Question

12.
Anideal gas having adiabatic constant yis enclosed in a adiabatic cylindrical container
having area of cross-section A. A piston of mass M can move freely in the cylinder is
attached with ideal springs k, and k, and light pulleys P., P, and P, as shown in figure.
When the piston is in equilibrium the volume of the gas in the cylinder is Vo and its
pressure is Po. The piston is slightly displaced from the equilibrium position and
released. Find the time period of oscillation. Pulley P, is fixed.
Pat
& k2
M
M
Za
(B)
PAQ
1 kqk2
12k, tkz
PA²
Vo
M
(A) 21
kzkz PoA²
V ky tkz. Vo
IM
13k₃ k2 VPA²
ky + kz
(C) 21
(D*) 256
M
4kgk2 & YPA²
ki tkz. Vo
Vol​

Answer 1

▶ Solution :

✏ First, see the attachment for better understanding.

✴ Calculation :

▪ From the given figure, it is clear that, two springs are in series connection.

$\star\bf\blue{K_{eq}=\dfrac{K_1K_2}{K_1+K_2}}\\ \\ \sf\:At\:equilibrium\:position\:of\:piston\\ \\ \implies\sf\:2T+P_oA=Mg\\ \\ \implies\sf\:2{\huge{[}}\dfrac{K_1K_2}{K_1+K_2}\times 2x{\huge{]}}+P_oA=Mg\\ \\ \because\bf\:T=Kx=\dfrac{K_1K_2}{K_1+K_2}(2x)\\ \\ \sf\:Displace\:it\:\Delta{x}\:from\:eq\\ \\ \implies\sf\:\dfrac{4K_1K_2}{K_1+K_2}(x+\Delta{x})+PA-Mg=F_{net}\:(Upward)\\ \\ \implies\sf\:\dfrac{4K_1K_2x}{K_1+K_2}+\dfrac{4K_1K_2\Delta{x}}{K_1+K_2}+PA-Mg=F_{net}\\ \\ \implies\sf\:Mg-P_oA+\dfrac{4K_1K_2\Delta{x}}{K_1+K_2}+PA-Mg=F_{net}\\ \\ \implies\sf\:\dfrac{4K_1K_2\Delta{x}}{K_1+K_2}+PA-P_oA=F_{net}\rightarrow(1)$

⏭ For adiabatic process, we know that

$\mapsto\bf\:PV^{\gamma}=constant\\ \\ \mapsto\sf\:P_o{V_o}^{\gamma}=P{V_o-A\Delta{x}}^{\gamma}\\ \\ \mapsto\bf\:P_o{\huge{(}}\dfrac{V_o}{V_o-A\Delta{x}}{\huge{)}}^{\gamma}=P\\ \\ \rightarrow\sf\:PA-P_oA=P_o{\huge{(}}\dfrac{V_o}{V_o-A\Delta{x}}{\huge{)}}^{\gamma}A-P_oA\\ \\ \rightarrow\sf\:PA-P_oA=P_oA{\huge{[}}{\huge{(}}1-\dfrac{A\Delta{x}}{V_o}^{-\gamma}{\huge{)}}-1{\huge{]}}\\ \\ \rightarrow\sf\:PA-P_oA=P_oA{\huge{[}}1+\dfrac{\gamma A\Delta{x}}{V_o}-1{\huge{]}}\\ \\ \rightarrow\sf\:PA-P_oA=\dfrac{\gamma P_oA^2\Delta{x}}{V_o}\\ \\ \dag\sf\:substituning\:value\:in\:(1)$

$\leadsto\sf\:\dfrac{4\Delta{x}K_1K_2}{K_1+K_2}+\dfrac{\gamma P_oA^2}{V_o}\Delta{x}=F_{net}\\ \\ \leadsto\sf\:{\huge{[}}\dfrac{4K_1K_2}{K_1+K_2}+\dfrac{\gamma P_oA^2}{V_o}{\huge{]}}\Delta{x}=F_{net}\\ \\ \dag\bf\:comparing\:with\:F=Kx\\ \\ \dashrightarrow\bf\purple{K=\dfrac{4K_1K_2}{K_1+K_2}+\dfrac{\gamma P_oA^2}{V_o}}\\ \\ \bigstar\bf\:\orange{T=2\pi\sqrt{\dfrac{M}{K}}}\\ \\ \underline{\boxed{\bf{\red{T=2\pi\sqrt{\dfrac{M}{\dfrac{4K_1K_2}{K_1+K_2}+\dfrac{\gamma P_oA^2}{V_o}}}}}}}$

✒ Option D) is correct !!

✌ Nice question...