The motion of a forced damped spring system is: ?(?) = 2 sin 4? + ?...

The motion of a forced damped spring system is:

?(?) = 2 sin 4? + ? ^-? cos 6?.

A) What is the driving frequency?

B) What is the transient part?

C) What is the steady-periodic part?

Could someone provide explanations as to how to obtain these solutions as I am lost?

Solutions

Expert Solution

The solution to the differential equation of a forced damped spring system driven by an external force has two parts, a transient part and a steady-state part, which must be used together to fit the physical boundary conditions of the problem.

The general solution is a sum of a transient solution that depends on initial conditions, and a steady state that is independent of initial conditions and depends only on the driving amplitude F0, driving frequency ω.

The transient part must contain a damping part in terms of a negative exponential.

The solution given to us is

?(?) = 2 sin 4? + ? ^-? cos 6?.

Clearly , the second part on the RHS contains the damping part ( e^-t) and hence , it is the transient part. (PART B)

The first part on the RHS is independent of the initial conditions and does not contain any damping part, and hence, it is the steady-periodic part. ( 2 sin 4?) (PART C)

The steady periodic part is due to the driving force. Thus, comparing 2 sin 4? with A sin (wt) gives the angular driving frequency w=4.

Hence, Driving frequency = w/(2*) = 4/(2*) = 2/ (PART A)

genius_generous answered 1 year ago

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