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

T = 1.2 s. `x_(1)=4sinomegat_(1)`
`therefore omegat_(1)=sin^(-1).((x_(1)))/(4)=sin^(-1)((2)/(4))=(pi)/(6)`
`x_(2)=4sinomegat_(2)`
`therefore omegat_(2)=sin((4)/(4))=sin^(-1)(1)=(pi)/(2)`
`t_(2)-t_(1)=((pi)/(2)-(pi)/(6))(1)/(omega)=(pi)/(2pi)T[(1)/(2)-(1)/(6)]`
`=(T)/(2)[(3-1)/(6)]=(T)/(2)[(2)/(6)]=(T)/(6)=(1.2)/(6)`
`=0.2 s`
`therefore` Therefore time taken by teh particle to move from x = 2cm to x =4cm and back again is
`0.2+0.2=0.4s`.