Question

A particle executing simple harmonic motion, obeys the equation for its position as $x=6\sin [4t-\frac{\pi }{6}]cm$.
The velocity of the particle when its position is $3cm$ is

• A. $2\sqrt{3}cm{s}^{-1}$
• B. $3\sqrt{3}cm{s}^{-1}$
• C. $4\sqrt{3}cm{s}^{-1}$
• D. $12\sqrt{3}cm{s}^{-1}$

Answer 1

The correct option is D $12\sqrt{3}cm{s}^{-1}$

Given $x=6\sin [4t-\frac{\pi }{6}]cm$
comparing with standard S.H.M equation,
$x=A\sin [\omega t+\varphi ]$
We get, $A=6cm$ and $\omega$ = $4rad{s}^{-1}$

At $x=3cm$
Speed, $V=\omega \sqrt{A^2-x^2}=4\sqrt{6^2-3^2}=4\sqrt{27}\Rightarrow V=12\sqrt{3}cm{s}^{-1}$