← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q6c — Step-by-Step Solution

20 marks · Section B

Lagrange's equations · Mechanics & Fluid Dynamics · asked 9× in 13 yrs · Read the full method →

Question

Suppose the Lagrangian of a mechanical system is given by

L=12m(ax˙2+2bx˙y˙+cy˙2)−12k(ax2+2bxy+cy2),L=\tfrac12 m(a\dot x^2+2b\dot x\dot y+c\dot y^2)-\tfrac12 k(ax^2+2bxy+cy^2),

where a,b,c,m(>0),k(>0)a,b,c,m(>0),k(>0) are constants and b2≠acb^2\ne ac. Write down the Lagrangian equations of motion and identify the system.

Technique

Euler–Lagrange equations; recognize TT and VV share the same matrix MM; invertibility from b2≠acb^2\ne ac decouples mMr¨=−kMrmM\ddot{\mathbf r}=-kM\mathbf r into mr¨=−krm\ddot{\mathbf r}=-k\mathbf r.

Solution

Setup. The configuration variables are x,yx,y. The Euler–Lagrange equations are

ddt ⁣(∂L∂x˙)−∂L∂x=0,ddt ⁣(∂L∂y˙)−∂L∂y=0.\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot x}\right)-\frac{\partial L}{\partial x}=0,\qquad \frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot y}\right)-\frac{\partial L}{\partial y}=0.

Both the kinetic and potential parts share the same symmetric matrix

M=(abbc),T=12m r˙ ⁣⊤Mr˙,V=12k r ⁣⊤Mr,r=(xy).M=\begin{pmatrix}a&b\\ b&c\end{pmatrix},\qquad T=\tfrac12 m\,\dot{\mathbf r}^{\!\top}M\dot{\mathbf r},\quad V=\tfrac12 k\,\mathbf r^{\!\top}M\mathbf r,\quad \mathbf r=\binom{x}{y}.

Step 1 — Generalized momenta and forces

∂L∂x˙=m(ax˙+by˙),∂L∂y˙=m(bx˙+cy˙).\frac{\partial L}{\partial\dot x}=m(a\dot x+b\dot y),\qquad \frac{\partial L}{\partial\dot y}=m(b\dot x+c\dot y). ∂L∂x=−k(ax+by),∂L∂y=−k(bx+cy).\frac{\partial L}{\partial x}=-k(ax+by),\qquad \frac{\partial L}{\partial y}=-k(bx+cy).

Step 2 — Equations of motion

  m(ax¨+by¨)=−k(ax+by),m(bx¨+cy¨)=−k(bx+cy).  \boxed{\;m(a\ddot x+b\ddot y)=-k(ax+by),\qquad m(b\ddot x+c\ddot y)=-k(bx+cy).\;}

In matrix form,

mMr¨=−kMr.mM\ddot{\mathbf r}=-kM\mathbf r.

Step 3 — Decouple using b2≠acb^2\ne ac

Because det⁡M=ac−b2≠0\det M=ac-b^2\ne0, the matrix MM is invertible. Multiply mMr¨=−kMrmM\ddot{\mathbf r}=-kM\mathbf r on the left by M−1M^{-1}:

mr¨=−kr ⟹   x¨+km x=0,y¨+km y=0.  m\ddot{\mathbf r}=-k\mathbf r\ \Longrightarrow\ \boxed{\;\ddot x+\frac{k}{m}\,x=0,\qquad \ddot y+\frac{k}{m}\,y=0.\;}

The two coordinates decouple completely.

Step 4 — Identify the system

Each equation is simple harmonic motion with the same angular frequency

ω=km,Tperiod=2πmk.\omega=\sqrt{\frac{k}{m}},\qquad T_{\text{period}}=2\pi\sqrt{\frac{m}{k}}.

General solution:

x(t)=A1cos⁡ωt+B1sin⁡ωt,y(t)=A2cos⁡ωt+B2sin⁡ωt.x(t)=A_1\cos\omega t+B_1\sin\omega t,\qquad y(t)=A_2\cos\omega t+B_2\sin\omega t.

Answer

  The system is two independent (uncoupled) simple harmonic oscillators of equal frequency ω=k/m.  \boxed{\;\text{The system is two independent (uncoupled) simple harmonic oscillators of equal frequency }\omega=\sqrt{k/m}.\;}
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