A particle is perfroming simple harmoic 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 sec`
B. `0.4 sec`
C. `0.3 sec`
D. `0.2 sec`

Answer

B

Explanation

Time taken by to move `x = 0` (mean position ) to `x = 4` (extreme position) `= (T)/(4) = (1.2)/(4) = 0.3s` Let `x` be the time taken by the particle as move from `x= 0` to `x = 2cm`
`y = a sin omega rArr 2 = 4 sin'(2pi)/(T) rArr (1)/(2) = sin (2pi)/(1.2)`
`rArr (pi)/(6) = (2pi)/(1.2) rArr t = 0.1s` Hence time to move from `x`
`= 2 to x = 4` will be equal to `0.3 - 0.1 = 0.2s`
Hence tatal time to move from `x = 2` to `x = 4` and back again `= 2 xx 0.2 = 0.4sec`