JEE MAIN - Physics (2013 (Offline) - No. 20)
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $$M.$$ The piston and the cylinder have equal cross sectional area $$A$$. When the piston is in equilibrium, the volume of the gas is $${V_0}$$ and its pressure is $${P_0}.$$ 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 frquency
$${1 \over {2\pi }}\,{{A\gamma {P_0}} \over {{V_0}M}}$$
$${1 \over {2\pi }}\,{{{V_0}M{P_0}} \over {{A^2}\gamma }}$$
$${1 \over {2\pi }}\,\sqrt {{{A\gamma {P_0}} \over {{V_0}M}}} $$
$${1 \over {2\pi }}\,\sqrt {{{M{V_0}} \over {A\gamma {P_0}}}} $$
Explanation
$${{Mg} \over A} = {P_0}$$
$$Mg = {P_0}A\,\,\,\,...\left( 1 \right)$$
$${P_0}V_0^\gamma = P{V^\gamma }$$
$$P = {{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^y}}}$$
Let piston is displaced by distance $$x$$
$$Mg - \left( {{{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right)A = {F_{restoring}}$$
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$${P_0}A\left( {1 - {{x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right) = {F_{restoring}}$$
$$\left[ {{x_0} - x \approx {x_0}} \right]$$
$$F = - {{\gamma {P_0}Ax} \over {{x_0}}}$$
$$\therefore$$ Frequency with which piston executes $$SHM.$$
$$f = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}A} \over {{x_0}M}}} = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}{A^2}} \over {M{V_0}}}} $$
$$Mg = {P_0}A\,\,\,\,...\left( 1 \right)$$
$${P_0}V_0^\gamma = P{V^\gamma }$$
$$P = {{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^y}}}$$
Let piston is displaced by distance $$x$$
$$Mg - \left( {{{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right)A = {F_{restoring}}$$
$${P_0}A\left( {1 - {{x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right) = {F_{restoring}}$$
$$\left[ {{x_0} - x \approx {x_0}} \right]$$
$$F = - {{\gamma {P_0}Ax} \over {{x_0}}}$$
$$\therefore$$ Frequency with which piston executes $$SHM.$$
$$f = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}A} \over {{x_0}M}}} = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}{A^2}} \over {M{V_0}}}} $$
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