Posted by **James** on Monday, July 9, 2012 at 9:00pm.

A 3-kg 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.

- Physics -
**Elena**, Tuesday, July 10, 2012 at 5:33am
(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.

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