For the coupled pendulums of mass

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Mechanics & Dynamical Systems · Linear algebra and vector spaces

Action and Symmetry, Ch Normal modes as a generalised eigenvalue problem

Problem

m_1=1.0\ \mathrm{kg}

m_2=3.0\ \mathrm{kg}

described above,

\mathbf T=\mathrm{diag}(1.0,3.0)\ \mathrm{kg\,m^2}

, with modes

\mathbf a^{(1)}=(1,1)

\omega_1^2=9.80\ \mathrm{rad^2\,s^{-2}}

and

\mathbf a^{(2)}=(3,-1)

\omega_2^2=26.47\ \mathrm{rad^2\,s^{-2}}

, satisfying

\mathbf a^{(r)\mathrm T}\mathbf T\mathbf a^{(s)}=0

for

r\ne s

. Define normal coordinates by

\boldsymbol\theta=\hat{\mathbf a}^{(1)}\eta_1+\hat{\mathbf a}^{(2)}\eta_2

, where

\hat{\mathbf a}^{(r)}

are the mode vectors normalised so that

\hat{\mathbf a}^{(r)\mathrm T}\mathbf T\hat{\mathbf a}^{(s)}=\delta_{rs}

. ((a)) Starting from

T=\tfrac12\dot{\boldsymbol\theta}^{\mathrm T}\mathbf T\dot{\boldsymbol\theta}

V=\tfrac12\boldsymbol\theta^{\mathrm T}\mathbf V\boldsymbol\theta

, and the mode equation

\mathbf V\mathbf a^{(r)}=\omega_r^2\mathbf T\mathbf a^{(r)}

, derive, for any number of degrees of freedom, that

T=\tfrac12\sum_r\dot\eta_r^2

and

V=\tfrac12\sum_r\omega_r^2\eta_r^2

. ((b)) Find

\hat{\mathbf a}^{(1)}

and

\hat{\mathbf a}^{(2)}

for the pendulum system, and verify by direct substitution that

T=\tfrac12(\dot\eta_1^2+\dot\eta_2^2)