For the damped oscillator shown, the mass m of the block is 200 g, k = 90 N mā1 and the damping constant b is 40 g sā1. What is the period of oscillation, time taken for its amplitude of vibrations to drop to half of its initial value.
Here km =90×0.2=18 kg2 s−2
⇒√km=4.283 kg s−1
Given value of b=0.04 kg s−1
We can see that b<<√km
So we can say the angular frequency of oscillation
ω≈√km
⇒T≈2π√mk=2×π×√0.290=0.3s
The amplitude depends on time as
A(t)=A0e−bt2m
So when amplitude drops to half it's initial value we have
A(t)=A02 ⇒A02=A0e−bt2m
⇒12=e−bt2m
⇒−ln2=−bt2m
⇒t=2m ln2b
=2×0.2×ln20.04
=6.93s