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. `0.3s`
B. `0.4s`
C. `0.5s`
D. `0.2s`

Answer

B

Explanation

Here, `a=4cm, T=1.2s`
As `x=asin omegat =asin ((2pi)/(T))t`
`:. T=(T)/(2pi) sin ^(-1) ((x)/(a))`
`a=4c, At x=2, t=(T)/(2pi)sin^(-1)((2)/(4))`
`t=(T)/(2pi)xx(pi)/(6)=(T)/(12)=(1.2)/(12)=(1)/(10)s`
At `x=4, t=(T)/(2pi)sin^(-1)((4)/(4)) or t=(T)/(2pi)xx(pi)/(2)=(T)/(4)`j
`t=(1.2)/(4)=(3)/(10)s`
Time taken in going from `x=+2cm` to `x=+4cm` will be `=(3)/(10)-(1)/(10)=0.2s` and same time will be taken for coming back.
So total time taken `=0.2+0.2=0.4s`