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
(a) The period of oscillation
(b) Time taken for its amplitude of vibrations to drop to half of its initial values
(c) The time for the mechanical energy to drop to half initial values

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`