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