GATE PH 2022 • Q41 MCQ • 2 Marks

Section 2: Classical Mechanics

Lagrangian & Hamiltonian Dynamics

Subtopic: Euler-Lagrange Equations of Motion

Question Statement:

A particle of mass 1 kg is released from a height of 1 m above the ground. When it reaches the ground, what is the value of Hamilton's action for this motion in J s? (g is the acceleration due to gravity; take gravitation potential to be zero on the ground)

Options:

  • (A) - 2 3 ඥ2g
  • (B) 5 3 ඥ2g
  • (C) 3ඥ2g
  • (D) - 1 3 ඥ2g

Official Answer Key:

D

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.41)** **1. Classical Mechanics Principles (Hamilton's Principle & Action)**: - Mass $m = 1$ kg released from rest from height $h = 1$ m. - Hamilton's principal function (action): $S = \int_0^{t_f} L\, dt = \int_0^{t_f} (T - V)\, dt$ - Trajectory under gravity (taking $y=0$ at ground, upward positive): $y(t) = h - \frac{1}{2}gt^2, \quad v(t) = \dot{y}(t) = -gt$ $t_f = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2}{g}}$ **2. Detailed Step-by-Step Integration**: - Kinetic and potential energies: $T = \frac{1}{2}m v^2 = \frac{1}{2}(1)(-gt)^2 = \frac{1}{2}g^2 t^2$ $V = mgy = 1 \cdot g\left(h - \frac{1}{2}gt^2\right) = gh - \frac{1}{2}g^2 t^2$ - Lagrangian $L = T - V$: $L(t) = \frac{1}{2}g^2 t^2 - \left(gh - \frac{1}{2}g^2 t^2\right) = g^2 t^2 - gh$ - Evaluate the action integral: $S = \int_0^{t_f} (g^2 t^2 - gh)\, dt = \left[ \frac{1}{3}g^2 t^3 - gh t \right]_0^{t_f} = \frac{1}{3}g^2 t_f^3 - gh t_f = t_f \left(\frac{1}{3}g^2 t_f^2 - gh\right)$ - Substitute $t_f^2 = \frac{2h}{g}$ and $h = 1$: $\frac{1}{3}g^2 \left(\frac{2h}{g}\right) - gh = \frac{2}{3}gh - gh = -\frac{1}{3}gh$ $S = \sqrt{\frac{2h}{g}} \left(-\frac{1}{3}gh\right) = -\frac{1}{3}h \sqrt{2gh} = -\frac{1}{3}(1)\sqrt{2g(1)} = -\frac{1}{3}\sqrt{2g}$ **3. Official Verification**: The verified official IIT Master Answer is **Option (D)** (`-1/3 \sqrt{2g}`).

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