Driven SHM k m Last time added damping. Got this kind of solution that oscillates due to initial conditions, but then decays. This is an important concept. Whenever you disturb a free oscillator, it will respond by oscillating at (near) the natural frequency, and the oscillations will damp out over time. Whether you start from displacement at rest, give it initial velocity (kick), any force impulse will set off an oscillation that decays, we call this a “transient” response. **We will use this setup to demo oscillators today, it is a “mass on a shim” resonator. This laser bounces off a mirror at the end of the shim to show the motion up on the screen. We can’t follow the actual motion or measure the amplitude, but we can see the amplitude from the length of the line. Note that when I give it an impulse (offset or kick), I get a response that decays away. Now lets add a driving force. You will see that this is a fundamentally different situation. Let’s drive the oscillator at a frequency w. It is NOT necessarily at the natural frequency or at the frequency of the damped free oscillator. It could be at any frequency. If you drive it this way, you get a complex response that might look something like this. There is some complex behavior, then it settles into a sinusoid. Let’s consider the response at long times first. We call this the “steady state” motion. At steady state, the oscillator will move with the same frequency as the drive. Resistance is futile. I will not prove this to you, but perhaps it is intuitive. If not, think of it as our “guess” to the differential equations that are coming. Whatever it takes, just accept it: the steady state response is a sinusoid at the same frequency as the driving force. Now what is happening at the beginning? We call this the “transient”. Remember that any time you give an oscillator a kick, it will respond by oscillating at (near) its natural frequency, but eventually come to rest due to damping. Well, turning on the sinusoidal driving force is a form of an impulse. The oscillator was just minding its own business, then received a jolt when the force was applied. So the complex transient part is just the free damped oscillator response we derived in the last class, PLUS the steady state response which starts right away. So Superposition is at work here! In this case, the drive frequency is near the natural frequency, so we are seeing beats between the free oscillator response and the steady state. This is why transients look complex! Now let’s do it mathematically! b

(Resistance is futile!!!) Driven SHM steady state: (Resistance is futile!!!) w = steady state freq. = the drive frequency So start with Newton’s 2nd, and just add the drive force in with the other forces. Dividing by m and using the substitutions from last time, it takes a simple form. This is the driven EOM. To solve it, lets start by just considering the steady state behavior. Here is a guess in the complex plane. The “ss” means that this is the steady state solution only. It is the standard guess, but note that the frequency w is not adjustable, we are assuming it is equal to the drive freqency. A is the steady state amplitude, and d is the phase lag between the drive and the response. Ass = steady state amplitude dss = steady state phase lag between drive and motion

Real Imaginary ratio: (steady state) So plug it into the equation of motion, and cast the driving force in the complex plane as well. Cancel the common factors as usual to get this equation which gives conditions for the parameters A and d. Set Real equal to Real and Imag equal to Imag. Real simplifies to this, Imag to that. Hmm… not too helpful! D only shows up in sine and cosine, so ratio equations to get tan(d). Now we have and expression for phase in terms of tan-1! This is the steady state phase lag between the drive force and the response. (steady state)

Another manipulation will give us the amplitude. Sin2 plus cos2 is one Another manipulation will give us the amplitude. Sin2 plus cos2 is one. Whenever you need to combine two equations and one has sine and one cosine, try those two: the ratio=tan and s2 + c2=1! This results in this function for the amplitude in terms of the drive frequency and the natural frequency, and the damping parameter. (steady state)

steady state solution: w0 Phase lag through 90 deg exactly at natural frequency. Amplitude peaks just below natural frequency: wm w0 90 180 A d So if we put it all together, it looks like this. Remember, this is the steady state solution only! Here is the amplitude, and here is that phase lag, now written as an inverse tangent. Notice a few things about this equation: There were no arbitrary parameters! For the free oscillators (damped and un-damped), we guessed Aejwt+f, and we got w in terms of the parameters and A and f arbitrary, defined by initial conditions. Now, using the same guess, we defined w right off the bat, then we solved for A and d, and we got parameters in terms of the parameters of the system. This is because it is a physically different problem. The free motion of an oscillator can be started in an arbitrary way: initial displacement, velocity, both, etc. However, our steady state motion is completely defined. The force that drives it is defined, so there can only be one response. This is why our answer has no arbitrary parameters. 2. Look at the amplitude function. If the drive frequency is near the response, this term goes to zero. If the damping is light, that greatly reduces this term. So we get a denominator that is very small, and therefore a large amplitude. Here is a plot of amplitude versus drive frequency. The peak is an effect called resonance. Note that it occurs just below the natural frequency, like the free motion of the damped oscillator. IT is actually a little lower, and for high Q it is essentially right at the natural frequency. There are lots of examples of simple resonances from everyday life. When you pump a swing, you have to do it at just the right frequency to get a good amplitude. When you turn off a fan and it slows down, it will shake briefly at some point. This is the blades slowing and passing through a mechanical resonance in the fan. If you have ever tuned an analog radio, you have worked with a electrical analog of this resonance. The assymetry of this curve can also be understood. Imagine very low frequency, like 0.1 Hz. At this frequency its just a DC deflection, and constant amplitude. At very high frequency, the oscillator eventually can’t “keep up”, so it goes to zero. Our mass on a shim is set up to show us resonance. We send an amplified sign wave to drive this speaker, which shakes the oscillator. As I sweep the frequency, we see this peak in amplitude. We will do more. w w Steady state: Ass and dss are not adjustable parameters, they are set by the properties of the system just like wo!!!

Steady state: Ass and dss are not adjustable parameters, they are set by the properties of the system just like wo!!!

For the missing part, add our result for the damped oscillator: (last lecture) + Here is the complete ss solution Remember its only steady state: its not the complete solution What we are missing is transient solution: physical and math reasons given Transient solution was determined before Like always: guess to add?

Damped x t Driven & Damped x t transient steady state

w A Q High Q = “sharp” resonance (for high Q only)

A simple harmonic oscillator driven by a sinusoidal force oscillates at the frequency of the force once the transient motion at the natural frequency decays. The steady state amplitude goes through a resonance near the natural frequency.