# Oscillator/spring quality vs frequency...

I apologize in advance for not being physics savy. My question stems from my interest in loudspeaker behavior, but I think I need to clear up some more basic physics.

I'm trying to understand what determines the Quality factor of an oscilator/resonator.

Given 2 system's with identical spring rate (k=F/x) and an equal attached mass... (m)... the two systems should have identical resonant frequency... (f=(1/2π)×(√k/√m))

am I right so far...I think I am. And I am aware I used big F for Force and little f for resonance...I've seen w some places for resonance, maybe I should have used w instead.

then my question is; is it possible for the 2 systems to have different quality factors and/or damping factors all while maintaing equal mass, spring rate, and resonance? if so, how?

Also, any recommendations for reading on the subject would be welcomed.

Thanks,

-Brett

Hesch
Gold Member
is it possible for the 2 systems to have different quality factors and/or damping factors all while maintaing equal mass, spring rate, and resonance?

The quality of the steel, that the spring is made of, will influence the dampening ratio, and a resonant frequency is lowered by dampening to a dampened frequency.
If you bend some steel (many times) it will heat up due to losses that will dampen the oscillation of a spring/mass. A high quality spring will have less losses.

But a loudspeaker has no spring? The membrane is attacked by som textile at the edge, having a behaviour that differs a lot from a spring. If a membrane seems springy, it is due to the oscillating airpressure inside the cabinet. But you should dampen this "air-spring" by filling the cabinet with rockwooll, glasswooll, whatever, not letting this resonance frequency interfere with Mozart. If you push the bass-speaker inwards, waits a few seconds until the air(pressure) has oozed out, and let the membrane go, you will see the membrane very slowly coming back into correct position. This speed has nothing with resonance frequency to do.

DonkeyB
tech99
Gold Member
The quality of the steel, that the spring is made of, will influence the dampening ratio, and a resonant frequency is lowered by dampening to a dampened frequency.
If you bend some steel (many times) it will heat up due to losses that will dampen the oscillation of a spring/mass. A high quality spring will have less losses.

But a loudspeaker has no spring? The membrane is attacked by som textile at the edge, having a behaviour that differs a lot from a spring. If a membrane seems springy, it is due to the oscillating airpressure inside the cabinet. But you should dampen this "air-spring" by filling the cabinet with rockwooll, glasswooll, whatever, not letting this resonance frequency interfere with Mozart. If you push the bass-speaker inwards, waits a few seconds until the air(pressure) has oozed out, and let the membrane go, you will see the membrane very slowly coming back into correct position. This speed has nothing with resonance frequency to do.
The quality of the steel, that the spring is made of, will influence the dampening ratio, and a resonant frequency is lowered by dampening to a dampened frequency.
If you bend some steel (many times) it will heat up due to losses that will dampen the oscillation of a spring/mass. A high quality spring will have less losses.

But a loudspeaker has no spring? The membrane is attacked by som textile at the edge, having a behaviour that differs a lot from a spring. If a membrane seems springy, it is due to the oscillating airpressure inside the cabinet. But you should dampen this "air-spring" by filling the cabinet with rockwooll, glasswooll, whatever, not letting this resonance frequency interfere with Mozart. If you push the bass-speaker inwards, waits a few seconds until the air(pressure) has oozed out, and let the membrane go, you will see the membrane very slowly coming back into correct position. This speed has nothing with resonance frequency to do.
I also think a good approach is to read about an electrical analogy of the mechanical system, such as an LCR series resonant circuit. In this case, if R is very small, it gives high "Q". This does not alter the centre frequency but makes the circuit very sharply tuned. The high Q also causes the circuit to continue oscillating for many cycles after it has been given an impulse, just like a 'speaker. On the other hand, if we add resistance (rock wool etc) to make the circuit more broadband and to have a better pulse response, the losses are then increased.

DonkeyB
sophiecentaur
Gold Member
just like a 'speaker.
Not like the speaker. IF the speaker is working properly, it is transferring energy into the air and is acting like a very significant load on any resonance. The restoring force of a speaker suspension will be very small and the mass will also be small. The loading of the air on the cone is in fact high (that's what it's designed for). When used in a proper enclosure, the matching of the speaker mechanics to the outside air is even greater and the damping is higher. For a speaker with a good frequency range, the Q will be very low - even critically damped.

DonkeyB
The quality of the steel, that the spring is made of, will influence the dampening ratio, and a resonant frequency is lowered by dampening to a dampened frequency.
If you bend some steel (many times) it will heat up due to losses that will dampen the oscillation of a spring/mass. A high quality spring will have less losses.

But a loudspeaker has no spring? The membrane is attacked by som textile at the edge, having a behaviour that differs a lot from a spring. If a membrane seems springy, it is due to the oscillating airpressure inside the cabinet. But you should dampen this "air-spring" by filling the cabinet with rockwooll, glasswooll, whatever, not letting this resonance frequency interfere with Mozart. If you push the bass-speaker inwards, waits a few seconds until the air(pressure) has oozed out, and let the membrane go, you will see the membrane very slowly coming back into correct position. This speed has nothing with resonance frequency to do.

What qualities of steel would cause a spring to dampen more or less without affecting its spring rate? Could 2 different springs be constructed such as they have the same spring rate, yet have a differing damping quality?

My understanding is that speakers are very much like springs, and they do in fact have a spring suspension attached to the speaker behind it's cone. This spring is called a spider and is usually attached where the cone and voice coil meet. It may look like a piece of textile, but under the textile are several bent rods connecting the voice coil and cone of the speaker to the frame/basket of the speaker.

Manufacturer's usually publish measured behaviors of their speakers, measured in free air, not mounted in an enclosure. Among them are Fs(resonant frequency), Cms(compliance of speaker suspension), and Mms (Mass of the speaker diaphram). These values fit, into the formula, (f=(1/2π)×(√k/√m)), realizing that compliance(CMS) is the reciprocal of springrate(K).

Manufacturer's also publish a few Q values of their speakers, Qes, Qms, and Qts; but I'm having difficulty figuring what factors determine these values as the formula for resonance seems to be independent of Q.

I also think a good approach is to read about an electrical analogy of the mechanical system, such as an LCR series resonant circuit. In this case, if R is very small, it gives high "Q". This does not alter the centre frequency but makes the circuit very sharply tuned. The high Q also causes the circuit to continue oscillating for many cycles after it has been given an impulse, just like a 'speaker. On the other hand, if we add resistance (rock wool etc) to make the circuit more broadband and to have a better pulse response, the losses are then increased.

R would then drive Q significantly and must be an overall obstruction to the osculation... so in a speaker probably energy lost flexing the speaker surrounds, or likely more significantly, the speaker magnet inducing voltages into the voice coil...maybe... What would Inductance and Capacitance be analogues to?

tech99
Gold Member
R would then drive Q significantly and must be an overall obstruction to the osculation... so in a speaker probably energy lost flexing the speaker surrounds, or likely more significantly, the speaker magnet inducing voltages into the voice coil...maybe... What would Inductance and Capacitance be analogues to?
If you imagine a hack saw blade clamped in a vice at one end, so it can be pulled to one side and then released, this is a good mechanical version. The mass of the blade is like the inductance, and the elasticity (ease of bending) is like the capacitance. The resistance is mainly due to air drag. I agree with sophiecentaur that part of the speaker resistance represents the "wanted" load and the Q is very low. My understanding is that the commonly used "infinite baffle" type of small enclosure has very low efficiency, maybe because the loss resistance is high due to the use of rock wool to dampen the resonance. It is very interesting to connect a speaker in free air to a signal generator and measure the voltage across it as you change the frequency from say 25 to 100 Hz, as the bass resonance peak is very easily seen. Then compare it with a speaker in an enclose.

DonkeyB
sophiecentaur
Gold Member
A loudspeaker works in a very complicated way andI think it adds a lot of extra imponderables to a straightforward discussion of how steel behaves in a spring.
The losses in a spring are shown in the Hysteresis curve of Load / Displacement. The area in the curve represents the energy lost per cycle. So I guess you would be after a 'low hysteresis' form of steel if you want a high Q steel resonator of any form.
I searched for tuning fork steel alloy (randomly chosen terms) and this link suggests that aluminium alloy could be the way to go (but it's only an advert, basically).

DonkeyB
sophiecentaur
Gold Member
Apart from the basic differential equations involved, the spring and the loudspeaker are about as far apart in the world of resonance as you could want. An ideal loudspeaker will have no resonances at all; it would just consist of a diaphragm that moves proportionally with the current in the voice coil (or, for an electrostatic speaker, the displacement should be proportional to the voltage) It is designed purposely, to lose energy (sound) to the air. Practicalities of the necessary suspension will always introduce some resonance somewhere in the loudspeaker mechanism. A spring is seldom made with damping properties but additional components are used to achieve damping, for instance the dampers in motocar suspension. Clock springs are made to have as little loss as possible because that defines the oscillation frequency best.