_0}\right)}$

Subtracting $\sf{P_0}$ from both sides,

$\longrightarrow\sf{P-P_0=P_0\left(1+\dfrac{A\gamma\,\Delta x}{V_0}\right)-P_0}$

$\longrightarrow\sf{P-P_0=\dfrac{A\gamma P_0\,\Delta x}{V_0}}$

Hence (2) becomes,

$\longrightarrow\sf{k\,\Delta x=\dfrac{A^2\gamma P_0\,\Delta x}{V_0}}$

$\longrightarrow\sf{k=\dfrac{A^2\gamma P_0}{V_0}}$

Finally, the frequency of the piston is,

$\longrightarrow\sf{f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{M}}}$

$\longrightarrow\sf{\underline{\underline{f=\dfrac{1}{2\pi}\sqrt{\dfrac{A^2\gamma P_0}{MV_0}}}}}$

Hence (D) is the answer.

Solution 2:-

We simply check the dimensions of the options mentioned here, to check which one has the same dimension as that of frequency.

$\longrightarrow\sf{[f]=T^{-1}}$

Avoiding the dimensionless constants $\sf{2\pi}$ and $\gamma,$

$\longrightarrow\sf{\left[\sqrt{\dfrac{MV_0}{AP_0}}\right]=\sqrt{\dfrac{M\,L^3}{L^2\,ML^{-1}T^{-2}}}}$

$\longrightarrow\sf{\left[\sqrt{\dfrac{MV_0}{A\gamma P_0}}\right]=LT}$

Hence (A) is wrong.

$\longrightarrow\sf{\left[\dfrac{AP_0}{V_0M}\right]=\dfrac{L^2\,ML^{-1}T^{-2}}{L^3\,M}}$

$\longrightarrow\sf{\left[\dfrac{AP_0}{V_0M}\right]=L^{-2}T^{-2}}$

Hence (B) is wrong.

$\longrightarrow\sf{\left[\dfrac{V_0MP_0}{A^2}\right]=\dfrac{L^3\,M\,ML^{-1}T^{-2}}{L^4}}$

$\longrightarrow\sf{\left[\dfrac{V_0MP_0}{A^2}\right]=M^2L^{-2}T^{-2}}$

Hence (C) is wrong. So we can be sure that (D) is the answer.

$\longrightarrow\sf{\left[\sqrt{\dfrac{A^2P_0}{MV_0}}\right]=\sqrt{\dfrac{L^4\,ML^{-1}T^{-2}}{M\,L^3}}}$

$\longrightarrow\sf{\underline{\underline{\left[\sqrt{\dfrac{A^2P_0}{MV_0}}\right]=T^{-1}}}}$

Hence (D) is the answer.