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

(a) We see that `km = 90 xx 0.2 = 18 kg N m^(-1) = kg^(2) s^(-2)`, therefore `sqrt(km) = 4.243 kg s^(-1)`, and `b = 0.04 kg s^(-1)`. Therefore, b is much less than `sqrt(km)`. Hence, the time period T from Eq. (14.34) is given by
`T = 2pi sqrt((m)/(k))`
`= 2pi sqrt((0.2 kg)/("90 N m"^(-1)))`
`= 0.3 s`
(b) Now, from Eq. (14.33), the time `T_(1//2)`, for the amplitude to drop to half of its intial value is given by,
`T_(1//2)=(ln(1//2))/(b//2m)`
`= (0.693)/(40)xx 2xx200 s`
`= 6.93 s`
(c) For calculating the time `t_(1//2)`, for its mechanical energy to drop to half its initial value we make use of Eq. (14.35). From this equation we have,
`E (t_(1//2))//E(0) = "exp"(-bt_(1//2)//m)`
Or `1//2= "exp" (-bt_(1//2)//m)`
`ln (1//2) = -(bt_(1//2)//m)`
Or `t_(1//2) = (0.693)/(40 "g s"^(-1)) xx 200 g`
`= 3.46 s`
This is just half of the decay period for amplitude. This is not suprising because, according to Eqs. (14.33) and (14.35), energy depends on the square of the amplitude. Notice that there is a factor of 2 in the exponents of the two exponentials.