JEE Main2013PhysicsOscillationsActual
An ideal gas enclosed in a vertical 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 V 0 and its pressure is M 0 . The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston execut
Options
- A1 2 π A 2 γ P 0 MV 0 .
- B1 2 π MV 0 A γ P 0 .
- C1 2 π A γ P 0 V 0 M
- D1 2 π V 0 MP 0 A 2 γ .
Correct answer
A. 1 2 π A 2 γ P 0 MV 0 .
Step-by-step solution
For an adiabatic process, Pv γ = constant Differentiating with respect to v , dP dv v γ + P γ v γ - 1 = 0 dP = - γP v dv = - γP v Adx Thus, the acceleration, a = dF M = γ PA 2 MV dx Comparing with , a = - ω 2 dx , we get, ω = γ P 0 A 2 M V 0 = 2 π f f = 1 2 π γ P 0 A 2 M V 0