# Undamped harmonic oscillator

1. Aug 12, 2017

### Richardbryant

1. The problem statement, all variables and given/known data
An undamped harmonic oscillator (b=0) is subject to an applied force Focos(wt). Show that if w=wo, there is no steady- state solution. Find a particular solution by starting with a solution for w=wo+#, and passing to the limit #->0, it will blow up. Try starting with a solution which fits the initial condition xo=0, so that i cannot blow up at t=0.

2. Relevant equations

3. The attempt at a solution
d^2x/dt^2+(wo^2)x=Fo cos(w+#)t/m
d^2y/dt^2+(wo^2)y=Fo sin(w+#)t/m
d^2z/dt^2+(wo^2)z=Foe^i(w+#)t/m (1)
Let Z=Ce^i(wo+#)t, plug in (1)
C=Fo/,[wo^2-(w+#)^2]

thus X= Fo cos(w+#)t/m[wo^2-(w+#)^2]
Xtr (trasient term )=Acos(wot-$)$= phase difference
After a couple of steps the final solution will blow up when limit #->0

2. Aug 12, 2017

### Dr.D

With no damping and sinusoidal excitation at the undamped natural frequency, the solution grows linearly with time. This is the reason there is no steady state.

You need to obtain the characteristic equation and look at the roots. That will get you started toward the proper results.

3. Aug 13, 2017

### vela

Staff Emeritus

4. Aug 14, 2017

### Richardbryant

Thanks for reply, i had been guessing the solution is also a trigonometric function , but it seems to be not working

5. Aug 14, 2017

### Richardbryant

The question is to find a x(t) satisfying the given condition

6. Aug 14, 2017

### vela

Staff Emeritus
Obviously, that's what the question is asking of you. What is YOUR specific question? You seem to be on the right track.

7. Aug 15, 2017

### Richardbryant

Oh, yeah i got the correct answer, but i didn't notice, thank you about that!

8. Aug 15, 2017

### vela

Staff Emeritus
Did you manage to show the system didn’t have a steady state solution?