The x-t graph of a particle performing simple harmonic motion is … - Vidyadip

MCQ

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

✓
$-\frac{\pi^2}{16}\,ms ^{-2}$
B
$\frac{\pi^2}{8}\,ms ^{-2}$
C
$-\frac{\pi^2}{8}\,ms ^{-2}$
D
$\frac{\pi^2}{16}\,ms ^{-2}$

✓

Answer

Correct option: A.

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

a
$x = A \sin (\omega t )$

$\frac{ dx }{ dt }= v = A \omega \cos (\omega t )$

$\frac{ dv }{ dt }= a =-\omega^2 A \sin (\omega t )$

$a =-\left(\frac{2 \pi}{8}\right)^2 \times 1 \sin \left(\frac{2 \pi}{8} \times 2\right)$

$\Rightarrow a =-\frac{\pi^2}{16} \times \sin \left(\frac{\pi}{2}\right)$

$\therefore a =\frac{-\pi^2}{16}\,m / s ^2$

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