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 slightly compressed isothermally from its equilibrium position, it oscillates simple harmonically, the period of oscillation will be

• A. $T=2\pi \sqrt{\left(\frac{Mh}{PA}\right)}$
• B. $T=2\pi \left(MA/Ph\right)$
• C. $T=2\pi \left(M/PAh\right)$
• D. $T=2\pi \left(MPhA\right)$

Answer 1

Answer: (A)

$T=2\pi \sqrt{\left(\frac{Mh}{PA}\right)}$

$\text{Given:-}$ Mass of the cylindrical piston = M

$\text{To find:-}$ The period of oscillation

$\text{Solution:-}$

Suppose that the initial pressure of the gas is P and initial volume is V = Ah. The piston is moved isothermally from C to D through a distance x. The gas inside the cylinder will be compressed and it will try to push the piston to its original position. When the piston is at D let the pressure of the gas be P + $\delta P$ and volume = V- $\delta V$ = V - Ax. Since the process is isothermal, we have

$PV=(P+\delta P)(V-\delta V)\approx PV-P\delta V+V\delta P$

or

$\delta P=\frac{P\delta V}{V}=\frac{PAx}{V}$

The return force acting on the piston is

$A\delta P=\frac{A^{2}Px}{V}$

The acceleration a of the piston is proportional to x and directed opposite to the direction of increasing x:

$a=-\frac{A^{2}Px}{MV}$

Thus, the motion of the piston is simple harmonic and its time period is

$T=2\pi \sqrt{\frac{MV}{A^{2}P}}=\frac{2\pi }{A}\sqrt{\frac{MV}{P}}=2\pi \sqrt{\frac{Mh}{PA}}$

$\text{Hence the correct option is A}$