==> \({v_{\max }} = a\omega \)
==> \(\omega = \frac{{100}}{{10}} = 10\,rad/sec\)
Hence \(v = \omega \sqrt {{a^2} - {y^2}} \)
==>\(50 = 10\sqrt {{{(10)}^2} - {y^2}} \)
==> \(y = 5\sqrt 3 \,cm\)
$(A)\;y= sin\omega t-cos\omega t$
$(B)\;y=sin^3\omega t$
$(C)\;y=5cos\left( {\frac{{3\pi }}{4} - 3\omega t} \right)$
$(D)\;y=1+\omega t+{\omega ^2}{t^2}$