Forced Harmonic Motion


by dangus
Tags: forced, harmonic, motion
dangus
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#1
Apr5-12, 09:43 AM
P: 12
I have a pendulum swinging with frequency w in a frictionless environment and I apply a periodic harmonic driving force also at frequency w but out of phase with the swinging pendulum. What happens exactly? Does the driving force bring the pendulum into phase with the force and then we see the resonance phenomena where the amplitude of motion will just spike? Or do we not see resonance?

Thanks
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Bob S
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#2
Apr5-12, 09:21 PM
P: 4,664
the attached solution to an undamped harmonic oscillator (mass on a spring) with a sinusoidal driving force is very similar to your problem. The driving force is given in [2] and the complete solution is given in [9]. You will have to change the initial conditions.

When the driving force frequency is close to the natural frequency, you will get very large amplitudes with a beat frequency.
Attached Files
File Type: pdf Forced oscillator problem4.pdf (18.0 KB, 23 views)
Ken G
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#3
Apr5-12, 10:14 PM
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And if you want the answer in words, after a good amount of time goes by it won't matter whatever was the initial phase of the oscillator-- its motion will be completely determined by the driving force and all vestiges of its initial state are lost. The final phase relationship with the driving force is interesting and surprising-- the displacement will be 180 degrees out of phase with the driving if the driving frequency is greater than the "natural" frequency of the oscillator (what it oscillates at if it is not driven at all), and it will be in phase with the driving if the driving frequency is less than the natural frequency. These are the two possibilities that exchange no net energy with the oscillator, as required by it having reached its steady-state oscillation amplitude. But note again that its initial phase is still irrelevant-- if it started out 180 (or even 0) degrees out of phase with the driving, but did not start out with its full steady-state amplitude, then it will need to get closer to 90 degrees out of phase with the driving while energy is pumped into it, and will only later end up being 180 (or 0) degrees out of phase. When the displacement is 90 degrees out of phase with the driving (which means that the force is maximal when the oscillator is at its equilibrium point and has maximum velocity to get worked on by the force), this is where you get the most rapid energization of the oscillation.

Bob S
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#4
Apr6-12, 10:41 PM
P: 4,664

Forced Harmonic Motion


The two attached plots are for two simulations of the forced undamped oscillator, using the solution of the differential equation shown in the attachment tto post #2. The first shows a natural frequency of 1 Hz, and a driven frequency of about 0.95 Hz. The second plot is the same natural frequency, but the driven frequency is now about 0.999999 Hz.

In the two plots, the fast frequency is half the sum frequency, and the beat frequency is the difference frequency.
Attached Thumbnails
Forced OscillatorA.jpg   Forced OscillatorB.jpg  
Ken G
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#5
Apr8-12, 07:44 PM
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Note also that such strong beating is not actually "generic", the phase of the driving was chosen very carefully to be at zero when t=0. Any other choice will lead to less beating. More importantly, one normally assumes some essentially negligible damping that won't affect the steady-state behavior but suffices to "damp out" the information of the initial state, and then there's no beating at all at large times and the response frequency is the driving frequency, not an average of the two. It's interesting though-- I never realized that having absolutely zero damping would make it so the initial state was not "forgotten" and the beating continues forever.
Bob S
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#6
Apr8-12, 09:34 PM
P: 4,664
Quote Quote by Ken G View Post
Note also that such strong beating is not actually "generic", the phase of the driving was chosen very carefully to be at zero when t=0. Any other choice will lead to less beating. More importantly, one normally assumes some essentially negligible damping that won't affect the steady-state behavior but suffices to "damp out" the information of the initial state, and then there's no beating at all at large times and the response frequency is the driving frequency, not an average of the two. It's interesting though-- I never realized that having absolutely zero damping would make it so the initial state was not "forgotten" and the beating continues forever.
In the solution to the differential equation I posted in post #2 attachment, both x(t)=0 and dx(t)/dt= 0 at t=0, meaning that the mass was motionless at t=0 (i.e., H=T + V =0), so there was no "phase" of the driving force at t=0. So the observed beating is not the result of the driving force beating wih some initial motion of the mass. The driving force caused both of the signals that were beating against each other.
Ken G
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#7
Apr9-12, 09:22 PM
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Quote Quote by Bob S View Post
In the solution to the differential equation I posted in post #2 attachment, both x(t)=0 and dx(t)/dt= 0 at t=0, meaning that the mass was motionless at t=0 (i.e., H=T + V =0), so there was no "phase" of the driving force at t=0.
Yes there is-- the driving is also taken to be zero at t=0 (it's taken to be a sine function), that's a special phase choice. Try the solution again if the driving is at its maximum at t=0 (a cosine function), I believe you will find no beating at all in that case. Intermediate choices will produce intermediate amounts of beating, but the "normal" approach assumes some tiny damping, and the initial conditions are completely lost after enough time-- so no beating there either. Still, your point is well taken that it's interesting what happens when the damping is exactly zero, that is rarely considered in steady-state solutions to a driven oscillator.
Bob S
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#8
Apr10-12, 02:06 PM
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Quote Quote by Ken G View Post
Yes there is-- the driving is also taken to be zero at t=0 (it's taken to be a sine function), that's a special phase choice. Try the solution again if the driving is at its maximum at t=0 (a cosine function), I believe you will find no beating at all in that case............
I solved the differential equation for a driving force of the form F(t)=0 (t<=0), F(t)=Aocos(ωt) (t>0), vs. the F(t)=Aosin(ωt) used for the plots in post #4. The result is
[tex] x(t)=A_o \frac{cos(\omega t)-cos(\omega_ot)}{\omega_o^2-\omega^2} [/tex]
The new plot for ω≈0.95 Hz and ωo=1 Hz is attached. It is nearly identical to the first plot in post #4, except that the initial phase is shifted earlier by about 90 degrees.
Attached Thumbnails
Forced_OscZ.jpg  
Ken G
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#9
Apr10-12, 05:36 PM
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You are correct, I thought switching to a cosine would mean you didn't need the two terms to be the same magnitude when you apply the dx/dt=0 initial constraint, but I missed that now you still need them to be the same magnitude to get the x=0 initial constraint. So it's true that you always get strong beating when the damping is exactly zero. I should still make the point, however, that this is rarely the physically desired solution, because usually we are interested in systems with some nonzero damping, even if the value of the damping is not important to the steady-state solution. So in the most common case of interest, there is no beating, and no memory of the initial condition, in the long-term solution, even when the damping is otherwise negligible. I feel this is an important point to make, though it does not contradict the zero-damping solution you are talking about (which I am glad to have heard about, it is generally not made mention of).


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