For the damped oscillator shown in Fig, the mass of the block is `200 g, k = 80 N m^(-1)` and the damping constant `b` is `40 gs^(-1)` Calculate
The period of oscillation

Answer

(a) `0.3 s` , (b) `6.93 s` , (c) `3.4 s`

Explanation

(a) As the damping constant, `b`
`(= 0.04 kg s^(-1)) ltlt sqrt(km)`
the time period `T` from the equation
`omega' = sqrt((k)/(m) - (b^(2))/(4m^(2)))` is given by.
`T = 2pisqrt((m)/(k)) = 2pisqrt((0.2kg)/(80Nm^(-1))) = 0.314 s`
(b) From the equation `x(t) = x_(m)ɵ^((bt)/(2m))` the time `T_(1//2)` for the amplitude to drop to drop to half of its initial value is
`(1)/(2) = ɵ^((-bt_(1//2))/(2m))`
or `log_(e) 2 = (bT_(1//2))/(2m)`
or `0.693 = (bT_(1//2))/(2m)`
or `T_(1//2) = (0.693l xx 2 xx 0.2)/(40 xx 10^(-3)) = 6.93 s`
(c) `E (t) = E(0)e^(-(bt_(1//2))/(m))`,
`(1)/(2) = e^(-(bt_(1//2))/(m)) , log_(e) 2 = (bt_(1//2))/(m)`
`t_(1//2) = (0.693 xx 0.2)/(40 xx 10^(3)) =m 3.4 sec`