The displacement of an object attached to a spring and executing simple harmonic motion is given by ${\text{x=2×10}}^{\text{-2}}\text{cos(πt)m}$ . The time at which the maximum speed first occurs is

detailed solution

Correct option is A

Given displacement x=2×10-2cos(πt)  differentiate above equation with respect to time velocity =v=dxdt=−2×10−2πsin(πt) For the first time when V=Vmax velocity is maximum when sinπt=1=sin π2 πt= π2 t=12=0.5second***Alternate solution: The given equation shows that the particle is at extreme position .  since πt=ω ⇒πt=2πT  time period is equal to 2 sec and the particle will take a time equal to   T4=0.5sec                                                                                                                            to reach the mean position where the velocity is maximum.