The x-t graph of a particle undergoing simple | vedclass

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Question

(D) The correct option is $D$. From the $x-t$ graph, at $t = 0$, the particle is at $x = 0$. Hence, the particle starts its simple harmonic motion $(S

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The correct option is $D$. From the $x-t$ graph, at $t = 0$, the particle is at $x = 0$. Hence, the particle starts its simple harmonic motion $(SHM)$ from the mean position. The general equation for $SHM$ is $x = A \sin(\omega t)$. From the graph, the time period $T = 8 \,s$ and the amplitude $A = 1 \,cm$. The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \,rad/s$. Thus, the equation of motion is $x = 1 \sin\left(\frac{\pi t}{4}ight)$. At $t = 4/3 \,s$, the displacement is $x = \sin\left(\frac{\pi}{4} 	imes \frac{4}{3}ight) = \sin\left(\frac{\pi}{3}ight) = \frac{\sqrt{3}}{2} \,cm$. The acceleration $a$ is given by $a = -\omega^2 x$. Substituting the values, $a = -\left(\frac{\pi}{4}ight)^2 	imes \frac{\sqrt{3}}{2} = -\frac{\pi^2}{16} 	imes \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}\pi^2}{32} \,cm/s^2$.

The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = 4/3 \,s$ is

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