For the damped oscillator shown in Fig. 14.19, 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.

For the damped oscillator shown in Fig. 14.19, the mass m of the block

1.	For the damped oscillator shown in Fig. 14.19, 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.
Answer» Solution :(a) We see that KM = `90x×0.2 = 18 kg Nm^(-1)=kg^(2)s^(-2),` therefore `sqrt(km)=4.243kgs^(-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=2pisqrt(m/k)` `=2pisqrt((0.2kg)/(90Nm^(-1)))` = 0.3 s (b) Now, from Eq. (14.33), the time, `T_(1//2)`, for the AMPLITUDE to drop to half of its initial value is given by `T_(1//2)=("ln"(1//2))/(b//2m)` `=(0.693)/(40)xx2xx200s` `=6.93s` (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)/(40gs^(-1))xx200g` =3.46 s This is just half of the DECAY period for amplitude. This is not surprising, 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.