Question

The velocity of a particle in S.H.M. at displacement y from mean position is ( $a=$ amplitude $\omega =$ angular frequency)

• A. $\omega \sqrt{a^{2}+y^{2}}$
• B. $\omega \sqrt{a^{2} - y^{2}}$
• C. $\omega y$
• D. $\omega^2 (a^2 -y^2)$

Answer 1

Answer: (B)

$\omega \sqrt{a^{2} - y^{2}}$

Answer 2

In S.H.M., velocity of particle at displacement $y$ from mean position is $v=\omega \sqrt{a^{2}-y^{2}}$