For the damped oscillator shown in previous Figure, the mass m of the block is 400 g, `k=45 Nm^(-1)` and the damping constant b is `80 g s^(-1)`. Calculate .
(a) The period of osciallation ,
(b) Time taken for its amplitude of vibrations to drop to half of its initial value and
(c ) The time taken for its mechanical energy to drop to half its initial value.

Explanation

(a) `km = 45 xx 0.4 = 18 kg Nm^(-1) =18 kg^(2) s^(-2)`
`:. sqrt(km) = 4.243 kgs^(-1)`
`b= 0.08 kg s^(-1) ` is much less than `sqrt(km)` . Hence, the time period T from equation ( 49) is given by ,brgt `T = 2pisqrt((m)/(k))`
`= 2pi sqrt((0.4)/( 45)) = 0.6 s `
(b) From equation ( 48), the time `T_(1//2)` for amplitude to drop to half of its initial value of given by `T_(1//2)= (log(2))/( (b)/(2m))`
`= ((0.692)/( 0.08))/( 2 xx 0.4)`
`= 6.93 s`
(c ) For calculateing the time `t_((1)/(2))` for its mechanical energy to drop to half its initial value we make use of equation `( 50 )`
`E(t_((1)/(2))) //E(0)= `exp. `(- (bt_((1)/(2)))/(m))`
` (1)/(2) = `exp` (-(bt_((1)/(2)))/(m))`
`ln((1)/(2))= (-bt_((1)/(2)))/(m)`
or `t_((1)/(2)) = ( 0.692)/( 80gs^(-1)) xx 400g`
`= 3.46 s `
This is just half of the decay period for amplitude, which is not surprising, because, according to equations ( 48 ) and ( 50) , energy depends on the square of the amplitude. Notice that there is a factor of 2 in the exponents of the two exponentials.