A particle executes SHM of period `1.2 s` and amplitude `8cm`. Find the time it takes to travel `3cm` from the positive extremity of its oscillation. `[cos^(-1)(5//8) = 0.9rad]`

Options

A. `0.28s`
B. `0.32 s`
C. `0.17 s`
D. `0.42 s`

Answer

C

Explanation

`X = A cos omega t`

`5 = 8 cos omega t`
`omega t = cos^(-1)(5//8) = 0.9`
`:. t = (0.9)/(omega) = (0.9)/((2pi//T))`
`= (0.9T)/(2pi) = (0.9 xx 1.2)/(2pi)`
`= 0.17 s`