For a particle executing simple harmonic moti | vedclass

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Question

(A) The displacement-time graph shown in the figure is a sine wave, so the equation of displacement is $x = A \sin(\omega t)$.From the graph, the ampl

Vedclass · Accepted Answer

$-\frac{\sqrt{3}}{32} \pi^2 \,cm \,s^{-2}$

Vedclass · Answer

(A) The displacement-time graph shown in the figure is a sine wave, so the equation of displacement is $x = A \sin(\omega t)$.From the graph, the amplitude $A = 1 \,cm$ and the time period $T = 8 \,s$.Therefore, the angular frequency $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{8} = \frac{\pi}{4} \,rad/s$.The equation for displacement is $x = 1 \sin\left(\frac{\pi}{4} tight)$.The acceleration $a$ in simple harmonic motion is given by $a = -\omega^2 x = -\omega^2 A \sin(\omega t)$.Substituting the values at $t = \frac{4}{3} \,s$:$a = -\left(\frac{\pi}{4}ight)^2 	imes 1 	imes \sin\left(\frac{\pi}{4} 	imes \frac{4}{3}ight)$$a = -\frac{\pi^2}{16} 	imes \sin\left(\frac{\pi}{3}ight)$$a = -\frac{\pi^2}{16} 	imes \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{32} \pi^2 \,cm \,s^{-2}$.Thus, the correct option is $A$.

$A \ (mm \ s^{-2})$	$16$	$8$	$0$	$-8$	$-16$
$x \ (mm)$	$-4$	$-2$	$0$	$2$	$4$

For a particle executing simple harmonic motion, the displacement-time $(x-t)$ graph is as shown in the figure. The acceleration of the particle at $t=\frac{4}{3} \,s$ is

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