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

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(A) From the graph,the amplitude $A = 1 \ m$ and the time period $T = 8 \ s$.The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\

Vedclass · Accepted Answer

$-\frac{\pi^2}{16} \ m/s^2$

Vedclass · Answer

(A) From the graph,the amplitude $A = 1 \ m$ and the time period $T = 8 \ s$.The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$.The equation of motion is $x(t) = A \sin(\omega t) = 1 \sin\left(\frac{\pi}{4} tight)$.The acceleration $a(t)$ is given by $a = \frac{d^2x}{dt^2} = -\omega^2 x = -\omega^2 A \sin(\omega t)$.Substituting the values at $t = 2 \ s$:$a = -\left(\frac{\pi}{4}ight)^2 	imes 1 	imes \sin\left(\frac{\pi}{4} 	imes 2ight)$$a = -\frac{\pi^2}{16} 	imes \sin\left(\frac{\pi}{2}ight)$Since $\sin\left(\frac{\pi}{2}ight) = 1$,we get:$a = -\frac{\pi^2}{16} \ m/s^2$.

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:

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