A particle of mass m = 2kg executes SHM in xy plane between points A and B under the action - Physics - Oscillations

Question

​​A particle of mass m = 2kg executes SHM in xy plane
between points A and B under the action of force Image not present
= Fxî + Fyĵ Minimum time taken by the particle to move
from A to B is 1s At t = 0 the particle is at x = 2 and y = 2 Then
Fx as function of time is :-
a) -4pi​​2​sinpit
b) -4pi2cospit
c) 4pi2cospit
​d) None

Pintu B. · Accepted Answer

Dear Student ,
Here in this case ,
let the equation is x=a cosπtso , dxdt=-aπ sinπt⇒d2xdt2=-aπ2 cosπtNow , Fx=maxx=2=-2×2π2 cosπt=-4π2 cosπtSimilarly if we take y=a cosπt then also at y=2 ,Fy=-4π2 cosπtSo here correct option is b 
Regards