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(B) Comparing the given equation $y = 2\sin \left( {\frac{{\pi t}}{2} + \phi } \right)$ with the standard $SHM$ equation $y = A\sin (\omega t + \phi )

Vedclass · Accepted Answer

$\frac{{{\pi ^2}}}{2} 	ext{ cm/s}^2$

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(B) Comparing the given equation $y = 2\sin \left( {\frac{{\pi t}}{2} + \phi } ight)$ with the standard $SHM$ equation $y = A\sin (\omega t + \phi )$:We get the amplitude $A = 2 	ext{ cm}$ and angular frequency $\omega = \frac{\pi }{2} 	ext{ rad/s}$.The maximum acceleration $a_{max}$ is given by the formula $a_{max} = \omega^2 A$.Substituting the values: $a_{max} = \left( \frac{\pi }{2} ight)^2 	imes 2 = \frac{\pi^2}{4} 	imes 2 = \frac{\pi^2}{2} 	ext{ cm/s}^2$.

What is the maximum acceleration of a particle performing $SHM$ given by the equation $y = 2\sin \left( {\frac{{\pi t}}{2} + \phi } \right)$,where $y$ is in $cm$?

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