VC1.6

From the given equation, $\frac{dr}{dt}=-a\omega \sin \omega t+b\omega \cos \omega t\,$ [Let this be equal be 1]

Therefore, $r\times\frac{dr}{dt}=(a\cos \omega t + b\sin \omega t)\times(-a\omega \sin \omega t+b\omega \cos \omega t)\,$

$a\times b \omega \cos^2 \omega t-b\times a \omega \sin^2 \omega t \,$ [As $a\times a=b\times b=0\,$ ]

$\omega (\cos ^2 \omega t+\sin^2 \omega t)a\times b=\omega a\times b\,$

Again differentiating the equation 1.

$\frac{d^2 r}{dt^2}=-a \omega ^{2} \cos \omega t-b\omega^{2} \sin \omega t=-\omega^{2}(a\cos \omega t+b\sin \omega t)=-\omega^{2} r \,$

Hence $\frac{d^2 r}{dt^2}=-\omega^{2} r\,$