NTA Abhyas JEE Main2020PhysicsOscillationsPractice
A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of gas. The cylinder is kept with its axis horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be (Assume isothermal process)
Options
- AT = 2 π M h P A
- BT = 2 π M A P h
- CT = 2 π M P A h
- DT = 2 π M P h A
Correct answer
A. T = 2 π M h P A
Step-by-step solution
Let the piston be displaced through distance x towards left, then volume decreases, pressure increases. If Δ P is an increase in pressure and Δ V is a decrease in volume, then considering the process to take place slowly (i.e. Isothermal) P 1 V 1 = P 2 V 2 ⇒ P V = ( P + Δ P ) ( V - ΔV ) ⇒ P V = P V + Δ P V - P Δ V - Δ P Δ V ⇒ Δ P . V - P . Δ V = 0 ( n e g l e c t i n g Δ P . Δ V ) Δ P ( A h ) = P ( A x ) ⇒ Δ P = P ⋅ x h F = - P A h x ω = P A M h T = 2 π M h P A