Hi,
while I won't be addressing my posts to any particular individual (as much as I may appreciate any particular individual's interest) in order to make it clear that I'm directing them at anybody who is interested in them, I would like to give a quick shout-out to Dale in thanks for the effort that he's been putting into this thread. Apologies for the 3-day delay in my response.
Dale said:
This counter force is very confusing and completely unnecessary. Perhaps it is the root of your confusion. I would recommend completely abandoning it and any conclusions associated with it.
To derive the equations of motion, all that is needed is a free body diagram. In the case of SHM there is one force (the spring) and in the case of DHM there are two (the spring and the drag). By Newton’s 2nd law the net force is equal to the mass times the acceleration. That is all you need for this derivation.
Yes, I see this now. I made the mistake of thinking ##k<0##. But ##k>0##.
Dale said:
Protea Grandiceps said:
falsely assumed what the above rules out, namely thatd2xdt2SHM=d2xdt2DHMd2xdt2SHM=d2xdt2DHMwhich isn't the case.
That is certainly not an assumption that is made, and I am not sure at all why you would believe this. Can you explain? (If you look at the equations you can clearly see that they are not the same so that assumption is definitely not made, so why do you believe it was)
I'm still grappling with this.
Beginning with
$$-k.x=F$$ and $$-k.x-b.v=F$$
Just to stress aspects of these equations that are important here I want to rewrite them - even though this is redundant - as:
$$-k.x_{SHM}=F_{SHM}$$ and $$-k.x_{SHM}-b.v_{SHM}=F_{DHM}$$
It follows that if ##b.v\neq0##, which is observed in DHM except at the peaks and troughs of the oscillating motion (assuming sinusoidal characteristics), $$F_{SHM}\neq F_{DHM}$$
Implicit in our discussion of the spring force equations is also our agreement that we are refining our modelling of a
single given oscillating system by progressing from where we are describing it as SHM to where we are describing it as DHM. Therefore, ##m## is the same in $$F_{SHM}=m.a~~and~~F_{DHM}=m.a$$
Now if ##F_{SHM} \neq F_{DHM}##, as we have established as indicated further above, and ##m## is the same in ##F_{SHM}=m.a## and ##F_{DHM}=m.a##, then it follows that$$a_{F_{SHM}} \neq a_{F_{DHM}}$$
I'd like to write this as
$$-k.x_{SHM}-.v_{SHM}=m.a_{DHM}=F_{DHM}$$
But $$a=\frac {d^2x}{dt^2}$$
Specifically here $$a=\frac {d^2x_{SHM}}{dt^2}$$
and $$a=\frac {d^2x_{DHM}}{dt^2}$$
and as shown above $$a_{F_{SHM}} \neq a_{F_{DHM}}$$
then $$\frac {d^2x_{SHM}}{dt^2}\neq \frac {d^2x_{DHM}}{dt^2}$$
Now we recall that the equation describing the motion of an oscillating system in DHM is derived from the equation for the spring force ##F_{DHM}##. To this end we write
$$-k.x_{SHM}-b.v_{SHM}=m.a_{DHM}=F_{DHM}$$as
$$\frac {d^2x_{DHM}}{dt^2}+(\frac{b}{m})\frac{dx_{SHM}}{dt}+(\frac{k}{m})x_{SHM}=0$$
And now comes the part with which I struggle to come to terms: There is a method by which second-order equations of the type ## x'' + ax' + bx = 0## are solved by making substitutions of ##x=e^{\lambda t}## for ##x## - I understand that. But the above equation is not of the type ##x'' + ax' + by = 0## since ##d^2x## refers to the DHM term and ##\frac {dx}{dt}## and ##x## refer to the SHM terms and therefore I wonder how it can be solved using this method. The above equation is really of the type ## y'' + ax' + bx = 0##, I'd say. There obviously is a way to solve it that I've overlooked.
So ignoring my question for a moment, the above equation is solved by making the following substitutions:
$$x_{SUBST}=e^{\lambda t}~~~for~~~x_{SHM}$$
$$x_{SUBST}'=\frac{dx_{SUBST}}{dt}= \lambda {e^{ \lambda t}}~~~for~~~\frac{dx_{SHM}}{dt}$$
$$x_{SUBST}''=\frac{d^2x_{SUBST}}{dt^2}=\lambda^{2}{e^{ \lambda t}}~~~for~~~\frac{d^2x_{DHM}}{dt^2}$$
The equation describing the motion of an oscillating system in DHM ##x_{DHM}## is then, following further steps that I find fairly straightforward, determined to be
$$x_{DHM}=A~e^{(-b/2m)t}(\cos\sqrt{\omega^2-(\frac{b}{2m})^2}+\theta)$$
Personally, I find this form more commonly cited:
$$ x_{DHM}=A~e^{-\gamma t}(\cos\omega' t + \theta)$$
where ##\omega=\sqrt{\frac {k} {m}}~~~##,##~~~\omega'=\sqrt{\omega^2-\gamma^2}~~~##,##~~~\gamma=\frac {b} {2m}##
Thanks for listening as always and have a great weekend.
PG
