Question

The x-t graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at t=2s is:

• A. $-\frac{\pi^2}{8}$ ms-2
• B. $\frac{\pi^2}{16}$ ms-2
• C. $\frac{-\pi^2}{16}$ ms-2
• D. $\frac{\pi^2}{8}$ms-2

Answer 1

Answer: (C)

$\frac{-\pi^2}{16}$ ms-2

Answer 2

The correct option is (C): $\frac{-\pi^2}{16}$ ms-2
$x=A\,sin(\omega t)$
$\frac{dx}{dt}=v=A\omega cos(\omega t)$
$\frac{dv}{dt}=a=-\omega^2 Asin(\omega t)$
$a=-(\frac{2\pi}{8})^2\times 1 sin(\frac{2\pi}{8}\times 2)$
$\Rightarrow a=\frac{\pi^2}{16}\times sin (\frac{\pi}{2})$
$\therefore a=-\frac{\pi^2}{16}$ m/s2