Question

A 2kg mass attached to a spring with k = 120 N / m is oscillating in an oil tub, which dampens oscillations. A) If the oil damping constant is b = 10kg / s, how long will it take for the amplitude of the oscillations to decrease to 2% of its original value? B) What should be the damping constant to reduce the amplitude of the oscillations by 99% in 1 second?

Answer 1

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Given

Spring constant k= 120 N/m

Damping constant b = 10 kg/s

Mass m = 2 kg

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a) The displacment equation for the damped oscillator is

                    x = A e^( - b / 2m ) t

Given that x /A =   0.02 ( that is 2.0% of the original )

                          = 0.02

Substitute this value in the above equation,
       e^(-b/2m)t = 0.02

         (b / 2m) t = ln (0.02)

                      t = - (2m/ b) ln (0.02)

                        = - [2( 2 kg) / (10 kg/s) ] ln (0.02)

                        = 1.5648 s

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b) The displacment is given as

       x = A e^ (- b / 2m ) t

The damping constant is

   b = -(2m/t) ln (x /A)

Here x/A =1-0.99

              =0.01

         t = 1 s

On substituion,

       b = -(2m/t) ln (x /A)

          = - [ 2(2 kg) /(1.0s)) ln[(0.01)   

          = 18.421 kg/s

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