An object with mass 0276 kg is hung on a spring whose spring

An object with mass 0.276 kg is hung on a spring whose spring constant is 85.4 N/m. The object is subject to a resistive force given by -bv, where v is its velocity. The ratio of the damped frequency to the undamped (natural) frequency) is 0.822. If this system is subjected to a sinusoidal driving force given by

F(t)=(8.68 N)sin(0.938_o t) ,

what is the (steady-state) amplitude (in cm) of the resulting sinusoidal motion?

Solution

Damped frequency = 0.938*₀ = 0.938* [ k/m] = 0.938* [ 85.4/0.276] = 16.5 rad/s

amplitude is calculated as follows:

    A = (Fo/m)/[ (^2-k/m)^2 + [(b/m)*]^2]^0.5

         =(8.68/0.276)/[(16.5^2- [85.4/0.276])^2 +(0.822*16.5)^2]^0.5

         = 0.794 m

         = 79.4 cm