In damped oscillatory motion a block of mass `20kg` is suspended to a spring of force constant `90N//m` in a medium and damping constant is `40g//s`. Find (a) time period of oscillation (b) time taken for amplitude of oscillation to drop to half of its intial value (c) time taken for its mechanical energy to drop to half of its initial value.

Explanation

Mass `m = 200g = 0.2kg`, Force constant `K = 90 N//m` damping constant `b = 40g//s = 0.04 kg//s`
`sqrt(Km) = sqrt(90xx0.2) = sqrt(18) kg//s, b lt ltsqrt(Km)`.
(a) time period
`T = 2pi sqrt((m)/(K)) = 2pi sqrt((0.2)/(90)) = 0.3s`
(b) amplitude `A = A_(0)e^(-bt//2m)` Let amplitude is dropped to half of its initial value after the time `T_(1//2)`.
`A_(0)e^((-bT_(1//2))/(2m))=(A_(0))/(2), e^((-bT_(1//2))/(2m)) = (1)/(2)`
Take netural logarithm on both sides
`(-b(T_(1//2)))/(2m) = In ((1)/(2)) rArr T_(1//2) = (In(2))/(b//2m)`
`= 2.303 xx 0.3010 xx 2m//b`
`T_(1//2) = 0.693 xx (2m)/(b) = 0.693 xx (2xx0.2)/(0.04) = 6.93s`
(c) Let the energy is dropped to half of its intial value after a time `t_(1//2)`.
Initial energy `E_(0) = (1)/(2)KA^(2)`, At time `t_(1//2)`, energy `= (1)/(2)E_(0)`
`= (1)/(2) KA^(2) e^((-bt_(1//2))/(m)) =(1)/(2)((1)/(2)KA^(2))`
`e^((-bt_(1//2))/(m)) = (1)/(2) rArr t_(1//2) = In (2) xx (m)/(b) = 0.693 xx (m)/(b)`
`t_(1//2) = 0.693 xx (0.2)/(0.04) =(0.2)/(0.04) = 3.46 s`