← 2023 Paper 2
UPSC 2023 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
A mechanical system with 2 degrees of freedom has the Lagrangian
L = 1 2 m ( x ˙ 2 + y ˙ 2 ) − 1 2 m ( ω 1 2 x 2 + ω 2 2 y 2 ) + k x y L=\tfrac12 m(\dot x^2+\dot y^2)-\tfrac12 m(\omega_1^2 x^2+\omega_2^2 y^2)+kxy L = 2 1 m ( x ˙ 2 + y ˙ 2 ) − 2 1 m ( ω 1 2 x 2 + ω 2 2 y 2 ) + k x y where m , ω 1 , ω 2 , k m,\omega_1,\omega_2,k m , ω 1 , ω 2 , k θ \theta θ
x = q 1 cos θ − q 2 sin θ , y = q 1 sin θ + q 2 cos θ x=q_1\cos\theta-q_2\sin\theta,\;y=q_1\sin\theta+q_2\cos\theta x = q 1 cos θ − q 2 sin θ , y = q 1 sin θ + q 2 cos θ the Lagrangian in terms of q 1 , q 2 q_1,q_2 q 1 , q 2 q 1 q 2 q_1q_2 q 1 q 2 q 1 q_1 q 1 q 2 q_2 q 2 θ \theta θ
Technique
Principal-axis transformation — orthogonal rotation diagonalises a symmetric quadratic form; the eigenvalues are the new Ω 2 \Omega^{2} Ω 2
Solution
Step 1 — Kinetic term is invariant
The rotation ( x , y ) ↦ ( q 1 , q 2 ) (x,y)\mapsto(q_1,q_2) ( x , y ) ↦ ( q 1 , q 2 )
x ˙ 2 + y ˙ 2 = q ˙ 1 2 + q ˙ 2 2 . \dot x^{2}+\dot y^{2}=\dot q_1^{2}+\dot q_2^{2}. x ˙ 2 + y ˙ 2 = q ˙ 1 2 + q ˙ 2 2 . Using sin 2 θ = 2 sin θ cos θ \sin 2\theta=2\sin\theta\cos\theta sin 2 θ = 2 sin θ cos θ cos 2 θ = cos 2 θ − sin 2 θ \cos 2\theta=\cos^{2}\theta-\sin^{2}\theta cos 2 θ = cos 2 θ − sin 2 θ
x 2 = q 1 2 cos 2 θ + q 2 2 sin 2 θ − q 1 q 2 sin 2 θ , x^{2}=q_1^{2}\cos^{2}\theta+q_2^{2}\sin^{2}\theta-q_1q_2\sin 2\theta, x 2 = q 1 2 cos 2 θ + q 2 2 sin 2 θ − q 1 q 2 sin 2 θ , y 2 = q 1 2 sin 2 θ + q 2 2 cos 2 θ + q 1 q 2 sin 2 θ , y^{2}=q_1^{2}\sin^{2}\theta+q_2^{2}\cos^{2}\theta+q_1q_2\sin 2\theta, y 2 = q 1 2 sin 2 θ + q 2 2 cos 2 θ + q 1 q 2 sin 2 θ , x y = 1 2 ( q 1 2 − q 2 2 ) sin 2 θ + q 1 q 2 cos 2 θ . xy=\tfrac{1}{2}(q_1^{2}-q_2^{2})\sin 2\theta+q_1q_2\cos 2\theta. x y = 2 1 ( q 1 2 − q 2 2 ) sin 2 θ + q 1 q 2 cos 2 θ . Substitute into the potential-side of L = T − V L=T-V L = T − V
− 1 2 m ( ω 1 2 x 2 + ω 2 2 y 2 ) + k x y . -\tfrac{1}{2}m(\omega_1^{2}x^{2}+\omega_2^{2}y^{2})+kxy. − 2 1 m ( ω 1 2 x 2 + ω 2 2 y 2 ) + k x y . Coefficient of q 1 q 2 q_1q_2 q 1 q 2 L L L
− 1 2 m [ − ω 1 2 sin 2 θ + ω 2 2 sin 2 θ ] + k cos 2 θ = 1 2 m ( ω 1 2 − ω 2 2 ) sin 2 θ + k cos 2 θ . -\tfrac{1}{2}m\bigl[-\omega_1^{2}\sin 2\theta+\omega_2^{2}\sin 2\theta\bigr]+k\cos 2\theta=\tfrac{1}{2}m(\omega_1^{2}-\omega_2^{2})\sin 2\theta+k\cos 2\theta. − 2 1 m [ − ω 1 2 sin 2 θ + ω 2 2 sin 2 θ ] + k cos 2 θ = 2 1 m ( ω 1 2 − ω 2 2 ) sin 2 θ + k cos 2 θ . Step 3 — Cross-term condition
Set the coefficient of q 1 q 2 q_1q_2 q 1 q 2
1 2 m ( ω 1 2 − ω 2 2 ) sin 2 θ + k cos 2 θ = 0 , \tfrac{1}{2}m(\omega_1^{2}-\omega_2^{2})\sin 2\theta+k\cos 2\theta=0, 2 1 m ( ω 1 2 − ω 2 2 ) sin 2 θ + k cos 2 θ = 0 , ⟹ tan 2 θ = − 2 k m ( ω 1 2 − ω 2 2 ) = 2 k m ( ω 2 2 − ω 1 2 ) . \Longrightarrow\;\tan 2\theta=\frac{-2k}{m(\omega_1^{2}-\omega_2^{2})}=\frac{2k}{m(\omega_2^{2}-\omega_1^{2})}. ⟹ tan 2 θ = m ( ω 1 2 − ω 2 2 ) − 2 k = m ( ω 2 2 − ω 1 2 ) 2 k . θ = 1 2 arctan 2 k m ( ω 2 2 − ω 1 2 ) . \boxed{\;\theta=\tfrac{1}{2}\arctan\!\frac{2k}{m(\omega_2^{2}-\omega_1^{2})}.\;} θ = 2 1 arctan m ( ω 2 2 − ω 1 2 ) 2 k . (If ω 1 = ω 2 \omega_1=\omega_2 ω 1 = ω 2 cos 2 θ = 0 \cos 2\theta=0 cos 2 θ = 0 θ = π / 4 \theta=\pi/4 θ = π /4
Step 4 — Diagonalised Lagrangian
After the rotation, the potential becomes
1 2 m Ω 1 2 q 1 2 + 1 2 m Ω 2 2 q 2 2 , \tfrac{1}{2}m\,\Omega_1^{2}q_1^{2}+\tfrac{1}{2}m\,\Omega_2^{2}q_2^{2}, 2 1 m Ω 1 2 q 1 2 + 2 1 m Ω 2 2 q 2 2 , where Ω 1 2 , Ω 2 2 \Omega_1^{2},\Omega_2^{2} Ω 1 2 , Ω 2 2
M = ( ω 1 2 − k / m − k / m ω 2 2 ) M=\begin{pmatrix}\omega_1^{2}&-k/m\\-k/m&\omega_2^{2}\end{pmatrix} M = ( ω 1 2 − k / m − k / m ω 2 2 ) (read off from the quadratic form ω 1 2 x 2 + ω 2 2 y 2 − ( 2 k / m ) x y \omega_1^{2}x^{2}+\omega_2^{2}y^{2}-(2k/m)xy ω 1 2 x 2 + ω 2 2 y 2 − ( 2 k / m ) x y 2 V / m 2V/m 2 V / m
Eigenvalues:
Ω 1 , 2 2 = ω 1 2 + ω 2 2 2 ± ( ω 1 2 − ω 2 2 2 ) 2 + k 2 m 2 . \Omega_{1,2}^{2}=\frac{\omega_1^{2}+\omega_2^{2}}{2}\pm\sqrt{\!\left(\frac{\omega_1^{2}-\omega_2^{2}}{2}\right)^{2}+\frac{k^{2}}{m^{2}}}. Ω 1 , 2 2 = 2 ω 1 2 + ω 2 2 ± ( 2 ω 1 2 − ω 2 2 ) 2 + m 2 k 2 . Both expressions are independent of θ \theta θ m , ω 1 , ω 2 , k m,\omega_1,\omega_2,k m , ω 1 , ω 2 , k
Step 5 — Lagrange’s equations (independent of θ \theta θ
In the decoupled form,
L = 1 2 m ( q ˙ 1 2 + q ˙ 2 2 ) − 1 2 m ( Ω 1 2 q 1 2 + Ω 2 2 q 2 2 ) . L=\tfrac{1}{2}m(\dot q_1^{2}+\dot q_2^{2})-\tfrac{1}{2}m(\Omega_1^{2}q_1^{2}+\Omega_2^{2}q_2^{2}). L = 2 1 m ( q ˙ 1 2 + q ˙ 2 2 ) − 2 1 m ( Ω 1 2 q 1 2 + Ω 2 2 q 2 2 ) . The Euler–Lagrange equations
d d t ( ∂ L ∂ q ˙ i ) − ∂ L ∂ q i = 0 \frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot q_i}\right)-\frac{\partial L}{\partial q_i}=0 d t d ( ∂ q ˙ i ∂ L ) − ∂ q i ∂ L = 0 give the two decoupled SHM equations:
Answer
q ¨ 1 + Ω 1 2 q 1 = 0 , q ¨ 2 + Ω 2 2 q 2 = 0 , \boxed{\;\ddot q_1+\Omega_1^{2}q_1=0,\qquad\ddot q_2+\Omega_2^{2}q_2=0,\;} q ¨ 1 + Ω 1 2 q 1 = 0 , q ¨ 2 + Ω 2 2 q 2 = 0 ,