Question

A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of a gas. The cylinder is kept with its axis horizontal. If the piston is disturbed slightly from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be:

A.\[T=2\pi \sqrt{\dfrac{Mh}{PA}}\]
B.\[T=2\pi \sqrt{\dfrac{MA}{Ph}}\]
C.\[T=2\pi \sqrt{\dfrac{M}{PAh}}\]
D.\[T=2\pi \sqrt{MPhA}\]

Answer 1

Hint: To find the period of oscillation, we have to use the isothermal equation. We can equate the acceleration of the piston to the simple harmonic motion equation. So, we can find out the time period.

Formula used:
\[PV=\text{a constant}\], where P is the pressure and V is the volume.
\[T=\dfrac{2\pi }{\omega }\]

Complete step by step answer:

Suppose the piston of mass M slightly pushed inward to the cylinder for a small displacement x. Since the cylinder contains gases, they will oppose the movement of the piston. Therefore, it will push back the piston to its original position. We can assume the temperature of the gases remains constant even after the pushing of the piston. So, we can consider here the isothermal process. According to the isothermal process,
\[PV=\text{a constant}\], where P is the pressure and V is the volume.
We can differentiate this equation.

\[PdV+VdP=0\]

From this, we can find out the change in pressure in the system.
\[dP=-\dfrac{PdV}{V}\], where the negative sign is arising since the pressure change is acting in the opposite direction of the applied force.

This change in pressure can be written as,

\[dP=\dfrac{F}{A}\], where F is the force acting on the piston and A is the cross-sectional area.
\[\dfrac{F}{A}=\dfrac{-PdV}{V}\]
\[\dfrac{F}{A}=\dfrac{-PdV}{V}=\dfrac{-P(A\times x)}{A\times h}\],

where x is the displacement happening to the piston due to the applied force and h is the total distance of the cylinder at which the gas occupied.

\[F=-\left( \dfrac{PA}{h} \right)x\]

As we know, the force is the product of mass and acceleration. We can substitute this as the force.

\[ma=-\left( \dfrac{PA}{h} \right)x\], where m is the mass and a is the acceleration. Here, we are using a piston of mass M.

\[Ma=-\left( \dfrac{PA}{h} \right)x\]
\[a=-\left( \dfrac{PA}{h} \right)\dfrac{x}{M}\]

It is similar to the acceleration of simple harmonic motion. In the question itself, it mentioned. The slight disturbance will make the simple harmonic motion to the system. Therefore we can compare the obtained acceleration with the acceleration of simple harmonic motion.

\[-{{\omega }^{2}}x=-\left( \dfrac{PA}{h} \right)\dfrac{x}{M}\], where \[\omega \] is the angular frequency.

\[{{\omega }^{2}}=\left( \dfrac{PA}{h} \right)\dfrac{1}{M}\]
\[\omega =\sqrt{\left( \dfrac{PA}{Mh} \right)}\]

From the angular frequency, we can find out the time period of oscillations.

\[T=\dfrac{2\pi }{\omega }\]
\[T=\dfrac{2\pi }{\sqrt{\left( \dfrac{PA}{Mh} \right)}}\]
\[T=2\pi \sqrt{\dfrac{Mh}{PA}}\]

Therefore, the correct option is A.

Note: Here we are dealing with the isothermal system. Then only temperature remains constant. The excess heat will go through the surface of the cylinder. If we are dealing with the adiabatic system, the heat transfer will not occur. All given options look similar. So special care is required to find the final answer.