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

Options

A. `0.6sec`
B. `0.4 sec`
C. `0.3 sec`
D. `0.2 sec`

Answer

B

Explanation

Time taken by period to move from `x = 0` (mean pisition) to `x = 4` (extermeposition) `= (T)/(4) = (1.2)/(4) = 0.3 sec`
Let `t` be the time taken by the particle to move from `x = 0 to x= 2 cm`
`y = a sin omega t rArr 2 = 4 sin"(2pi)/(T)t rArr (1)/(2) = sin"(2pi)/(1.2)t`
`rArr (pi)/(6) = (2pi)/(1.2) t rArr = 0.1 sec`
Hence time to move from `x = 2 to x= 4` will be equal to `0.3 - 0.1 = 0.2 sec`
Hence total time to move from`x = 2 to x= 4`and back again ` 2 xx 0.2 = 0.4sec`