VC1.6

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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\,

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