h2 physics - damping

absol SG

we learn in oscillations chapter that increased damping can be achieved through using an oscillator of bigger surface area(for damping provided by air) or a more viscous liquid(in which the oscillator is immersed in).

what if...your oscillator's motion goes up and down because its a spring with a mass attached to it? does increasing the spring constant of the spring cause an increase in damping? if so, why do you consider it to be so(as in what aspects of the spring with a higher spring constant)?

like, i dont know, using a spring with the higher spring constant, does it take a longer time to return to eqm position or something for example? and therefore increased damping?

tut4nkh4m3n SG

i don't really understand yr question.

but as far as i know, the force constant or spring constant is just a proportionality constant. in SHM, force constant, k, is given by mw^2 (m = mass, w = angular freq)

tut4nkh4m3n SG

OHHH

i THINK i understand what you mean.

F = kx

with a larger k value, the force needed to extend a certain length, x, is greater. so, if i'm not wrong, it will result in a shorter time for the object to become stationary at equilibrium position.

absol SG

as in im wondering if changes in spring constant(eg an increase) will lead to an increase in damping, and if so, why....

tut4nkh4m3n SG

damping is the result of some external force acting on whatever that's oscillating, it has nothing to do with the force constant.

e.g. it would be wrong to say that a spring's oscillation is damped when it has a higher spring constant

eagle SG

OHHH

i THINK i understand what you mean.

F = kx

with a larger k value, the force needed to extend a certain length, x, is greater. so, if i'm not wrong, it will result in a shorter time for the object to become stationary at equilibrium position.

Just some terms usage. It will result in a shorter period for the SHM, and not for the object to become stationary. As you said, spring constant has nothing to do with damping. So the object won't become stationary ;)

w = sqrt (k/m) = 2π / T

So higher k will result in lower T
Hence shorter period.

tut4nkh4m3n SG

yup. but if you take into acc energy loss, air resistance etc, it will take less time to become stationary at eqm position, am i right?

tut4nkh4m3n SG

talking abt damping, resonance etc...did you just feel the tremors? o.O

FrozernFire SG

yup. but if you take into acc energy loss, air resistance etc, it will take less time to become stationary at eqm position, am i right?

yup. i'm doing h2 physics as well. we learn 3 types of damping, light, critical and heavy.

light - oscillation amplitude decays exponentially

critical - object does not oscillate, returns to eqm position in shortest time possible

heavy - object does not oscillate, returns to eqm position after a relatively long time, longer compared to critical damping

only if the question mention damping then you take energy loss into account, else like always, assume no energy loss

absol SG

so in short, is the idea of "a higher spring constant spring leads to increased damping" the same type of misconception as "using a liquid of higher density will lead to increased damping"?

because in both situations only the period of the oscillator will be affected. damping remains....constant.

eagle SG

For liquid, viscosity is the property to be considered in damping

absol SG

yep that i know haha.