The acceleration-displacement graph of a part | vedclass

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(A) For a particle executing $SHM$,the acceleration $a$ is given by $a = -\omega^{2}x$.The slope of the acceleration-displacement graph is $m = \frac{

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$\frac{4\pi}{\sqrt{3}} \, s$

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(A) For a particle executing $SHM$,the acceleration $a$ is given by $a = -\omega^{2}x$.The slope of the acceleration-displacement graph is $m = \frac{a}{x} = -\omega^{2}$.From the given graph,the angle made by the line with the negative $x$-axis is $37^{\circ}$. The slope of the line is $	an(180^{\circ} - 37^{\circ}) = -	an 37^{\circ} = -\frac{3}{4}$.Equating the slopes: $-\omega^{2} = -\frac{3}{4} \Rightarrow \omega^{2} = \frac{3}{4}$.Since $\omega = \frac{2\pi}{T}$,we have $\omega^{2} = \frac{4\pi^{2}}{T^{2}} = \frac{3}{4}$.Solving for $T$: $T^{2} = \frac{16\pi^{2}}{3} \Rightarrow T = \frac{4\pi}{\sqrt{3}} \, s$.

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

The acceleration-displacement graph of a particle executing $SHM$ is shown in the figure. The time period of the simple harmonic motion is:

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