alleryIf displacement of a particle in SHM is $x = A\sin(\omega t)$, velocity at $x = A/2$ is:
A$A\omega$
B$A\omega/2$
C$A\omega\sqrt 3/2$
D$0$
Answer & Solution
Correct answer: C. $A\omega\sqrt 3/2$
$v = \omega\sqrt{A^2 - x^2}$.
At $x = A/2$: $v = \omega\sqrt{A^2 - A^2/4} = \omega A\sqrt{3}/2$.
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