The displacement of an oscillating particle v | vedclass

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(D) The given equation for displacement is $y = \sin \left( \frac{\pi t}{4} + \frac{\pi}{6} \right)$.Comparing this with the standard $SHM$ equation $

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$0.62$

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(D) The given equation for displacement is $y = \sin \left( \frac{\pi t}{4} + \frac{\pi}{6} ight)$.Comparing this with the standard $SHM$ equation $y = A \sin(\omega t + \phi)$,we get amplitude $A = 1 \ cm$ and angular frequency $\omega = \frac{\pi}{4} \ rad/s$.The acceleration $a$ of a particle in $SHM$ is given by $a = -\omega^2 y$.The maximum acceleration is given by $a_{\max} = \omega^2 A$.Substituting the values: $a_{\max} = \left( \frac{\pi}{4} ight)^2 	imes 1 = \frac{\pi^2}{16}$.Using $\pi^2 \approx 9.869$,we get $a_{\max} = \frac{9.869}{16} \approx 0.6168 \ cm/s^2$.Rounding to two decimal places,we get $a_{\max} \approx 0.62 \ cm/s^2$.

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

The displacement of an oscillating particle varies with time $t$ (in seconds) according to the equation $y (cm) = \sin \left[ \frac{\pi}{2} \left( \frac{t}{2} + \frac{1}{3} \right) \right]$. The maximum acceleration of the particle is approximately ..... $cm/s^2$.

Similar Questions

Values of the acceleration $A$ of a particle moving in simple harmonic motion as a function of its displacement $x$ are given in the table below:
$A \ (mm \ s^{-2})$ $16$ $8$ $0$ $-8$ $-16$
$x \ (mm)$ $-4$ $-2$ $0$ $2$ $4$

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The displacement of a particle varies with time as $x = 12 \sin \omega t - 16 \sin^3 \omega t$ (in $cm$). If its motion is $S.H.M.$,then its maximum acceleration is

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