For a particle executing simple harmonic moti | vedclass

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

Vedclass · Accepted Answer

$0.62$

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(D) The given equation is $y = \sin \left( \frac{\pi t}{4} + \frac{\pi}{6} ight)$.Comparing this with the standard $SHM$ equation $y = A \sin(\omega t + \phi)$:The amplitude $A = 1 \ cm$ and the angular frequency $\omega = \frac{\pi}{4} \ rad/sec$.The maximum acceleration $a_{max}$ is given by the formula $a_{max} = \omega^2 A$.Substituting the values: $a_{max} = \left( \frac{\pi}{4} ight)^2 	imes 1$.$a_{max} = \frac{\pi^2}{16} \approx \frac{9.8696}{16} \approx 0.6168 \ cm/sec^2$.Rounding to two decimal places,we get $0.62 \ cm/sec^2$.

For a particle executing simple harmonic motion given by $y(cm) = \sin \frac{\pi}{2} \left( \frac{t}{2} + \frac{1}{3} \right)$,what is the maximum acceleration in $cm/sec^2$?

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