1.

If y = 2 (cm) sin[(pi t)/(2)+phi] then the maximum acceleration of the particle doing the S.H.M. is

Answer»

`(PI)/(2)"cm"//"s"^(2)`
`(pi^(2))/(2)" cm"//"s"^(2)`
`(pi^(2))/(4)" cm"//"s"^(2)`
`(pi)/(4)" cm"//"s"^(2)`

Solution :`y=2sin((pi t)/(2)+PHI)""v=(DY)/(dt)=2xx(pi)/(2)COS((pi t)/(2)+phi)`
`a=(d^(2)y)/(dt)=-(pi^(2))/(2)sin((pi t)/(2)+phi)impliesa_("MAX")=(pi^(2))/(2)`


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