A body of mass 5 g is executing SHM about a fixed point O. with an amplitude of 10 cm, its maximum velocity is $100\ cms^{- 1}.$ Its velocity will be ${50\ cms}^{- 1}$ at a distance (in cm)

Question

Dr. Abubacker · Accepted Answer

(c) $v_{\max} = 100 = a\ \omega;$
$\omega = \frac{100}{a} = \frac{100}{10} = 10\ rads^{- 1}.$
$v^{2} = \omega^{2}(a^{2} - y^{2})$
or $50^{2} = 10^{2}\left( 10^{2} - y^{2} \right)$ or $25 = 100 - y^{2}$
or $y = \sqrt{75} = 5\sqrt{3}\ cm$.

Q. A body of mass 5 g is executing SHM about a fixed point O. with an amplitude of 10 cm, its maximum velocity is $100\ cms^{- 1}.$ Its velocity will be ${50\ cms}^{- 1}$ at a distance (in cm)

Explanation

Reviewed by Dr. Abubacker

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