An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M . The piston and cylinder have equal cross sectional area A . When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency:
(A)2π1A2γV0MP0
(B)2π1MV0A2γP0✓ Correct
(C)2π1AγP0MV0
(D)2π1V0MAγP0
Answer: (B)
Solution
The piston rests where the gas pressure holds up its weight, and a small downward push x changes the volume by Ax. Because the system is isolated the compression is adiabatic, PVγ=const, so differentiating gives a pressure change dP=−V0γP0Ax. That produces a restoring force F=−AdP=−V0γP0A2x, the signature of simple harmonic motion with ω2=MV0γP0A2. The frequency is therefore f=2π1MV0A2γP0.
JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option