The displacement of a particle executing a S.h.M. at any ins"tan"t t is `x=4 "sin" (2pi t)`. The acceleration produced in the particle is given by

Options

A. `-16 pi^(2) "sin" 2pi t`
B. `-16 pi^(2) "cos" 2pi t`
C. `12 pi "cos" 2pi t`
D. `16 pi "sin" (2pi)^(2)t`

Answer

A

Explanation

`x=4 "sin" (2pi t)`
`therefore (dx)/(dt)=4 "cos" (2pi t)xx2pi =8 "cos" (2pi t)`
`a=(d^(2)x)/(dt^(2))=8pi xx2pi (-"sin" 2pi t)= - 16pi^(2)("sin" 2pi t)`