What Is the Steady-State Oscillation of the Mass-Spring System?

[John Michael]

Homework Statement

Find the steady-state oscillation of the mass–spring system
modeled by the given ODE. Show the details of your
calculations.

Homework Equations

1. y'' + 6y' + 8y = 130 cos 3t

2. 4y’’ + 8y’ + 13y = 8 sin 1.5t

The Attempt at a Solution

1. cos(3t) at the end means the basic angular frequency is 3 radians per second. Hence the steady-state oscillation frequency is 3/2pi Hz = 0.477 Hz.

2. solve for homogeneous differential equation,

4y'' + 8y' + 13y = 0

propose y = e^(ct)

y' = ce^(ct)

y'' = c²e^(ct)

substitute into mass-spring motion equation,

4y'' + 8y' + 13y = 0

4c²e^(ct) + 8ce^(ct) + 13ce^(ct) = 0

e^(ct)(4c² + 8c + 13) = 0

of course for unique solution, it must be e^(ct) ≠ 0

4c² + 8c + 13 = 0

c = -1 ± (3i)/2homogeneous solution is

y(t) = e^(-t) (A sin (3t/2) + B cos (3t/2))

where A and B is an arbitrary constants which dependent to boundary conditions

 

[rock.freak667]

The Attempt at a Solution

You will need to post your attempt before we can help you.

 

[John Michael]

please help me

 

[rock.freak667]

please help me

Well you need to get the particular integral for the first one. At steady state means that as t→∞. So you can see that in your calculations, as t→∞, the homogeneous part tends to zero.

The PI for cos(3t) would be y=Ccos(3t)+Dsin(3t), you need to get 'C' and 'D'.

 

[klondike]

please help me

You can solve the particular integral as rock.freak suggested.

A more physically intuitive way of doing it may look like this:

The problem is basically the steady-state sinusoidal response of a second order LTI (Linear Time Invariant)system. If the system is stable(bounded input bounded output), the output is an attenuated sinusoid which has the same freq as the excitation and a constant phase lag.

Firstly you need to confirm if the system is stable otherwise there is no stead-state response because it's unbound.

It's convenient to use complex exponential to express the sinusoidal excitation and the solution. because sine and cos are special case of it.

The ode
y''+2\zeta \omega_{n} y'+\omega_{n}^{2}y=Ae^{j\omega t}

Stability test:
Given the natural freq \omega_{n} >0, the system is stable if and only if \zeta>0. It means the homogeneous solutions die as t approaches infinity.
The output, or solution y=Me^{j(\omega t+\theta)}

You then plug y into the ODE, the magnitude M and the phase lag theta can be very easily found. In engineering, M(\omega) is called frequency response, \theta(\omega) is called phase response.

 

[John Michael]

thank you