# Damping: stiff spring vs. less stiff spring

hi, this is a question in my textbook that i don't get:

Does a stiff spring undergo slower or faster damping than a less stiff spring of the same legth?

i think the stiff one would undergo a faster damping because it tends not to move even though there's a mass attached to it, doesn't it?
Also, i think we are supposed to relate energy when we are answering the question.

i will appreciate any help i can get.

## Answers and Replies

If the spring is vibrating at a frequency the same as the vibration it is damping out then they are synchonized and the spring force doen't oppose the driving vibration.

However if they are at different frquency the spring ends up opposing the vibration. So it depends on the frequency of vibration.

w^2=k/m
I guess making w as small as possilbe is the best way because then there is the least movment even when the spring does start vibrating. W is the smallest with a loose spring and large mass.

hi, this is a question in my textbook that i don't get:

Does a stiff spring undergo slower or faster damping than a less stiff spring of the same legth?

i think the stiff one would undergo a faster damping because it tends not to move even though there's a mass attached to it, doesn't it?
Also, i think we are supposed to relate energy when we are answering the question.

i will appreciate any help i can get.

what is meant by damping in this context? I'm thinking its related to the energy dissipated (lost) by heat, which just like the friction in a pendulum will eventually cause the system to come to rest.

what is meant by damping in this context? I'm thinking its related to the energy dissipated (lost) by heat, which just like the friction in a pendulum will eventually cause the system to come to rest.
ya. what you said is right. it's like how a pendulum comes to rest. in this case, though, it's not the pendulum, but two springs with different stiffness. Which one will undergo a faster damping? a stiff one or less stiff one?

tuff problem, if air resistance is a factor, I would say its the less stiff spring which will undergo greater displacements and higher vees. I'd be inclined to say as much for the same reason re internal friction and heating, but as it turns out to be a knotty problem with some non frequency dependent mechanisms at play.

tuff problem, if air resistance is a factor, I would say its the less stiff spring which will undergo greater displacements and higher vees. I'd be inclined to say as much for the same reason re internal friction and heating, but as it turns out to be a knotty problem with some non frequency dependent mechanisms at play.

it's true that a less stiff spring will undergo greater displacement. i don't know what 'vees' are, though.
anyway, so which one do you think will undergo faster damping? I think it's the stiff one because it tends to stay at its equilibrium state and does not easily affected by stretching/compressing. It's the same reason as to why a stiff spring is harder to stretch. anyone agrees?

sorry,
vee=velocity;gee or g's=acceleration

I still stand by my previous post.

One reason is that is the way it is done. I want to play around with holograms. I have to make a vibration free table. The way to damp the table enough for an interferometer (1/10th the wavelength of red light vibration tolerance) is put something heavy on a weak spring...ie....box of sand on some partially inflated innertubes.

The small frequency is more important than the small amplitude in this case and stiffer springs have a larger frequency

NotMrX, i don't understand what you are saying.

What does frequency have to do with my question??

NotMrX, i don't understand what you are saying.

What does frequency have to do with my question??
Suppose a spring isn't driven by some other oscillator or vibrator. The it vibrates at its own frequency. That is the frequency that an oscillator can drive it at most effectively.

Frequency is how many times the spring will osscilate per second. It depends on how stiff the spring is.

To answer you question I typed "spring frequency" into google and got many pages back. here is one:
http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html
That w symbol I mentioned before was the angular frequency of a spring. I assumed your book already mentioned that to you. Out of curiousity what book are you using?

Maybe I misunderstood. You know springs and frequency are realated but you don't see how the two relate to the question. Here is a simple experiment you can do if you have a few bucks to spare on parts and a little time. Take a small board like a trim for a house. Attach some springs of different strenght to the board on one side and a mass to the other side. hang the board horizontaly so the springs hang down. Give rhytmic tap. Depending on the tempo of your beat (how many beats per minute) different springs are going to vibrate. This means that springs mainly absorb near their natural harmonic frequency w^2= k/m If you want to damp it out then you don't want the spring to absorb so you want the springs natural frequency far away from the driving frequency.

The thing is ideal springs aren't actually dampeners. There has to be either friction or air resistance to absorb the heat.

http://en.wikipedia.org/wiki/Shock_absorbers
wikipedia says:
Spring-based shock absorbers commonly use coil springs or leaf springs, though torsion bars can be used in torsional shocks as well. Ideal springs alone, however, are not shock absorbers as springs only store and do not dissipate or absorb energy. Vehicles typically employ both springs or torsion bars as well as hydraulic shock absorbers. In this combination, "shock absorber" is reserved specifically for the hydraulic piston that absorbs and dissipates vibration.

Wikipedia talk about damping in more detail:
http://en.wikipedia.org/wiki/Damping#System_behavior

looking at the graph on that page, the critical damping seems to damp it out quicker so the conditions for critical damping are optimal. The conditions are that where c is the dampening coefficient
$$1=\frac{c}{2\sqrt{km}}$$
or that
$$k=\frac{c^2}{4*m}$$
I might have tried to answer a problem a little to involved for me. I guess that is part of reason why I am not an administrater:tongue2: Also I have only taken basic physics classes.
Anyways looks like it depends on the natural frequency and the damping constant. The situation I desribed before had a small k and large M and was probably underdamping which would make it vibrate just at really low frequncy which is okay for holograms and I guess confused me.

To rephrase, there are three types of damping. Critical damping works the fastest and is the best damper. The optimal spring constant depends on the amount of friction and the amount of mass attached to spring. Sorry for any confusion. Thanks for asking the question. I enjoyed working through the problem. As it turns out it is related to the natural frequency but not in the exact way I initially thought.

it's true that a less stiff spring will undergo greater displacement. i don't know what 'vees' are, though.
anyway, so which one do you think will undergo faster damping? I think it's the stiff one because it tends to stay at its equilibrium state and does not easily affected by stretching/compressing. It's the same reason as to why a stiff spring is harder to stretch. anyone agrees?

you can guess all you want and very reasonably at that, and still be wrong, as the problem turns out to be somewhat intractable. Generally introductory physics course would treat any internal dampening (if at all) as a viscous phenomenon, where the braking would be similar to a dashpot in parallel with the spring, and therefore proportional to the velocity of the springs motion. As it turns out things are not that simple, but the math works out beautifully via a second order ordinary differential equation with the solution an exponentially decaying sinusoid along the lines of
A*cos(wt)*e^(-kt) Unfortunately experiments do not corroborate this type of behavior.

Let us know what the prof has to say.

same issue

let me put this problem (for which i'm also searching for an answer to) in a different light: for 2 equal lengthed springs, stretched the same amount, with the same mass attached on the end, which one stops first: the stiffer or the looser spring? aka which one looses energy to the environment faster?
*any* help is appreciated.

A stiff spring will undergo faster dampening than a less stiff spring of the same length because it has a higher spring constant. This means that it requires more force to displace it and therefore will not undergo the same displacement as the looser spring. Because the mass attached to it will have less displacement, this means that the gravitational potential energy, as well as the kinetic energy will be lower then the loose spring. If there is less energy in the spring then there is less to be lost due to friction in the spring and the air.
Assuming that both springs encounter the same amount of friction, it will take longer for this friction to slow down the loose spring. This is due to the fact that less force is needed to extend it the same length as the stiff spring. If the same mass is attached then the loose spring will extend further because of it's lower spring force constant (k) therefore creating a higher gravitational potential energy and kinetic energy, as it has a greater displacement. Because there is more energy to be lost, the looser spring will take a longer time to stop completely, thus meaning that the stiffer a spring is, the faster the damping.

(I'm doing the same lab write-up right now, it's brutal, none of my results add up properly)

lol me too

thats so jokes, my lab data isn't adding up either lol my values for the force constant didn't fit the equation 1/ktotal=1/k1 + 1/k2 (for springs in a series, the second part in the lab) i've been doing this lab for DAYS... can anyone show me how to calculate the force constant given:
the mass on the spring
the time for 10 cycles (aka vibrations, up and downs, w/e)
the displacement from equilibrium position at each cycle

thats so jokes, my lab data isn't adding up either lol my values for the force constant didn't fit the equation 1/ktotal=1/k1 + 1/k2 (for springs in a series, the second part in the lab) i've been doing this lab for DAYS... can anyone show me how to calculate the force constant given:
the mass on the spring
the time for 10 cycles (aka vibrations, up and downs, w/e)
the displacement from equilibrium position at each cycle

What I did was use Hooke's Law: F=kx
F = mg
x = displacement of the spring, NOT the mass

from there you can solve for k to find the force constant. I figured out my data and it's all pretty close now, I only had 2 days to do it though, which sucked.

You could use Simple harmonic motion formula,

somewhere you should have been introduced to eqn like

x(t)=Asin(sqrt[k/m]*t) where sqrt(k/m)= omega,

and T (the period for one complete oscillation)=1/f where f=omega/2pi

{omega being the w looking greek letter}

BTW the displacement of mass and spring from equilibrium should always be th same but the equilibrium point may vary depending on whether say spring is horizontal or vertical.

got it.

actually I think I was looking for T= (m/k)^(1/2) since the mass is the same for both springs and the springs r of equal length, the spring constant determines the periodic time therfore a stiffer spring will stop vibrating first.

thanks 4 everyones help