A piston of mass $$20kg$$ cylindrical in shape has a base area of $$10$$ $$cm2$$. It slides smoothly inside a long cylinder with gas of certain mass on the other side. The cylinder is kept with its axis horizontal. The piston is at equilibrium for length of $$20cm$$ of the gas column. If the piston is disturbed from its equilibrium, it executes simple harmonic motion. Then the time period of oscillation of the cylinder is (Atmospheric$$ pressure $$=$$ $$105Pa$$ $$)

Question

A piston of mass $$20kg$$ cylindrical in shape has a base area of $$10$$ $$cm2$$. It slides  smoothly inside a long cylinder with gas of certain mass on the other side. The  cylinder is kept with its axis horizontal. The piston is at equilibrium for length of  $$20cm$$ of the gas column. If the piston is disturbed from its equilibrium, it executes  simple harmonic motion. Then the time period of oscillation of the cylinder is  (Atmospheric$$ pressure $$=$$ $$105Pa$$  $$)

Infinity Learn · Accepted Answer

The restoring force responsible for simple harmonic motion is the excess pressure.  Considering the process to be isothermal,$$PV=P+$$ΔPV−ΔV . Simplifying this  $$0=$$ΔP.V−P.ΔV        (Neglecting$$  ΔP.Δ$$V) .as the term is very smallHere,$$volume=$$ $$V=Ah$$ and  Δ$$V=Ax$$ Substituting these, we getΔ$$P=P$$.ΔVV  i.e.$$=$$>  Δ$$P=P$$.AxAhBy using the definition of force, $$F=$$ΔP.A  ,ΔP.$$A=F=P$$.Axh .This can be now equated to restoring force   $$F=kxTherefore$$, $$kx$$$$=P$$.Axh ,Or,  $$k=P$$.Ah. For a simple harmonic motion, we know that  $$k=m$$ω$$2$$ Therefore, mω$$2=P$$.Ah  Or,  ω$$2=P$$.Amh, i.e. ω$$=P$$.Amh ,Using the relation  ω$$=2$$$$π$$T , we $$get2$$$$π$$$$T=P$$.Amh Or  $$T=2$$$$π$$mhP.A. Here $$m=20$$ kg, $$h=20$$ $$cm=0.2m$$, $$A=10$$ square $$cm=10-3$$ $$m2$$, $$P=105Pa$$ ,Substituting these values we get,$$T=2$$$$π$$$$20$$×$$0.2105$$×$$10$$−$$3$$ $$T=2$$$$π$$$$4102=2$$×π×$$210=2$$$$π$$$$5s$$

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