A vertical cylinder of radius r contains a quantity of the idealgas and is fitte

Question

A vertical cylinder of radius r contains a quantity of the idealgas and is fitted with a piston with mass m that is free to move.The piston and the walls of the cylinder is placed in aconstant-temperature bath. The outside air pressure is P. Inequilibrium, the piston sits at a height h above the bottom of thecylinder.
a) Find the absolute pressure of the gas trapped below the pistonwhen in equilibrium.
b) The piston is pulled up by a small distance and released. Findthe net force acting on the piston when its base is a distance h +y above the bottom of the cylinder, where y is much less thanh.
c) After the piston is displaced from equilibrium and released, itoscillates up and down. Find the frequency of these smalloscillations. If the displacement is not small, are theoscillations simple harmonic? How can you tell?

Explanation / Answer

This problem is tricky. I gave this one to my students lastsemester, but I also gave them an outline of how to do theproblem. . (a) The key here is to just do a free body diagram onthe piston. .    forces pushing up?    pressure ofgas * area of piston . forces pulling/pushing down?    pressure of outside air * area + mg . forces are balanced, so: .      p A = PA + mg          or .     p = P + mg/A . p = P + mg / r2           is the pressure of the gas . (b) Now... use ideal gas law. Since temperature isconstant, only pressure and volume change when the piston ismoved. .     Find the new pressure: .     initial   pV = final pV .     pi ( r2h) = pf ( r2 (h +y ) ) .     pf = pi [ h / (h + y ) ] . . Now... for the net force on the piston, you can use the sameidea from part (a). .    Net force =   up forces - down forces =    pressure of gas *A   - (pressure of air * area + mg) .    Net force = pi [ h /(h + y ) ] * A   -    P A - mg . Net force = (P + mg/A) [ h / (h + y)] * A - P A - mg . Net force = (PA + mg) [ h / (h+y)    -   1 ] . Net force = ( P r2 + mg ) [ ( h - (h+y) ) / (h+y)] . Net force = ( P r2 + mg) [ - y / (h+y) ] . Net force =   - y ( P r2 + mg) / (h+y) . Note that the minus sign simply means that when the piston ispulled up, the net force is down. . (c)   Now... if the displacement is "small", then wecan make an approximation:   h + y ˜ h    (because y is small compared to h). . So now...      net force = - y ( P r2 + mg) / h    . I will now make a simple algebraic step and replace all theconstant info with a letter:   = ( P r2 + mg) / h     ok? . So...           F = - y .        ma = -y .      a + (/m) y = 0 .      y"   + (/m)y = 0 .    where   y"   means the secondderivative of the position (i.e., the acceleration). This is now asimple differential equation that you should recognize: it isthe form of the harmonic oscillator. . The simple solution is given by:     y= A cost       where    2 = /m . The frequency then is just... .    f = /2 = [ / m ]1/2 / 2 .    f = [ ( P r2 + mg) / mh ]1/2 / 2  =    [ P r2 / 4 mh    +   g / 42 h]1/2 . For the lastquestion... if the displacement is not small, then we cannot usethe approximation at the beginning of part c, so we will not getthe same result... and so the oscillations would NOT be simpleharmonic. . (You only get simple harmonicoscillations when the force is proportional to the negative of thedisplacement, i.e. F = - y) . For the lastquestion... if the displacement is not small, then we cannot usethe approximation at the beginning of part c, so we will not getthe same result... and so the oscillations would NOT be simpleharmonic. . (You only get simple harmonicoscillations when the force is proportional to the negative of thedisplacement, i.e. F = - y)

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