# Forced Oscillations

1. Dec 12, 2013

### Jimmy87

Hi, why is it that as the frequency of the driver decreases below the natural frequency of the oscillator it reaches a fixed amplitude when the external frequency is zero whereas whenever you go to the other extreme and have a very high external frequency the amplitude of the oscillator approaches zero? The graph that displays this information shows that the amplitude of the oscillator has a value (Fo/k) when the driving frequency is zero (i.e. the graph doesn't go through the origin). How can you have an amplitude at zero frequency? Thanks for any help given!

2. Dec 12, 2013

### AlephZero

A frequency of zero is the same as doing statics, not dynamics. You apply a force to the spring and it stretches.

At high frequencies, the acceleration is approximately constant. From Newton's second law and ignoring the stiffness of the spring, the acceleration is close to $F/m$ so the amplitude is approximately $F/(m\omega^2)$.

3. Dec 13, 2013

### Jimmy87

Thanks AlephZero. So are you saying that at the instantaneous moment you turn the signal generator on, you don't have a frequency but you still have a force therefore you are dealing with a static case at that instant so therefore the mass on the spring will get an amplitude equal to that of the vibrator? In other words are you saying that you are dealing with a case before the vibrator has completed one cycle?

4. Dec 13, 2013

### AlephZero

Well, in real life you probably can't set the vibrator to "zero" frequency. But you can imagine that if the frequency is very low compared with the natural vibration frequencies of the object (e.g. one cycle per hour or whatever), you can ignore the inertia forces (i.e. mass x acceleration) because the acceleration is very small. So this is the same as applying a "static" load that slowly increases and decreases, and measuring the static deflections it produces (and the deflections are proportional to the load, of course).

The math still makes sense if the frequency is zero. A force $F_0 \sin \omega t$ is always $0$ if $\omega = 0$, but a force $F_0 \cos \omega t$ is a constant force $F_0$.

Note, this only makes sense for a structure that is constrained in some way. If the object is completely free (e.g. floating weightless in space) the displacements will become to "infinitely large" as the frequency goes to zero, because you are applying a force in one direction for a long time and that will move the object a large distance.