© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Equilibrium Condition
When the piston is in equilibrium, the force due to the gas pressure below the piston balances the weight of the piston. If the cross-sectional area of the cylinder (and piston) is $A$, the pressure of the gas in equilibrium is $P_0$, and the mass of the piston is $M$, then:
$P_0 \cdot A = M g \quad \Longrightarrow \quad P_0 = \frac{M g}{A}.
Step 2: Use the Adiabatic Relation
Because the process is assumed to be adiabatic (the system is isolated), the gas satisfies
$P\,V^\gamma = \text{constant} \quad \Longrightarrow \quad P_0\,V_0^\gamma = P\,V^\gamma,
where $\gamma$ is the adiabatic index (ratio of specific heats).
Step 3: Relate Piston Position to Volume
Let the equilibrium height of the gas column be $x_0$, so that $V_0 = A\,x_0$. If the piston is displaced a small distance $x$ downwards, the new volume becomes $V = A\,(x_0 - x)$ within a small approximation.
Step 4: Express the Pressure After Displacement
Using the adiabatic relation after a small displacement $x$,
$P \, (A\, (x_0 - x))^\gamma \;=\; P_0 \,(A\,x_0)^\gamma \;\Longrightarrow\; P = P_0 \,\frac{x_0^\gamma}{(x_0 - x)^\gamma}.
Step 5: Write the Restoring Force
Net force on the piston (taking downward as positive) can be written as:
$F_{\text{net}} \;=\; \big(Mg\big) - \big(P \cdot A\big).
Substituting the expression for $P$,
$F_{\text{net}} \;=\; Mg - \left(P_0 \,\frac{x_0^\gamma}{(x_0 - x)^\gamma}\right)A.
When $x$ is small relative to $x_0,$ one often approximates $(x_0 - x)^\gamma \approx x_0^\gamma \left(1 - \frac{\gamma x}{x_0}\right)^{-1}$. To first-order, the change in pressure leads to a restoring force proportional to $-\,x$:
$F_{\text{restoring}} \approx -\,\frac{\gamma\,P_0\,A}{x_0}\;x.
The negative sign indicates that the force always acts in the direction opposite to the displacement, a hallmark of simple harmonic motion.
Step 6: Identify the Effective Spring Constant
By comparing $F_{\text{restoring}} = -\,k_{\text{eff}}\,x$ with $F_{\text{restoring}} \approx -\,\frac{\gamma\,P_0\,A}{x_0}\;x,$ we see that the effective spring constant is
$k_{\text{eff}} = \frac{\gamma\,P_0\,A}{x_0} .
Step 7: Write the Formula for the Frequency
The frequency $f$ of a mass $M$ undergoing simple harmonic motion with an effective spring constant $k_{\text{eff}}$ is given by
$f = \frac{1}{2\pi} \sqrt{\frac{k_{\text{eff}}}{M}}.$
Substituting $k_{\text{eff}} = \frac{\gamma\,P_0\,A}{x_0}$,
$f = \frac{1}{2\pi} \sqrt{ \frac{\gamma\,P_0\,A}{x_0\,M}}.
Since $x_0 = \frac{V_0}{A}$, we get
$f = \frac{1}{2\pi} \sqrt{ \frac{\gamma\,P_0\,A}{\left(\frac{V_0}{A}\right)\,M}} = \frac{1}{2\pi} \sqrt{ \frac{\gamma\,P_0\,A^2}{V_0\,M}} = \frac{1}{2\pi}\,\sqrt{\frac{A\,\gamma\,P_0}{V_0\,M}}.
Step 8: Final Answer
Thus, the piston executes simple harmonic motion with frequency
$\displaystyle f \;=\; \frac{1}{2\pi} \,\sqrt{\frac{A\,\gamma\,P_0}{V_0\,M}}.
