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Let there be a cylindrical tube, closed at one-end, with a well-fitting but freely moving piston of mass $m$. [. . .] The piston has certain equilibrium position. If the piston is moved a distance $y$ , lengthening the air-column, the internal pressure drops& the result is to provide a restoring force on $m$. We can, in fact, write an equation of the form: $$F = A\Delta p$$ where $\Delta p$ is the change of pressure.

Why does the air-column impart restoring force on the piston? I mean to say when a system implies restoring force, it must have an associated change in potential energy. So, was there any increase in potential energy of the air-column when it was expanded? If so, why? What is the cause of increasing potential energy of the air-column?

1 Answers1

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The question says:

If the piston is moved a distance $y$, lengthening the air-column, the internal pressure drops

In other words, someone or something had to grab the piston and pull it upwards against the force caused by the difference in pressure between the air inside the coulmn and the air outside. This meant some work was done, and that work went into increasing the potential energy of the air/column/piston system.

John Rennie
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  • Hi sir! Ok , then the external agent does work to bring up the piston. But can you please tell how this work increases the potential energy of the air-column? Isn't the work stored as potential energy in the piston-agent system? –  May 27 '15 at 05:58
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    @user36790: if you pull on the end of a spring and stretch it, the work you do goes into increasing the potential energy of the spring. This is just the same. – John Rennie May 27 '15 at 06:00
  • Sorry sir, if I am bothering you but may I request you to please help me in this question? I got an answer that explained well mathematically but I had asked for the physical reason. I'll be grateful if you help. –  May 27 '15 at 06:33