Question

A 3.33-kg mass attached to a spring with k = 151 N
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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



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