For the damped oscillator, the mass m of the block is 200 g, k = 90 N m^-1 and the damping constant b is 40 g s^-1. Calculate (a) the period of oscillation, (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.

We see that k m = 90 × 0.2 = 18 kg N m^-1 = kg² s^-2. Therefore (km)^1/2 = 4.243 kg s^-1, and b = 0.04 kg s^-1. Therefore b is much less than km.

(a) The time period T is given by:

T = 2π √^m/_k

On putting the values we get T = 2π √^0.2/₉₀ = 0.3 s

(b) The time T_1/2 for the amplitude to drop to half of its initial value is:

T_1/2 = ln(1/2) / (b/2m)

= 0.693 / (40 × 2 × 200) s = 6.93 s

(c) For the mechanical energy to drop to half its initial value:

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)

⇒ t_1/2 = 0.693 / (40 g s^-1 × 200 g) = 3.46 s

This is half of the decay period for amplitude, since energy depends on the square of the amplitude. Notice the factor of 2 in the exponents of the two exponentials.