A piston of mass m is fitted to an airtight horizontal cylindrical jar. The cylinder
and piston have identical unit area of cross-section. The gas inside the jar has
volume V and is held at pressure P = P10. The piston is pushed inside the
jar very slowly over a small distance. On releasing, the piston performs an
undamped simple harmonic motion of low frequency. Assuming that the gas is
ideal and no heat is exchanged with the atmosphere, the frequency of the small
oscillations is proportional to
Options:
(A) $\sqrt{\frac{P}{\gamma m V}}$
(B) $\sqrt{\frac{P\gamma}{V m}}$
(C) $\sqrt{\frac{P}{m V^{\gamma-1}}}$
(D) $\sqrt{\frac{\gamma P}{m V^{\gamma-1}}}$
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.46)**
**1. Thermodynamics Principles & Small Oscillations**:
- Horizontal airtight cylinder of unit cross-sectional area ($A = 1$), initial volume $V$, initial pressure $P$.
- Piston of mass $m$ displaced by small distance $x$ from equilibrium.
- The gas undergoes rapid oscillations, modeled as an adiabatic process:
$P V^\gamma = \text{constant}$
**2. Detailed Step-by-Step Derivation**:
- Differentiate adiabatic relation:
$dP = -\gamma \frac{P}{V} dV$
- For displacement $x$, change in volume is $dV = A x = 1 \cdot x = x$:
$dP = -\gamma \frac{P}{V} x$
- Net restoring force acting on the piston:
$F = (dP) A = -\left(\frac{\gamma P}{V}\right) x$
- Equation of motion for mass $m$:
$m \ddot{x} + \left(\frac{\gamma P}{V}\right) x = 0 \implies \ddot{x} + \left(\frac{\gamma P}{m V}\right) x = 0$
- Angular frequency of small oscillations $\omega$:
$\omega = \sqrt{\frac{\gamma P}{m V}}$
**3. Official Verification**:
The verified official IIT Master Answer is **Option (B)** (`\sqrt{\gamma P / (V m)}`).
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