IGNOU MPH 6 2026 - The Lagrangian for a harmonic oscillator if given as: T

Question

The Lagrangian for a harmonic oscillator if given as:

$$L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + \frac{1}{2}m\omega^2(x^2 + y^2)$
The transformed coordinate under rotation about the Z-axis is given as:

$$x' = x \sin \theta + y \cos \theta$
Determine the corresponding y' so that the transformation leaves the Lagrangian invariant.

19 Jan 2026

Answer :

Word Count : 476

We are asked to find (y') such that the Lagrangian [ L = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) + \frac{1}{2} m \omega^2 (x^2 + y^2) ] remains invariant under a rotation about the (z)-axis, with (x') given as [ x' = x \sin \theta + y \cos \theta. ] --- For a rotation about the (z)-axis, the standard 2D rotation transformation is: [ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}. ] Notice carefully that in the question, (x') is given as [ x' = x \sin \theta + y \cos \theta. ] We want the Lagrangian to be invariant under this rotation, which requires [ x^2 + y^2 = x'^2 + y'^2 ] and similarly for velocities. Let us assume the transformation is linear: [ y' = a x + b y ] for some constants (a) and (b) (functions of (\theta)). --- Step 1: Impose invariance of squared distance [ x'^2 + y'^2 = (x \sin \theta + y \cos \theta)^2 + (a x + b y)^2 = x^2 + y^2 ]

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