Question

The time period of a particle executing SHM is $\frac{2\pi }{\omega }$ and its velocity at a distance $b$ from mean position is $\sqrt{3}b\omega$. Its amplitude is

• A. b
• B. 2b
• C. 3b
• D. 4b

Answer 1

The correct option is B 2b

$x=A\sin \omega t$
$v=A\omega \cos \omega t$
Thus, when $x=b,$ we have $\sin \omega t=\frac{b}{A}$
and hence, $\cos \omega t=\sqrt{1-{\left(\frac{b}{A}\right)}^2}$
Thus, $\sqrt{3}b\omega =A\omega \sqrt{1-{\left(\frac{b}{A}\right)}^2}$
or $3b^2=A^2(1-\frac{b^2}{A^2})$
or $3b^2=A^2-b^2$
Thus, $A^2=4b^2$ or $A=2b$