A particle is performing simple harmonic motion along `x-`axis with amplitude `4 cm` and time period `1.2 sec` .The minimum time taken by the particle to move from `x = 2 cm to x = +4 cm` and back again is given by

Options

A. 0.6 s
B. 0.4 s
C. 0.3 s
D. 0.2 s

Answer

B

Explanation

Time to go from O to B `=(T)/(4)=0.3 s`
Time to go from O to A,
`2=4 "sin" ((2pi t)/(T)) therefore (1)/(2)="sin" (pi)/(6) = "sin" (2pi t)/(T)`
`therefore t=(T)/(12)=0.1 s`
`therefore` Time to go from A to B =0.3-0.1=0.2 s
`therefore` Total time =0.4 s