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]`

Explanation

Given `T = 1.2 s, A = 87 cm`
`y = 8 - 3 = 5 cm`
The equation of simple harmonic motion, when motion starts from exterme position.
`y = A cos omega t = A cos ((2 pi t)/(T))`
`5 = 8 cos ((2 pi t)/(T))`
or ` cos ((2 pi t)/(T)) = (5)/(8) = 0.625`
`= cos^(-1) (0.625) = 51^(@) xx (pi)/(180)`
`t = (51 xx pi)/(180) xx (T)/(2 pi) = (51)/(360) = (51)/(360) xx 1.2 = 0.17 s`