The displacement of a particle performing S.H.M. is x = 0.1 sin `(5pit)+0.4cos(5pit)m`.
Then acceleration of the particle at 0.1 s is

Options

A. `10pi m//s`
B. `-2pi^(2)m//s^(2)`
C. `-2.5 pi^(2)m//s^(2)`
D. `4pi^(2)m//s^(2)`

Answer

C

Explanation

`x=0.1sin5pit+0.4cos5pit" at t"=0.1s`
`x=0.1sin5pixx0.1+0.4cos5pixx0.1`
`0.1sin.(pi)/(2)+0.4cos.(pi)/(2)`
`=0.1xx1+0.4xx0=0.1+0=0.1`
`a=-omega^(2)x`
`=-(5pi)^(2)xx0.1=-25pi^(2)xx0.1=-25pi^(2)`