Question

The equation of particle executing simple harmonic motion $x=10\text{sin}(\pi t+\frac{\pi }{6})$ where $x,t$ are in SI units. Find the velocity of the particle at $t=1$ seconds.

• A. $5\sqrt{3}\pi {\text{ms}}^{-1}$
• B. $-5\sqrt{3}\pi {\text{ms}}^{-1}$
• C. $10\sqrt{3}\pi {\text{ms}}^{-1}$
• D. $10\pi {\text{ms}}^{-1}$

Answer 1

The correct option is B $-5\sqrt{3}\pi {\text{ms}}^{-1}$

Given,
$x=10\text{sin}(\pi t+\frac{\pi }{6})$
Comparing it with $x=\text{A sin}(\omega t+\varphi )$
$\omega =\pi ,A=10,\varphi =\frac{\pi }{6}$

Finding $v$, by differentiating $x$
$v\left(t\right)=\frac{dx}{dt}=10\pi \text{cos}(\pi t+\frac{\pi }{6})$
$\left[v\right(t){]}_{t=1}=10\pi \text{cos}(\pi +\frac{\pi }{6})=10\pi \text{cos}\frac{7\pi }{6}=-10\pi \text{cos}\frac{\pi }{6}$
$\Rightarrow \left[v\right(t){]}_{t=1\text{sec}}=-10\pi \frac{\sqrt{3}}{2}=-5\sqrt{3}\pi {\text{ms}}^{-1}$