Physics - SHM Question with Solution | TestHub

In equilibrium, pressure inside the cylinder = pressure just outside it

or$\text{p}={\text{p}}_0+\frac{\text{Mg}}{\text{A}}$

When piston is displaced slightly by an amount x, change in volume,

$\text{dV}=-\text{Ax}$

Since, the cylinder is isolated from the surroundings, nature of the process is adiabatic. In adiabatic process,

$\frac{\text{dp}}{\text{dV}}=-\gamma \frac{\text{p}}{\text{V}}$




or increase in pressure inside the cylinder,

$\text{dp}=-\frac{\left(\gamma \text{p}\right)}{\text{V}}\left(\text{dV}\right)=\gamma \left(\frac{{\text{p}}_0+\frac{\text{Mg}}{\text{A}}}{{\text{V}}_0}\right)\left(\text{Ax}\right)$​

This increase in pressure when multiplied with area of cross-section A will give a net upward force (or the restoring force). Hence,

$\text{F}=-\left(\text{dp}\right)\text{A}=-\gamma \left(\frac{{\text{p}}_0{\text{A}}^2+\text{MgA}}{{\text{V}}_0}\right)\text{x}$

Or$\text{a}=\frac{\text{F}}{\text{M}}=-\gamma \left(\frac{{\text{p}}_0{\text{A}}^2+\text{MgA}}{{\text{V}}_0\text{M}}\right)\text{x}$

Since,$\text{a}\propto -\text{x}$, motion of the piston is simple harmonic in nature. Frequency of this oscillation is given by

$\text{f}=\frac{1}{2\pi }\sqrt{\left|\frac{\text{a}}{\text{x}}\right|}=\frac{1}{2\pi }\sqrt{\frac{\mathrm{\gamma }\left({\text{p}}_0{\text{A}}^2+\text{MgA}\right)}{{\text{V}}_0\text{M}}}$