Question

A 3.33-kg mass attached to a spring with k = 151 N/m is oscillating in a vat of oil, which damps the oscillations. If the damping constant of the oil is b = 10.3 kg/s, how long will it take the amplitude of the oscillations to decrease to 1.70 % of its original value? What should the damping constant be to reduce the amplitude of the oscillations by 98.3 % in 1.30 s?

Answer 1

The mass, m = 3… kg

      The spring constant, k = 151 N/m

      The damping constant of the oil, b = 10.3 kg/s

a)
      The displacment equation for the damped oscillator is

               x(t) = xme(-b/2m)t   

      Given that, x/xm = 0.01 ( that is 1% of the original )

      So the above equation changes to,

                   e(-b/2m)t = 0.017

      Apply the natural logerethms on both sides, we get

                  [-b/2m]t = ln(0.017)

      Therefore, the required time is

                       t = ln(0.017)/[-b/2m]

                         = -2mln(0.017)/b
                         = -(2)(3.33 kg)ln(0.017)/(10.3 kg/s)

                         = 2.63 s

b)
      The time t =1.30 s

      Then, x/xm = 100% - 98.3%

                         = 1.7%

                         = 0.017

      From equation (1), the damping constant is

                b = -(2m/t) ln (x /xm)

                   = -(2)(3.33 kg)ln(0.017)/(1.3 s)

                   = 20.9 kg/s

                   = 21 kg/s