Oscillatory motion of a spring-mass system

  • #1
Hello all,

Any help with this will be much appreciated. I've scoured the web and searched through textbooks, but I don't have a definite answer to my question as of yet.

First, here's the background on my question:
If I have a standard mass on a spring set into motion, All the textbooks say that the standard solution to the differential equation which describes the motion of the mass is Something of the x=ACos(wt + phi) form, where w is the angular frequency and phi is a phase angle.

However, there is a general equation that describes sinusoidal oscillatory motion that is in the form x=Acos(kx-wt), where k is the angular wave number 2pi/lambda, and where lambda is the wavelength. I've seen this equation typically ascribed to transverse waves, but not to longitudinal ones. So what I'm stuck on is... It seems to me that there should be an analogous angular wavenumber value for longitudinal waves, but there I'm not so sure what the "wavelength" would be. On some websites I've seen them define the wavelength of a longitudinal wave to be from compression to compression or rarefaction to rarefaction, but for a mass on a spring Would that just be the "initial displacement"? Also, assuming that friction damps the oscillation, wouldn't I have an angular wavenumber that's changing in time, delta k? If so, How would I factor that into an equation that describes the motion using that X=ACos(kx-wt) form? (k here representing the angular wavenumber and A including the damping decay exponential factor).

Thank you for your thoughts on this!!

Answers and Replies

  • #2
x=Acos(kx-wt) doesn't make any sense, because of the x's. Thats why it only applied to transverse waves where it takes the form y=Acos(kx-wt) (y or z or whatever).

The analogy to the wavenumber is essentially the wavelength... perhaps more precisely the frequency - but effectively the same thing.
The initial displacement is the amplitude of the oscillatory motion, not the wavelength. The wavelength is concerned with its temporal oscillations, not solely the spatial ones.

When you add friction, you can think of it in 2 equivalent ways: either you preserve the same form of the wave equation and add a damping term (e^-___) or you make your wavenumber time dependent (&&|| complex).
  • #3
Woops, that was a typo.....thanks for pointing that out lzkelly. I did mean to use a different variable, like Y=Acos(kx-wt)..

So do you mean....I can take the wavenumber k=2pi/lambda, which would be it's spacial component or I can take k=2pi/T? i.e. k is then angular frequency?
  • #4
for a non-damped wave, yes.
When the wave is damped, "k" becomes complex -> it has real(oscillating) and imaginary(decaying) components. Each piece becomes much more complicated than in the undamped case.
The differential equation can be found on wikipedia, or http://mathworld.wolfram.com/DampedSimpleHarmonicMotion.html
the general solution gets pretty ugly.

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