A particle executing SHM according to the equation `x=5cos(2pit+(pi)/(4))` in SI units. The displacement and acceleration of the particle at t=1.5 s is

Options

A. ` -3 m , 100 m//s ^ 2 `
B. ` + 2.54 m, 200 m//s ^ 2 `
C. ` - 3.54, 140 m//s ^ 2 `
D. `+ 3.55 ,120 m//s ^ 2 `

Answer

C

Explanation

The given equation of simple harmonic motion is
` x ( t ) = 5 cos (2pi t + ( pi )/( 4)) ` …(i)
The standard equation of SHM is
` x (t) = A cos ( omega t + phi ) ` …(ii)
compare Eqs. (i) and (ii), we get
` omega = 2 pi s ^( - 1 ) `
At ` t = 1.5 s `
Displacement
` x ( t ) = 5 cos ( 2pi + 1.5 + ( pi ) /( 4)) `
` = 5 cos ( 3pi + ( pi ) /( 4)) = - 5 cos (( pi )/( 4)) `
` = -5 xx 0.707 m = - 3.45 m `
Acceleration,
` a = -omega ^ 2 xx ` displacement
` = - (2pi s ^( - 1 ) ) ^ 2 xx ( -3.54 m ) = 140 m//s ^ 2 `