Physics
posted by James on .
A 3kg mass attached to a spring with k = 145 N/m is oscillating in a vat of oil, which damps the oscillations.
(a) If the damping constant of the oil is b = 14 kg/s, how long will it take the amplitude of the oscillations to decrease to 1% of its original value?
(b) What should the damping constant be to reduce the amplitude of the oscillations by 91% in 4 s?
The answers are 1.97s and 3.61 kg/s. How can I find these answers? Thank you in advance.

(a)
The amplitude of damped oscillations depends on time as
A=Aₒ•e^(β•t),
The damping coefficient β =b/2m, where b=14 kg/s is the damping constant,
β =b/2m =14/2•3=2.33 s^1
Given:
Aₒ/A=100%/1%= 100
Aₒ/A= Aₒ/ Aₒ•e^(β•t) = e^(β•t) =100,
β•t =ln100,
t=ln100/ β = 4.6/2.33=1.97 s.
(b)
Given:
Aₒ/A=100%/9%= 100/9
Aₒ/A= Aₒ/ Aₒ•e^(β•t) = e^(β•t) =100/9,
β•t =ln(100/9)=2.408,
β =2.408/4=0.602 s^1,
b=2m β =2•3•0.602=3.61 kg/s.