A particle is subjected to two simple harmonic motions `x_(1)=4 sin(100pit) and x_(2)==3 sin ^((100pi t + (pi)/(2))` Then maximum acceleration of the particle is

Options

A. `10^(4)` unit
B. `6.0 xx10^(5)` unit
C. `7.0xx10^(6)`unit
D. `4xx10^(3)` unit

Answer

B

Explanation

Given `A_(1) =4 " "omega =100pi `
`A_(2) =3" "Delta omega=(pi//3)/(60^(@))`
`A _(n et) = sqrt(A_(1)^(2) + A_(2)^(2) +2A_(1) A_(2)cos Delta omega)`
`A _(n et) =sqrt((4)^(3) +(3)^(2) +2xx4 xx3xxccos 60^(@) )`
`A _(n et) =sqrt(16 +9+12)=sqrt37 =6.1`
`A _(max)=A _(n et)omega^(2)`
`=6.1 xx(100pi) ^(2) =6.0 xx10^(5) ` unit