A partcle is performing simple harmonic motion along x- axis with amplitude `4cm` and time period `1.2sec` The minimum time period taken by the again is given by

Explanation

If x is the displacement of a particle in time t, then
`x=asinomegat=asin(2pi)/(T)t`
So, `t=(T)/(2pi)sin^(-1)((x)/(a)), ` where `a=4cm`.
At `x=2cm t=(T)/(2pi)sin^(-1)((2)/(4))=(T)/(2pi)xx(pi)/(6)=(T)/(12)`
`=(1.2)/(12)=(1)/(10)s`
At `x=4cm t=(T)/(2pi)sin^(-1)((4)/(4))`
`=(T)/(2pi)xx(pi)/(2)=(T)/(4)=(1.2)/(4)=(3)/(10)s`
So time taken in gooinf grom `x=+2cm` to `x=+4cm` will be
`=(3)/(10)-(1)/(10)=(2)/(10)=0/2s`
The same time will be taken for coming back.
So total time taken `=0.2+0.2=0.4s`