MCQ
A body of mass $5\, gm$ is executing $S.H.M.$ about a point with amplitude $10 \,cm$. Its maximum velocity is $100\, cm/sec$. Its velocity will be $50\, cm/sec$ at a distance
- A$5$
- B$5\sqrt 2 $
- ✓$5\sqrt 3 $
- D$10\sqrt 2 $
==> ${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}} $
==> $y = 5\sqrt 3 \,cm$
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